Physics Part Ii Chapter 10 Mechanical Properties Of Fluids
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    NCERT Solution For Class 11 Physics Physics Part Ii

    Mechanical Properties Of Fluids Here is the CBSE Physics Chapter 10 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Mechanical Properties Of Fluids Chapter 10 NCERT Solutions for Class 11 Physics Mechanical Properties Of Fluids Chapter 10 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 1
    CBSEENPH11017801

    Define interatomic forces and give an example?

    Solution
    The force between atoms of the molecule is called interatomic force.

    E.g.The nucleus and electron cloud of a molecule is binded by electrostatic force. 
    Question 2
    CBSEENPH11017803

    What do you understand by intermolecular forces?

    Solution
    Intermolecular force is the force between the molecules of the substance is called intermolecular force.
    There is a dipole-dipole interaction between the molecules.
    Question 3
    CBSEENPH11017804

    What is the force of adhesion?

    Solution

    Adhesive force is the attractive force between two unlike molecules. They are caused by forces acting between two substances like electrostatic force (attraction due to opposing charges), mechanical force etc. 

    E.g., Force of attraction between gum and paper.

    Question 4
    CBSEENPH11017806

    What is the force of cohesion?

    Solution
    Cohesive attraction or cohesive force is the force of attraction between same types of molecules sticking together. 
    Question 5
    CBSEENPH11017807

    What is the nature of inter-atomic or inter-molecular forces?

    Solution

    Inter-atomic or inter-molecular forces are electrostatic in nature. 

    Question 6
    CBSEENPH11017809

    Do intermolecular or inter-atomic forces follow inverse square law?

    Solution
    No. Intermolecular and inter-atomic forces do not obey the inverse square law. 
    Question 7
    CBSEENPH11017810

    Which force is more stronger-interatomic or intermolecular force?

    Solution
    Inter-atomic force is stronger than the intermolecular force.
    Interatomic force is caused as a result of sharing and transfer of electrons between the atoms. And, intermolecular forces arises because of partial charges present on different atoms within a molecule. Hence, this makes interatomic force stronger than intermolecular force.   
    Question 8
    CBSEENPH11017813

     Define rigid body.

    Solution
    A rigid body generally means a hard solid object having a definite shape and size. That is, the constituent particles do not change the relative position when deforming force is applied on a rigid body.


    Question 9
    CBSEENPH11017814

    Which state of matter possesses rigidity?

    Solution
    Solids posess rigidity. 
    Question 10
    CBSEENPH11017816

    What is elasticity?

    Solution

    The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity. 

    Question 11
    CBSEENPH11017818
    Question 12
    CBSEENPH11017819
    Question 13
    CBSEENPH11017820

    Why strain is dimensionless?

    Solution
    Strain is a ratio of change in dimension to the original dimension. So, it has no unit or dimensionless formula.


    Question 14
    CBSEENPH11017822

    What is stress?

    Solution
    The restoring force per unit area is called stress. This restoring force is equal in magnitude but opposite in direction to the applied force. 
    Question 15
    CBSEENPH11017823

    Name the two types of stress. 

    Solution
    The two types of stress are:
    1. Normal stress, and
    2. Tangential stress 
    Question 16
    CBSEENPH11017824

    What is tensile stress?

    Solution

    When a body is stretched by two equal forces applied normal to its cross-sectional area, the dimension of the body is increased.The restoring force per unit area in this case is called tensile stress. 

    Question 17
    CBSEENPH11017826

    What is compression stress?

    Solution

    If a body is compressed under the action of applied forces resulting in decrease in it's dimension. The restoring force per unit area is known as compressive stress. 

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    Question 18
    CBSEENPH11017827

    Define tangential stress.

    Solution

    When two equal and opposite deforming forces are applied parallel to the cross-sectional area of an object, there is relative displacement between the opposite faces of the body. The restoring force per unit area developed due to the applied tangential force is known as tangential stress.

    Question 20
    CBSEENPH11017830
    Question 22
    CBSEENPH11017834

    What do you mean by elastic limit? 

    Solution
    Elastic limit is the maximum stress which when removed after applied, the body regains its original state completely. 
    The body obeys Hook's law, if the applied stress is less than the elastic limit. 
    Question 23
    CBSEENPH11017836

    Does the linear relation between stress and strain (Hooke's law) valid for large values of strain?

    Solution
    No. Hooke's law is not valid for large values of strain. 
    Question 24
    CBSEENPH11017838

    Which of the two forces deforming or restoring is responsible for elastic behaviour of a substance?

    Solution
    Restoring force is responsible for the elastic behaviour of a substance. 
    Question 25
    CBSEENPH11017839

    What are elastic bodies?

    Solution
    Elastic bodies are bodies that regain their original state completely on removal of deforming force. 
    Question 26
    CBSEENPH11017841

    What are plastic bodies?

    Solution
    The bodies that get permanently deformed and which  do not regain their original state completely on removal of deforming force are called plastic bodies. 
    Question 27
    CBSEENPH11017843

    Give an example of nearly elastic material.

    Solution
    Quartz.
    Question 28
    CBSEENPH11017844
    Question 29
    CBSEENPH11017846

    Why do the liquids and gases have zero value of modulus of rigidity? 

    Solution
    Liquids and gases have zero value of modulus of rigidity because they do not have any definite shape themselves. Liquids and gases take the shape of the container they are in.
    Question 30
    CBSEENPH11017847

    Name two factors that change the elastic properties of a body.

    Solution
    Two factors that change the elastic properties of a body are:
    1. Change in temperature, and
    2. Presence of impurities 
    Question 31
    CBSEENPH11017848

    It is easy to break the wire by bending repeatedly in opposite direction instead of breaking it by stretching directly. Why?

    Solution
    It is easy to break the wire by bending repeatedly in opposite directioninstead of breaking it by stretching directly because by applying alternate force on the wire, the wire loses its strength and hence it is easy to break.
    Question 32
    CBSEENPH11017850

    What is elastomer?

    Solution
    The elastic substances that can be subjected to large strain without breaking are called elastomers. 
    Question 33
    CBSEENPH11017859

    What is ‘elastic after effect’? 

    Solution
    When deforming forces are removed, the body takes time to come back to its original state. The delay in coming back to it's original state is called 'elastic after effect'. 
    Question 34
    CBSEENPH11017860

    What is the relation between deforming force and restoring force at equilibrium?

    Solution
    When in stable equilibrium both these forces, deforming force and restoring force are equal and opposite.
    Question 35
    CBSEENPH11017871

    What type of stress will produce longitudinal strain, volumetric strain and shear strain?

    Solution
    Normal stress produces longitudinal and volumetric strains while shear strain is produced by tangential stress. 
    Question 36
    CBSEENPH11017872

    Do you think that a metal wire suspended freely from one end is under any stress?

    Solution
    Yes, wire is under stress due to its own weight. 
    Question 37
    CBSEENPH11017873

    A copper wire of length l has Young's modulus Y. If the wire is cut into two equal halves of each length l/2, what will be Young's modulus of each half?

    Solution
    The young's modulus of the wire will remain the same, that is Y because it is  independent of the dimensions of material.

    Young's modulus is dependent on the nature of the material. 
    Question 38
    CBSEENPH11017874

    How does modulus of elasticity change with temperature?

    Solution
    The modulus of elasticity is inversely proportional to temperature.
    That is it decreases with increase in temperature. 
    Question 39
    CBSEENPH11017875

    How is compressibility related with Bulk modulus?

    Solution
    The reciprocal of bulk modulus is called the compressibilty.
    It is denoted by k. 

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    Question 40
    CBSEENPH11017879

    Why do we use long wires to measure the Young's modulus of wires?

    Solution
    Long wires are used to measure Young's modulus. Extensions in long wires may be large and so young's modulus can be measured accurately. 
    Question 41
    CBSEENPH11017880

    What type of material is preferred for the manufacture of springs?

    Solution
    Material of high value of modulus of elasticity is used to manufacture springs.
    Young's modulus is large for metals, which requires a larger force to produce a small change in length. 
    Question 42
    CBSEENPH11017881

    Why do we prefer steel than copper in manufacturing springs?

    Solution
    Steel is preferred for manufacturing springs. Steel is more elastic than copper because steel has greater value of Young's modulus than copper. 
    Question 43
    CBSEENPH11017882

    Why do springs deform permanently after a prolonged use?

    Solution
    Springs deform permanently because of elastic fatigueness. The elasticity of the material of spring is lost and it deforms permanently. 

    Question 44
    CBSEENPH11017883

    Do we work in stretching a wire?

    Solution
    Yes. Work is done in stretching a wire. A force is applied to stretch a wire and thus, work is done. 
    Question 45
    CBSEENPH11017884

    Why we have to do work in stretching a wire?

    Solution
    When a wire is stretched, the inter-atomic separation increases and inter-atomic forces comes into play. To displace the atoms against interatomic forces, we need to do work to deform the spring. 
    Question 46
    CBSEENPH11017888

     Do elastomers follow Hook's law?

    Solution
    The substances which can be stretched to cause large strains are called elastomers.
    Hooke's law states that stress is directly proportional to the strain.
    Elastomers violate hook's law.
    Question 47
    CBSEENPH11017890

    A wire is cut to half of its original length. How will its Young's modulus change?

    Solution
    Young's modulus is independent of the dimensions of wire but depends on the material of wire. Therefore the Young's modulus of wire remains unchanged.
    Question 48
    CBSEENPH11017891

    Which is more elastic- steel or rubber?

    Solution
    Steel is more elastic than rubber. Elasticity is measured by how much time a body takes to come back to it's original state after being deformed.

    The Young's modulus of steel is greater than rubber. 
    Question 49
    CBSEENPH11017892

    Graphite consists of planes of carbon atom, which are held by weak forces. What expectations can you make about its rigidity?

    Solution
    The value of modulus of rigidity of the graphite will be small because the carbon atoms are held be weak forces. 
    Question 50
    CBSEENPH11017896

    Plot the graph between intermolecular forces. What is the range of these forces?

    Solution
    The distance upto which the influence of one molecule can be experienced by other is known as the range on intermolecular forces. 
    Range is different for different substances.
    It is of the order of 10-10 m. 
    Question 51
    CBSEENPH11017897

    Amorphous substances do not have sharp melting point but get softening range before melting, while crystalline substances have sharp melting point. Explain.

    Solution
    Amorphous substances possess short-range arrangement of atoms or molecules. These short and ordered arrangement disappears gradually on heating. Before melting, this gradual change makes the amorphous substance soft.
    While in crystalline substances, there is a long-range order arrangement of atoms or molecules. This arrangement disappears at a definite temperature and hence have sharp melting points. 
    Question 52
    CBSEENPH11017898

    Why solids have definite shape while liquids do not have definite shape?

    Solution
    Solids: Intermolecular forces are very strong and thermal agitations are not sufficiently strong to separate the molecules from their mean position. Solids are rigid and hence they have definite shapes.
    Liquids: In liquids intermolecular forces are not sufficiently strong to hold the molecules at definite sites, as a result they move freely within the bulk of liquid, therefore, do not possess definite shapes. Liquids take the same shape as that of the container. 
    Question 53
    CBSEENPH11017914

    What is stress and what is its origin?

    Solution
    When an external force is applied on the body there is a deformation in it's shape. The deformation in the shape is due to relative change in the position of atoms or molecules. This relative change in the position causes the attractive or repulsive interatomic or intermolecular forces to set up in the solid which is called restoring force. This restoring force is the origin of stress. The restoring force per unit area is called stress.

    Tips: -

     

    Question 54
    CBSEENPH11017916

    What is strain? Define different types of strains.

    Solution

    Whenever a deforming force is applied on a body, the shape of the body changes. Strain is defined as the ratio of change in configuration to the original configuration.

    Different types of strain are:

    (i) Longitudinal strain: The deformation produced in the length of the body as a result of the deforming force is longitudnal strain. 
    So, longitudinal strain is defined as the ratio of change in length (∆L) of body to its original length L.

                Longitudinal strain = ΔLL

    (ii) Volumetric strain: When the deforming force produces a change in volume only then the strain produced in the body is called volumetric strain. The volumetric strain is defined as the ratio of change in volume (∆V) of the body to its original volume V.

                  Volumetric strain = ΔVV

    (iii) Shearing strain: When only the shape of the body deforms without any change in it's volume, it is shear strain.
    So, shearing strain is defined as the ratio of the relative displacement of the opposite faces of the body to the length of the cylinder L.

                    shearing strain = xL

    All types of strains are dimensionless.

    Question 55
    CBSEENPH11017917

    What is yield point and why it is so named? 

    Solution
    Yield point is also known as elastic limit. Beyond the elastic limit, when the load is increased, a stress is reached where the material continues to deform without any further increase in load. This point is called yield point. Since at yield point the deformation continues to increase without increase in stress, thus it may be said that there is total surrender of material to stress. The synonym of surrender is yield, that is why this point is called yield point. 
    Question 56
    CBSEENPH11017919

    Describe the atomic theory of elasticity. 

    Solution
    The atoms or molecules are bounded by interatomic and intermolecular forces and are in stable equilibrium whenever a perfect state is acquired. In a perfect equilibrium state, the separation is such that the net interatomic intermolecular force is zero. When a body is heated, cooled or any physical factors leads to any change in its state, the distance between the two atoms or molecules increases and attractive force comes into play. When the separation between the atoms or molecules decreases the repulsive force comes into play.
    Question 57
    CBSEENPH11017920

    What is 'elastic after effect'?

    Solution
    When an external force is applied, the body gets distorted from it's original shape. The body takes some time to regain into it's original shape. Therefore, the delay or the time that is taken for the body to acquire it's original shape after the deforming force is removed is called 'elastic after effect'.
    Question 58
    CBSEENPH11017922

    Why the springs are made of steel and not of copper?

    Solution
    The Young's modulus of steel is greater than copper. Therefore, more restoring force is developed in steel than copper when deformed by strain. Due to this cause steel is preferred over copper in manufacturing the springs. 
    Question 59
    CBSEENPH11017923

    Why do spring balances show wrong readings after they have been used for a long time?

    Solution
    When spring balances have been used for a long time, elastic fatigue is developed. So, the spring balances will no longer show the right readings.

    Note: Elastic fatigue is the property of a body to lose it's elastic property by repeated action of restoring and deforming force. 
    Question 60
    CBSEENPH11017924

    How does the Young's modulus change with temperature? Explain.

    Solution
    As temperature increases, the Young's modulus decreases. On heating, the substance expands and distance between the atoms increases. As a result inter-atomic and inter-molecular force decreases. So, Young's modulus also decreases.  

    Question 61
    CBSEENPH11017929

    What are the factors that affect elasticity? In what way they affect its value?

    Solution

    The elasticity of a material is affected by the following factors: 

    (i) Effect of temperature: On heating, mostly the elasticity of materials decreases.

    (ii) Effect of impurities: Depending upon the nature of impurity, the elasticity of materials may increase or decrease.

    (iii) Annealing: Annealing decreases the elasticity of materials.

    (iv) Hammering and rolling: This process increases the elasticity of materials. 

    NoteAnnealing is a heat process where a metal is heated to a specific temperature /colour and then allowed to cool slowly. This softens the metal which means it can be cut and shaped more easily. Iron rod is heated till it turns red-hot.

    Question 62
    CBSEENPH11017930

    If we compress or elongate the solids, what happens to potential energy of solid? 

    Solution
    When a soild is compressed or elongated, it's P.E increases. Potential energy increases because work has to be done against force of repulsion during compression and against force of attraction during elongation.
    Question 63
    CBSEENPH11017931

    What is elastic fatigue and why bridges are declared unsafe after long use?

    Solution
    When continuous restoring and deforming force is applied, a body loses its strength and body is said to undergo elastic fatigueness. Bridges are regularly and continuously used. Hence, they are under alternate stress which causes elastic fatiguenes in steel. Therefore, bridges are declared unsafe for prologed use. 
    Question 64
    CBSEENPH11017932

    A piece of rubber under an ordinary stress can display 100% strain; yet when unloaded, it returns to its original length. Does this show that the restoring forces in a rubber piece are strictly conservative?

    Solution
    In case of rubber, the loading curve of stress versus strain graph does not coincide with unloading curve because of elastic hysteresis. Hence work done by restoring force over one complete cycle of loading and unloading is not zero. Hence restoring forces in a rubber piece are not conservative. 


    Question 65
    CBSEENPH11017936

    A rectangular beam is bent into arc of a circle. What type of strain is produced in it?

    Solution
    When rectangular beam is bent into arc of a circle, the layers on convex side of neutral plane undergo extensional stress and the layers that are on concave side undergo compressional stress.
    Question 66
    CBSEENPH11017938

    What is fluid?

    Solution
    Any material that can flow is a fluid. Liquids and gases are examples of fluid.


    Question 67
    CBSEENPH11018076

    What is hydrostatics?

    Solution
    Hydrostatics is the branch of fluid mechanics that studies incompressible fluids at rest. The study of fluids at rest or objects placed at rest in fluids is hydrostatics.
    Question 68
    CBSEENPH11018077

    What is hydrodynamics?

    Solution
    Hydrodynamics is the branch of science that studies about the force exerted by the fluids or acting on the fluids. 
    Question 69
    CBSEENPH11018078

    Can fluids withstand tangential stress?

    Solution
    The restoring force per unit area due to the application of equal and opposite deformation force applied parallel t the cross-sectional area of the cylinder is known as tangential stress.

    Fluids cannot withstand tangential stress. 
    Question 70
    CBSEENPH11018079

    Can fluids withstand normal stress?

    Solution
    Force is applied normal to it's cross-sectional area. The restoring force per unit area in this case is known as normal stress.

    Fluids can withstand normal stress. 
    Question 71
    CBSEENPH11018080

    Why the fluids do not have any modulus of rigidity?

    Solution
    Fluids cannot withstand tangential stress, therefore fluids do not have any modulus of rigidity.
    Question 72
    CBSEENPH11018081

    Which phase of fluid is more compressible?

    Solution
    The intermolecular force of attraction in gas molecules is weak. So, sas phase is more compressible.
    Question 73
    CBSEENPH11018082

    What is an ideal liquid?

    Solution
    A liquid is said to be ideal if:

    i) It is incompressible,
    ii) It is non-viscous, and 
    iii) It's flow is steady. 
    Question 74
    CBSEENPH11018083

    Do liquids exert pressure?

    Solution
    Yes, liquids exert pressure.
    Question 75
    CBSEENPH11018084

    Define pressure.

    Solution
    Pressure is defined as the normal force acting per unit area on the surface of liquid at rest.


    Question 76
    CBSEENPH11018085

    What is the direction of pressure?

    Solution
    Pressure is a scalar quantity and has no direction.
    Question 77
    CBSEENPH11018086

    What is the direction in which pressure exerts force on the walls of a container?

    Solution
    The pressure exerts the force perpendicular to the walls of the container.
    Question 78
    CBSEENPH11018087

    What is the SI unit of pressure?

    Solution
    The SI unit of pressure is N m-2 or Pascal.
    Question 79
    CBSEENPH11018088

    Plot the graph between intermolecular forces. What is the range of these forces? What is the range of these forces?

    Solution
    The graph for intermolecular attractive and repulsive forces is as shown below.
    It is of the order of 10-10 m. 

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    Question 80
    CBSEENPH11018089

    The interatomic forces between two atoms varies as under:


    straight F subscript alt proportional to negative straight alpha over straight r to the power of 6 space space space space and space space space space straight F subscript rep proportional to negative straight b over straight r to the power of 9

    Find the equilibrium separation between two atoms.

    Solution
    The net force between two atoms is
    straight F equals straight F subscript alt plus straight F subscript rep equals negative straight alpha over straight r to the power of 6 plus straight b over straight r to the power of 9
    Let r0 be the equilibrium separation between the two atoms. At equilibrium separation the net force between the two atoms is zero.
    therefore space space space space space space space space minus straight alpha over straight r subscript 0 to the power of 6 plus straight b over straight r subscript 0 to the power of 9 equals 0
rightwards double arrow space space space space space space space space space space straight r subscript 0 cubed equals straight b over straight a
rightwards double arrow space space space space space space space space space straight r subscript 0 equals cube root of straight b over straight a end root
    Question 81
    CBSEENPH11018090

     If α is the volumetric strain and l is the longitudinal strain, then find the Poisson’s ratio in terms of a and β.

    Solution

    Let a wire of length l, diameter d be loaded by force F.
    Let dl be the change in length and dV the change in volume, and dD be the change in the diameter.

    Volume of wire before the force is applied is,
                  
straight V apostrophe equals πr squared straight t equals πD squared over 4 straight ell 
    Volume of wire after the force is applied is,
                 straight V apostrophe equals straight pi over 4 open parentheses straight D plus dD close parentheses squared open parentheses straight ell plus straight d straight ell close parentheses
    Change in volume is,
    dV = V' - V
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rightwards double arrow space space space space space space space space dV over straight V equals 2 dD over straight D plus fraction numerator straight d straight ell over denominator straight ell end fraction space

rightwards double arrow space space space space space fraction numerator dVIV over denominator straight d straight ell straight I straight ell end fraction equals 2 space space fraction numerator dDID over denominator straight d straight ell straight I straight ell end fraction plus 1
    
rightwards double arrow space space straight alpha over straight beta equals negative 2 straight sigma plus 1 space space space space space space space space space space space space space space space space space space space space space space open square brackets because space straight sigma equals space space fraction numerator dDID over denominator straight d straight ell straight I straight ell end fraction close square brackets
rightwards double arrow space space space straight sigma equals 1 half open parentheses 1 minus straight alpha over straight beta close parentheses

    Question 82
    CBSEENPH11018091

    Show that isothermal bulk modulus of a gas is equal to the pressure of the gas.

    Solution
    Bulk modulus is given by,
                       straight K space equals space minus straight V space dP over dV
    Negative sign implies if pressure increases volume decreases.
    For isothermal condition, equations of state is given by, 
                         PV = constant 
    Now, differentiating this equation, we get 
    space space space space space space space Pdv space plus space VdP space equals space 0

rightwards double arrow space space dP over dV space equals space minus straight P over straight V
space space space space space space space space space space straight K space equals space minus space straight V space dP over dV
space space space space space space space space space space space space space space equals space minus space straight V space dP over dV space
space space space space space space space space space space space space space space equals space minus straight V space left parenthesis negative straight P over straight V right parenthesis
space space space space space space space space space space space space space space equals space straight P space
space space space space space space space space space space space space
    Therefore, bulk modulus of a gas is equal to the pressure of a gas is proved. 
    Question 83
    CBSEENPH11018092

    The stress-strain graph for a metal wire is shown in the figure. Up to the point E, the wire returns to its original state O along the curve EPO, when it is gradually unloaded. Point B corresponds to the fracture of the wire.


    (a) Upto what point on the curve is the Hooke’s law obeyed? (This point is sometimes called ‘Proportional limit’).

    (b) Which point on the curve corresponds to ‘elastic limit’ or yield point of the wire?

    (c) Indicate the elastic and plastic regions of the stress-strain graph.

    (d) Describe what happens when the wire is loaded upto a stress corresponding to the point A on the graph, and then unloaded gradually. In particular, explain the dotted curve.

    (e) What is peculiar(odd) about the portion of the stress-strain graph from C to B? Upto what stress can the wire be subjected without causing fracture?

    Solution

    (a) Hooke's law states that that stress is directly proportional to strain. From the graph, we can say that Hooke’s law is obeyed from point O to P. 

    (b) E point on the curve corresponds to elastic limit or the yield point.

    (c) Elastic region is the region in which the body regains its state after removal of stress and is from O to E. Plastic region is the region in which there is residue strain after removing the stress and that region is E to B. 

    (d) From O to P, the strain is directly proportional to stress, beyond P increase in strain is more for same increase in stress. Beyond E, when stress is increased, it does not retrace its path on decreasing the stress, but returns to point O' along path AO'. Therefore there is residue strain of 00' when stress is increased corresponding to the point A.

    (e) From C to B, strain increases even if the wire is being unloaded and at B, it fractures. Till point C, stress can be applied without causing the fracture. 



    Question 84
    CBSEENPH11018093

    The stress-strain graphs for materials A and B are as shown in figure. Which of the material:

    (i) has greater value of Young's modulus, 

    (ii) is more ductile or more brittle? 

     

    Solution

    (i) For graph A, slope of OE is more than that of graph B. Therefore, Young's modulus of material A is greater than that of B.

    (ii) Material A shows wider plastic region than that of B. Thus, A is ductile than B and B is more brittle than A. 

    Question 85
    CBSEENPH11018094

    A beam has one end fixed in a wall and loaded at the other end. Three locations A, B and C are shown in figure. Which location would you prefer to make a circular hole?


    Solution

    At point A, the beam undergoes tensile strain, therefore hole at A will increase in size along the length of the beam and becomes oblate. 

    At point C, the beam undergoes compression strain, therefore hole at C will decrease in size along the length of beam and becomes prolate. 

    The point B is in neutral zone, therefore the hole at B will be circular.

    Question 86
    CBSEENPH11018095

    Two wires of same material but different diameters d1 and d2 (d1 > d2) are loaded. Which can support more weight?

    Solution
    Breaking stress is defined as breaking load per unit area. Breaking stress is constant of a material.

    Therefore, 

    Breaking load = Breaking stress X Area

                            = Breaking stress x πd24

    Breaking load is directly proportional to the diameter. Therefore, thick wire can support more weight than the thin wire. 
    Question 87
    CBSEENPH11018096

    The figure shows the load extension graph of two wires of material A and B having same dimensions. Which of the two materials has greater value of Young’s modulus?


    Solution
    Young's modulus is given by, 

               Y = F×LA×L 

    Therefore, 
                F = YALL
    Thus the slope of load v/s extension graph is  YAL.

    For wires of same dimensions, the slope of graph is directly proportional to Young's modulus. Since slope of graph for A is greater than that of B, therefore the value of Young's modulus of material of A is greater than B. 
    Question 88
    CBSEENPH11018097

    A cable is cut into two equal lengths. How the breaking force will change, if two parts are connected in parallel?

    Solution
    Breaking force is given by breaking stress x Area of cross section
    If F is the force required to break the wire of area of cross section a, then breaking force becomes 2F. 
     
    Question 89
    CBSEENPH11018098

    Find the ratio of breaking force required to break the wire of radius r to that of 2r.

    Solution

    Radius of the wire is given by r and 2r. 
    The breaking force of wire is given by, 

                             F = σA

    where,
    σ is the breaking stress (constant of material) and A is the area of cross section.

    Let F1 and F2 be the breaking force for wires respectively.

        F1=σπr2and  F2=σπ(2r)2 

    Therefore, 
                          F2=4 F1

    Hence the breaking force required to break the thick wire is four times the thin wire. 

    Question 90
    CBSEENPH11018099

    Read the following two statements below carefully and state, with reasons, if it is true or false.

    (a) The Young’s modulus of rubber is greater than that of steel;

    (b) The stretching of a coil is determined by its shear modulus.

    Solution

    a) False, because the modulus of elasticity is inversely proportional to the strain for a given stress. If steel and rubber are under the same stress, then strain in steel is less than rubber. Hence steel is more elastic than rubber.

    (b) True, because the change takes place in the shape of the coil spring. Therefore, stretching is determined by its shear modulus. 

    Question 91
    CBSEENPH11018100

    Plot the stress versus strain curve for rubber and what does the area of the loop of the curve represent? 

    Solution
    When the rubber undergoes the cycle of loading and unloading some of the mechanical energy is lost in the cycle. The area under the graph is equal to the work done to complete the cycle of loading and unloading. 
    The below graph is a hysteresis curve. 

                               
    Question 92
    CBSEENPH11018101

    A rod of metal is clamped between two rigid supports and is heated by temperature θ. If wire is not allowed to bend then what type of stress will be set up in the wire and how does stress developed in wire depend on length of the wire? 

    Solution

    When a rod of metal is heated, compressive stress will be set up in the rod. The wire is not allowed to bend and the force is applied normal to it's cross sectional area.  
    Let 'L' be the length of wire, A be the area of cross section, Y be Young's modulus and 'α' be the coefficient of linear expansion, heated to a temperature 'θ'.
    If the rod was free then increase in length of wire due to rise in the temperature will be Lαθ.

    Here, since rod is not allowed to expand, therefore, compression in rod is equal to increase in the length of rod i.e. Lαθ. 

    If S is the compression stress in rod then,

    Y = SLLαθ=Sαθor     S=Yαθ

    Hence stress is independent of the length of the wire.

    Question 93
    CBSEENPH11018102

    An ideal gas of bulk modulus K is enclosed in a container of fixed walls. Find the stress set up in the gas when heated through temperature θ. Let γ be the temperature coefficient of cubical expansion of gas. 

    Solution

    Given, an ideal gas of bulk modulus 'K' and is heated to a temperature θ

    Increase of temperature of gas at constant volume is equivalent if the gas is first heated at constant pressure and then compressed isothermally to original volume.

    Let V be the volume of gas. When it is heated through a temperature θ at constant pressure, then the increase in volume of gas is,

    ∆V = Vγθ

    Now compress the gas isothermally so that the volume decreases by ∆V = Vγθ.

    If ∆P be the stress set up in the gas, then

    K=stressstrain=ΔPΔV/V=ΔPγθ         ΔP = Kγθ 

    Question 94
    CBSEENPH11018103

    Derive an expression for the increase in the length of a uniform rope hanging from the ceiling of a room, due to its own weight.

    Solution
    Let,
    Mass of the wire = M
    Area of cross section = A
    Length of the rope = L 
    Let, be the increase in the length of rope due to its own weight.
    Since the rope is uniform and whole of the weight of rope can supposed to be concentrated at the centre of gravity, therefore the effective length of the rope is L/2 loaded by weight Mg.
    Thus, Young's modulus is given by, 
    Error converting from MathML to accessible text.   
    Question 95
    CBSEENPH11018104

    Derive an expression for the radius of rope used in a crane to lift the mass M kg. The elastic limit of the material of rope is σ N/m2, and safety factor of crane is K.

    Solution
    Given mass = M kg
    Elastic limit of the material = σ N/m2
    Safety factor for crane = K

    The design of the rope is such that it should tolerate a load of KM kg.
    Let r be the radius of rope required to be used in the crane.
    So, using the formula, 

       Elastic limit =Maximum loadArea of cross sectioni.e.            σ=KMgπr2            r = KMgπσ 
    Question 96
    CBSEENPH11018105

    What is the relation between energy density and stress? Plot a graph between them.

    Solution
    Energy density in terms of strain and stress is given by, 
    Energy space density space equals space 1 half cross times stress cross times strain

rightwards double arrow space Energy space density space equals space 1 half cross times stress space cross times space stress over straight gamma space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half space fraction numerator left parenthesis stress right parenthesis squared over denominator straight gamma end fraction
    Graph between Energy density and Stress is as shown below:

    Question 97
    CBSEENPH11018106

    A load W is hanging from the string of length l and area of cross section A. The string suddenly breaks. Find the change in the temperature of the string. Take heat capacity of string as Q and Young’s modulus as Y.

    Solution
    Length of the string =l
    Heat capacity of the string = Q
    Young's modulus = Y

    The potential energy stored in the loaded string is,

                     U=12×stress×strain×volume   = 12(Stress)2YxVolume   =12W2A2YA   =12W2AY

    When the string breaks, the potential energy stored in the string is converted into heat.

    Let θ be the rise in temperature.
    Therefore,

             =12W2AY      θ=12W2QAY 
    is the required change in temperature of the string. 
    Question 98
    CBSEENPH11018107

    When a wire is loaded by weight M1, its length is Land when loaded by weight Mits length becomes L2. Find the original length of the wire. 

    Solution
    L1 wire is loaded by weight Mand when M2 is loaded length becomes M2

    Let L be the original length, A be the area of cross section of wire and Y be Young's modulus of the material of wire.
    Therefore, the extension in wire is (L
    1–L) when loaded by weight M1 and that is (L– L) when loaded by weight M2.

    Thus, Young's modulus is,

    Y = M1g×LA(L1 - L)= M2g×LA(L2 - L)        M1(L1-L) = M2(L2 - L) M1L2 - M1L1 = M2L1 -M1L2                      L = M2L1 - M1L2(M2-M1)

    Question 99
    CBSEENPH11018108

    A wire increases by 10–3 of its length, when a stress of 108N/m2 is applied on it. What is the Young's modulus of material of wire?

    Solution

    Given,
    Wire is increased of it's length, i.e., strain = 10-3 
    Stress = 108N/m2

     Young's modulus = StressStrain= 10810-3= 1011 Nm-2  

    Question 100
    CBSEENPH11018109

    Two wires P and Q of same length and material but radii in the ratio 2:1 are suspended from a rigid support. Find the ratio of strain produced in the wires when: (i) both are under same stress (ii) both are loaded by same weight.

    Solution

    Given the radii of two wires is of the ratio 2:1.
    Using the formula as per Hooke's law, 

    Strain  Stress =LoadArea=Fπr2and           rP : rQ=2 : 1

    i) When both the wires are under the same stress, strain produced will be the same.

    ii) When both the wires are loaded by the same weight, then

    StrainpStrainQ=(rQ)2(rP)2=14 




     

    Question 101
    CBSEENPH11018110

    A rod of length 2m and cross sectional area 8x 10–6m2 is clamped between two rigid supports. If the wire is heated through, 25°C, then find the thermal stress and compressional force set up in the wire. The Young's modulus of rod is 2x1011N/m2 and its coefficient of linear expansion is 1.6 x 10–5 per°C. 

    Solution
    Given, 
    Length of the rod = 2 m
    Cross-sectional area = 8× 10-6 m2
    Tempertaure, T = 25o C
    Young's modulus of the rod, Y = 2x1011N/m
    Coefficient of linear expansion =  s per°C.

    Therefore, 
    Thermal stress set up in the wire is,
                 
                  S = Yαθ    = 2x1011 ×1.6 x 105×25    = 8×107 N/m2 

    Compressional force is given by, 

             F = SA = 8×107×8x 106 = 640 N
    Question 102
    CBSEENPH11018111

    A wire loaded by weight of density 7800 kg/m3 is stretched to length of 1m. On immersing the weight in liquid of density 1300kg/m3, the length shortens by 6mm. Find the original length of the wire.

    Solution

    Given,
    Density of the load, ρ = 7800 kg/m3 
    Length of the wire, l = 1 m
    Density of liquid, = 1300 kg/m3
    Let, V be the volume of weight suspended from the wire.
     
    Therefore, tension in the wire when load is in the air is equal to, 
     
    Error converting from MathML to accessible text.

    Question 103
    CBSEENPH11018112
    Question 104
    CBSEENPH11018113

    A 4m long wire of radius 4 mm is elongated by 0.01% when loaded by a force of 80π newton. What is the stress in the wire and Young's modulus of wire?

    Solution

    Length of the wire, l = 4m
    Radius of the wire, r = 4 mm = 0.004 m
    Force on the wire, F = 80π N 
    Strain = 0.01% = 0.0001

    Now, using the formula of stress,

    Stress = Fπr2  = 80ππ(0.004)2 = 5× 106 N/m2

    Young's modulus of the wire is given by, 
    Y = StressStrain=5×1060.0001 = 5×1010 N/m2 




    Question 105
    CBSEENPH11018114

    A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

    Solution

    Length of steel wire, L= 4.7m
    Length of copper wire, Lc=3.5m
    Area of cross-section of steel wire, As=3.0 x 10-5 m2
    Area of copper wire, A= 4.0 x 10-5 m

    To find = the ratio of Young's modulus of steel to that of copper?

    ΔLs=ΔLc  and  Fs=FcYoung's modulus is given by, Y=FALΔL          YsYc=FsAsLsΔLsAcFcΔLcLc                   = LsLcAcAs                   =4.7×4.0×10-53.5×3.0×10-5                   =1.8

    Question 106
    CBSEENPH11018115

    Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

    Solution

    Given, 
    Diameter of the wires = 0.25 cm
    That is, d= d= 0.25 cm

    Therefore, 
    Radius of the wires will be given by, 

    rs = rB = 0.125  cm = 1.25 ×10-4 m

    Unloaded length of the steel wire, Ls=1.5 m
    Unloaded length of the brass wire, L= 1 m

    Young's modulus of steel, Ys = 2 × 1011 Pa
    Young's modulus of brass, YB0.91 × 1011 Pa
    m1 = 4 kg and m2 = 6 kg

    Brass wire is under a tension of load 6 kg.

    Therefore, increase in length is given by

    Ls = FAsLBYB= 6×9.8×10π (1.25×10-4)2 × 0.91×1011 m 
    Since, brass wire is under a tension of load 6 kg.

    "
    Steel wire is under a tension of 10 kg load. 

    Therefore, elongation of length of the wire, 

     LS = FAsLsYs               = 10×9.8×1.5π (1.25×10-4)2 × 1011m               = 1.5 × 10-4 m

    Question 107
    CBSEENPH11018116

    One meter long elastic wire is fixed at one end and at the other end a weight of 3kg is attached. The mass is pulled away from vertical line (passing through point of suspension) to a distance 0.6m and made to revolve with horizontal circle of radius 0-6m. Find the increase in length of the wire. Given that area of cross-section of wire 2x10–7m2 and Young's modulus of material of wire 2.5x1010 N/m2.

    Solution

    The given question is illustrated in the figure. below.


    From right angled ΔSOA,
    s i n theta equals fraction numerator 0.6 over denominator 1 end fraction equals 0.6
therefore space space space space space space c o s theta equals square root of 1 minus s i n squared straight theta end root equals 0.8
    The different forces acting on mass are:

    (i) Weight mg in vertically downward direction.
    (ii) Tension T in the string along AS.
    Now resolving the components of T as shown in fig. above.
    The component Tcosθ balances the weight mg of the mass and component Tsinθ provides the necessary centripetal force to whirl the mass.
    straight i. straight e. space space space space space space space straight T space c o s theta equals m g space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis

rightwards double arrow space space space space space space straight T equals fraction numerator m g over denominator c o s theta end fraction equals fraction numerator 3 cross times 9.8 over denominator 0.8 end fraction equals 36.75 straight N
N o w comma
space space space space space space space space space space space straight Y equals fraction numerator straight T divided by straight A over denominator capital delta l divided by straight l end fraction

rightwards double arrow space space space space space capital delta l equals fraction numerator Y l over denominator Y A end fraction
space space space space space space space space space space space space space equals fraction numerator 36.75 cross times 1 over denominator 2.5 cross times 10 to the power of 10 cross times 2 cross times 10 to the power of negative 7 end exponent end fraction

space space space space space space space space space space space space space equals 7.35 cross times 10 to the power of negative 3 end exponent straight m equals 7.35 m m

    Question 108
    CBSEENPH11018117

    A wire suspended vertically from a rigid support is loaded by 200 N weight. The load stretches the wire by 0.8mm. Find the elastic potential energy stored in the wire.

    Solution

    Given, 
    Weight of the load = 200 N
    Elongation caused in the wire is given by, l = 0.8 mm = 8×10-4 m 

    To find - Elastic potential energy stored in the wire = ?

    Therefore, energy stored in the wire is given by,

     U=12F×Δl    = 12×200×8×10-4   = 0.08 J  

    Question 109
    CBSEENPH11018118

    A piano wire of length 1.2m and area of cross section 4 x 10–7 m2 is under tension of 80N. Find the extension in the wire and potential energy stored in the wire. The Young's modulus of the material of wire is 2.1x1011N/m2.

    Solution
    Length of the wire, l = 1.2 m
    Area of cross-section, A = 4 x 10–7 m2
    Tension, T = 80 N
    Young's modulus of the wire, Y = 2.1 × 1011 N/m

    Now, using the formula, 

    Young's modulus is given by Y = FLA L

    Extension in the piano wire is given by, L = TLAY=80×1.2 2.1 × 1011 ×4×10-7 = 11.43 × 10-4 m

    Potential Energy stored in the wire is,

    U = 12YA(L)2L    = 12×2.1×1011×4×10-7×(11.43×10-4)21.2    = 0.0457 J
                

    Question 110
    CBSEENPH11018119

    The volume of a spherical body is decreased by10–3 % when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.

    (1 Atmosphere pressure =1.01x105 N/m2).

    Solution

    Volume of a spherical body decreased = 10-3 % 

    Pressure, P = 40 atm

    We know,
    Bulk modulus, K=-VdPdV


    Percentage change in volume is given by 10-3 %

        ΔVV=10-3100=10-5Therefore,              K=40×1.01×10510-5               =4.04×1011 N/m2 

    is the bulk modulus of the body.

    Question 112
    CBSEENPH11018121

    What is the value of one torr in Pascal?

    Solution
    1 torr = 133 Pascal
    Question 113
    CBSEENPH11018122

    What is bar?

    Solution
    Bar is the unit of pressure and is equal to 105 Pascal.
    Question 114
    CBSEENPH11018123

    You have two identical glass flasks. One is open mouthed and other is filled with helium and sealed. Which of the two will weigh more?

    Solution
    The open mouthed flask will weigh more because the atmospheric particles will get inside the flask which will be obviously heavier than the helium gas. 
    Question 115
    CBSEENPH11018124

    A cylinder of length l and area of cross section a is filled with liquid of density ρ to a height h. What is the pressure at the bottom due to height column of liquid?

    Solution
    Given, 
    Length of the cylinder = 

    Area of cross-section = A
    Density of the liquid = ρ 
    Height of the liquid column = h
    Therefore, pressure at the bottom due to height column of liquid is given by, 
                      F = mg
    ⇒              PA = mg
    and,
                     m = ρV = ρhA 
    Therefore, P = ρgh
    Thus, pressure due to height column of liquid is given by ρgh. 
     
     
    Question 116
    CBSEENPH11018125

    State Pascal’s law.

    Solution
    Pascal's law states that if some pressure is applied at any point of incompressible liquid then the same pressure is transmitted to all the points of liquid and on the walls of the container. 

    That is pressure exerted is same in all directions in a fluid at rest.
    Question 117
    CBSEENPH11018126

    Does the liquid exert any force on the walls of container?

    Solution
    Yes, liquid exerts pressure and hence force on the walls of the container due to the height of the column of liquid.
    Question 118
    CBSEENPH11018127

    Explain why?

    Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

    Solution
    When a force is applied at any point, hydrostatic pressure is transmitted equally in all directions and not just in a particular direction. 
    This the reason pressure is a scalar quantity even though force is a vector. 
    Question 119
    CBSEENPH11018128

    Explain why?

    The blood pressure in humans is greater at the feet than at the brain. 

    Solution
    The height column of blood is more at feet than at the brain. So, blood pressure is greater at feet than at the brain. 

    Sponsor Area

    Question 120
    CBSEENPH11018129

    On which law does the hydraulic system work?

    Solution
    The hydraulic system works on Pascal law.
    Question 121
    CBSEENPH11018130

    A balloon filled with helium does not rise in air indefinitely but halts after a certain height. Why?

    Solution
    When a balloon is filled with helium, there is an inner pressure because of which the balloon expands more and more as it goes higher. But as the balloon goes higher, the atmospheric pressure also decreases. This is why the balloon ultimately stops after reaching a certain height. 
    Question 122
    CBSEENPH11018131

    Atmospheric pressure at a height of about 6 km decreases to nearly half its value at the sea level, though the height of the atmosphere is more than 100 km. Explain why.

    Solution
    The density of air decreases with height. At sea level the atmospheric pressure is not always 760 mm of Hg. There is a decrease in the Hg level. 
    Question 123
    CBSEENPH11018132

    We have two barometers. One barometer contains mercury and second water. Which one will register more pressure? (Neglect the surface tension effect).

    Solution
    Mercury barometer will register more pressure than water barometer. This is because of the following reasons: 

    i) Mercury is around 14 times heavier than water i.e., it is denser than water.

    ii) The vapour pressure of mercury is low as compared to that of water. So, mercury will be more sensitive to changes in the atmospheric pressure and rises more quickly to record the changes in the atmospheric pressure. 
    Question 124
    CBSEENPH11018133

    A spring balance reads 10 kg when a bucket of water is suspended from it. What is the reading on spring balance when 2.2 kg icecube is put into it? (Density of ice is 0.920gm/cc).

    Solution
    When an ice cube is put into the bucket, then the total mass of the system suspended from the spring is given by 10 + 2.2 kg = 12.2 kg.

    That is, the spring balance will read 12.2 kg.



    Question 125
    CBSEENPH11018134

    What is the standard pressure in torr?

    Solution
     Standard pressure is 760 torr.
    Question 126
    CBSEENPH11018135

    Why storage tanks are made thick at the bottom than at the top?

    Solution
    Storage tanks are made thick at the bottom because the pressure at the bottom due to height column of liquid is greater than at the top.
    Question 127
    CBSEENPH11018136

    Does the pressure due to height column of liquid at the bottom of container depend on the shape of container?

    Solution
    No, the shape of the container does not determine the pressure due to height column of liquid at the bottom of container. 
    Question 128
    CBSEENPH11018137

    Name the factors on which the pressure due to height column of liquid at the bottom of container depends.

    Solution
    The pressure at the bottom of container depends on the following factors:

    i) acceleration due to gravity,
    ii) height column of liquid, and
    iii) density of liquid.
    Question 129
    CBSEENPH11018138

    If we exhaust the air from a sealed tin can, what will happen to it?

    Solution
    If the air from the sealed tin can is exhausted, the tin can will squeeze due to inward force exerted by atmospheric pressure.
    Question 130
    CBSEENPH11018139

    What is a barometer?

    Solution
    Barometer is a device used to measure the atmospheric pressure. Barometer keeps a track of the change in the atmospheric pressure. 
    Question 131
    CBSEENPH11018140

    Barometer tube of mercury barometer is one meter long. Would you require a longer or a shorter tube if the mercury were replaced by water in the tube?

    Solution
    If the mercury in the barometer tube is replaced with water, we need a longer barometric tube. Water is less denser tan mercury and so we need longer tube. 
    Question 132
    CBSEENPH11018141

    An air bubble is released from the bottom of tank. What happens to the size of air bubble as it rises upward in the tank?

    Solution
    When an air bubble is released from the bottom of the tank, as the bubble rises up, the pressure inside the bubble decreases and hence the size of bubble expands.
    Question 133
    CBSEENPH11018142

    State Archimedes’s principle.

    Solution
    Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. 

    The law is fundamental to fluid mechanics. 
    Question 134
    CBSEENPH11018143

    When does a body gets totally immersed in the liquid? 

    Solution
    A body gets totally immersed in the liquid when the density of the body is greater than the density of the liquid.


    Question 135
    CBSEENPH11018144

    When the body in the liquid floats?

    Solution
    When the density of the body is less than that of the liquid, a body in the liquid floats. 
    Question 136
    CBSEENPH11018145

    State laws of floatation.

    Solution

    The laws of floatation are:

    (i) The weight of body is equal to the weight of liquid displaced.

    (ii) Centre of gravity of body and that of liquid displaced are in same vertical line. 

    Question 137
    CBSEENPH11018146

    What is centre of buoyancy?

    Solution
    When a body is placed in a liquid, it displaces some volume of the liquid.
    Centre of the buoyancy is the centre of gravity of the liquid displaced by the body. 
    Question 138
    CBSEENPH11018147

    Where should be the centre of buoyancy of a floating object so that object is in stable equilibrium?

    Solution
    For an object to be in stable equilibroum, the centre of buoyancy should be vertically above the centre of gravity of body. If the body tilts in liquid the restoring torque brings the body to original state in this case.
    Question 139
    CBSEENPH11018148

    Does the force of buoyancy depend upon the mass of the body?

    Solution
    Force of buoyancy depends upon the mass of liquid displaced and does not depend upon the mass of the body.

    Question 140
    CBSEENPH11018149

    What are the factors on which the buoyant force depends?

    Solution
    The factors on which the buoyant force depends are:

    i) volume of body inside the liquid,
    ii) density of liquid, and
    iii) acceleration due to gravity.
    Question 141
    CBSEENPH11018150

    A body of density twice the density of water is first weighed in water and then in glycerine of density 1.3gm/cc. In which case the body appears to be lighter?

    Solution
    Density of glycerine is higher than water. So, glycerine exerts more buoyant force and hence the body will appear lighter in glycerine than in water. 
    Question 142
    CBSEENPH11018151

    Write an expression for buoyant force exerted by liquid when immersed in liquid.

    Solution

    The buoyant force on body is, 

                               U = Vρg 

    where,
    V is the volume of body immersed in liquid,and
    ρ is the density of liquid.

    Question 143
    CBSEENPH11018152

    Why does ice float over water?

    Solution
    Ice floats over water because the density of ice is less than that of water.
    Question 144
    CBSEENPH11018153

    A liquid exerts a buoyant force of 2.4 kg on a body of mass 6.4 kg when it is immersed in the liquid. What is the apparent weight of the body in the liquid?

    Solution
    Buoyant force exerted by the liquid = 2.4 kg
    Mass of the body when immersed in liquid, m = 6.4 kg 

    Therefore,

    Apparent weight of the body in the liquid is given by, = 6.4 –2.4 = 4 kg. 
    Question 145
    CBSEENPH11018154

    A cork is floating in water. What is the apparent weight of the cork?

    Solution
    Apparent weight = weight of cork - weight of displaced liquid. 

    When the cork is floating in the liquid, the weight of the cork is equal to the weight of the liquid displaced. 

    So, the apparent weight = 0
    Question 146
    CBSEENPH11018155

    What is the direction in which the buoyant force acts?

    Solution
    Bouyant force acts in the upward direction. That is in a direction opposite to the direction of gravity.
    Question 147
    CBSEENPH11018156

    When the floating bodies are in unstable equilibrium?

    Solution
    For floating bodies, the centre of buoyancy lies below the centre of gravity. Hence, they are in unstable equilibrium. 
    Question 148
    CBSEENPH11018157

    What the distance between meta centre and centre of buoyancy called?

    Solution

    The distance between meta centre and centre of buoyancy is called meta-centre height.

    Question 149
    CBSEENPH11018158

    When will the floating body be in neutral equilibrium?

    Solution
    When the centre of gravity of body coincides with the centre of buoyancy, the floating body will be in neutral equilibrium. 
    Question 150
    CBSEENPH11018159

    What is thrust? What is its direction?

    Solution
    The force exerted perpendicular by liquid on any surface in contact is known as thrust. 
    Question 151
    CBSEENPH11018160

    Define the terms thrust and pressure. Give the SI unit of pressure. Name three applications of the knowledge of pressure.

    Solution

    Thrust: The force exerted by liquid normal to any surface in contact with it, is called thrust.

    Pressure: The thrust exerted by a liquid per unit area of surface in contact with liquid is called pressure.

    The SI unit of pressure is N/m2 or Pascal (Pa).

    Application of pressure:

    i) Hydraulic machines
    ii) Blood pressure - Pressure exerted by the circulating blood upon the walls of the blood vessels. 
    iii) Variation of pressure with depth - submarines are designed to withstand such enormous pressure. 




    Question 152
    CBSEENPH11018161

    Show that the force acting on the liquid in equilibrium is perpendicular to the free surface of liquid.

    Solution
    Consider a liquid in a container. The liquid is in an equilibrium condition. Force F is acting on the liquid at angle θ with the free surface of liquid as shown in figure below.

     


    On resolving the components of force, we get
    Fcosθ - component of force along the surface
    Fsinθ - component of force ormal to the surface

    In the equilibrium state, component of force along the surface is 0. 

    Therefore, 

        Fcosθ = 0 
     cos θ = 0o i.e.,    θ = 90o

    So, the force must be normal to surface.
    Thus, the free surface of liquid in equilibrium adjusts normal to the direction of force acting on it. 


     
    Question 153
    CBSEENPH11018162

    State and prove Pascal's law.

    Solution

    PAscal's law states that, if some pressure is applied at any point of incompressible liquid then the same pressure is transmitted to all the points of liquid and on the walls of the container.

    Let us imagine an arbitrary right angled prismatic triangle in the liquid of density ρ. This prismatic element is very small so, every part is considered at the same depth from the liquid surface. Therefore, effect of gravity is the same at all these points. That, the small element is in equilibrium. 


    The area of faces ABFE, ABDC and CDFE are ad, bd and cd respectively. Let the pressure of liquid on faces ABFE, ABDC and CDFE be P1, P2 and P3respectively.

    The pressure of liquid exerts the force normal to the surface. Let us assume pressure P1exerts the force F1 on the surface ABFE, pressure P2 exerts force F2 on the surface ABDC and pressure P3 exerts force on the surface CDFE. 

    So, Force F1 is given by, 

    F1 = P1 × Area ABFE = P1ad F2 = P2 × Area ABDC = P2bd F3 = P3 × Area CDFE = P3cd Also, sin θ = ba and sin θ = ca 

    Since the prism is in equilibrium, so net force on the prism is zero. 

    Thus, 

    F1 sin θ = F2 , and F1 cos θ = F3 Therefore, P1 ad (b/a) = P2 bd , and    ... (1)P1 ad (c/a)= P3 cd               ... (2) So, from equations (1) and (2), we haveP1 = P2 , and P1 = P3  P1 = P2 = P3 So, Pascal's law is proved.




    Question 154
    CBSEENPH11018163

    Verify Pascal’s law by giving an example.

    Solution
    Verification of Pascal's law giving an example

    Given, a container having opening of different cross-section provided with frictionless piston containing incompressible liquid. 

     


    Let the area of cross-section of pistons be A, 2A and 3A respectively. 

    Now, applying force F on the piston of area A, keeping the other pistons in position, a force of 2F is required on piston of area 2A and 3F force is required on piston of area 3A.

    Therefore,

    Pressure on piston of area A is, 
                                 P1=FA
                           
    Pressure on piston of area 2A is, 
                       P2=2F 2A=FA

    Pressure on piston of area 3A is,

                      P3=3F3A=FA 

    Therefore, from the above 3 equations it is proved that pressure exerted is the same. 

    Hence, Pascal's law is proved. 
    Question 155
    CBSEENPH11018164

    What are practical applications of Pascal's law?

    Solution
    The practical application of Pascal's law are:

    (i) Hydraulic lift
    (ii) Hydraulic press
    (iii) Hydraulic brake.
    Question 156
    CBSEENPH11018165

    What is hydraulic lift? What is its principle? Discuss its working.

    Solution

    Hydraulic lift is a machine which is based on Pascal's law. In these devices fluids are used for transmitting presure. 

    Hydraulic lift consists of two cylinders of different cross-sectional areas connected with a pipe. The cylinders are filled with incompressible liquid and frictionless pistons are fitted in both the cylinders as shown in figure below. 



    Let 'a' and 'A' be the area of cross-section of smaller piston and bigger piston respectively. The load to be lifted is placed on bigger cross-section and effort is applied on smaller piston.
    Let a force 'f' be applied on the smaller piston.

    The pressure exerted by f on piston is,


                            P=fa    ... (1) 

    Now, according to Pascal's law, the pressure transmitted to bigger piston is also P.

    Thus, force on bigger piston is, given by, 
    F = PA = Aaf We have, A > a.Let, A = ηaSo, F = ηf 

    That is, the transmitted force gets multiplied by a factor η of applied force. 


     

    Question 157
    CBSEENPH11018166

    Obtain an expression for the pressure exerted by liquid column.

    Solution

    Consider a liquid of density ρ in a vessel as shown in figure.


    To find: Pressure difference between two points A and B separated by vertical height h.

    Consider an imaginary cuboid of area of cross-section a of liquid with upper and lower cap passing through A and B respectively in order to evaluate the pressure difference between points A and B. 

    Volume of the imaginary cylinder is, V = ah 

    Mass of liquid of imaginary cylinder, m = ρah

    Let, P1 and P2 be the pressure on the upper and lower face of cylinder.

    Forces acting on the imaginary cylinder are:

    (i) Weight, mg = ρahg in vertically downward direction.

    (ii) Downward thrust of F1 =P1a on upper cap.

    (iii) Upward thrust of F2 = P2a on lower face.

    As the imaginary cylinder in the liquid is in equilibrium, therefore the net force on the cylinder is zero. 

                 i.e., F1 + mg = F2  P1a + ρahg = P2a    P2 - P1 = ρgh 

    Thus, the pressure difference between two points separated vertically by height h in the presence of gravity is ρgh.

    Note: In the absence of gravity this pressure difference becomes zero.

     

    Question 158
    CBSEENPH11018167

    What is atmosphere and atmospheric pressure? Discuss the Torricelli’s experiment to measure the atmospheric pressure.

    Solution

    Atmosphere is a gaseous envelope surrounding the earth and the pressure exerted by the atmosphere is called atmospheric pressure.

    The cause of atmospheric pressure is motion of air molecules which are in continuous mtion.
    The molecules strike the surface of body placed in it and exert a huge force.

    To measure the atmospheric pressure,
    i) Torricelli took a meter long graduated tube and filled it with clean and dry mercury.
    ii) By closing the tube with thumb, he inverted the tube in a cistern ( a tub filled with mercury) as shown in the figure.


    iii) He observed that the level of mercury first fell down and finally stayed with a column of height 76cm in the tube above the free surface of mercury in the cistern leaving behind vacuum.
    iv) At point C, vacuum is formed. Therefore, pressure at this point is zero. Point B in the tube is 76cm below C. So, pressure at B is given by, 
    PB =PC + ρgh = 0 + ρgh =  ρgh 

    where ρ is density of mercury and h be the height column of mercury in the tube above point B.

    In the given figure, point A is at the interface of air and mercury, therefore it is both in air and as well as in mercury.
    Thus pressure at point A is equal to atmospheric pressure, i.e.

    PA = Atmospheric pressure as vertical height between A and B is zero, therefore pressure at A and B is same, i.e.

                                     PA=P
    Atmospheric pressure = pgh
                                     = 13600 x 9.8 x 0.76
                                     = 1.013 x 10N/m

    Question 159
    CBSEENPH11018168

    What is the approximate thrust exerted by air on the human body?

    Solution
    The average surface area of human body is about 1.5m2.
    The atmospheric pressure is about 1 x 10
    5N/m2.

    We have, 
    Pressure, P = FA  F = P×A

    Therefore,
    Thrust exerted by air = 1.5x10
    5N. 
    Question 160
    CBSEENPH11018169

    Due to atmospheric pressure, the air exerts a force of 1.5x105N. This force is equivalent to if we place 15000kg weight on our body. Why we do not get squeeze or get crushed under such a high force?

    Solution
    Given that, 
    Air exerts a force of 1.5 × 105 N 

    We do not get squeezed or crushed under such a high force because our body is able to bear such a heavy weight as body contains tiny pores. The air exerts pressure from inside. Also, there is blood pressure which is exerted from inside. 

    Both these pressure combine and balance the atmospheric pressure which is exerted, enabling us to tolerate such a huge pressure without getting crushed. 
    Question 161
    CBSEENPH11018170

    Why high blood pressure patients are not advised to go at high altitudes?

    Solution
    the atmospheric pressure decreases as we go higher. So, at higher altitudes, the blood pressure is greater than atmospheric pressure. If a person already has the problem of high pressure, the difference between blood pressure and atmospheric pressure becomes so high. This results in bleeding from the nose or ears and in some cases the veins burst. 


    Question 162
    CBSEENPH11018171

    Why the foundation of high building is made wide?

    Solution
    The whole weight of building is on the foundation. The foundation of building should be made wide enough otherwise, the stress on the earth will be greater than the stress at which the earth yields and hence building will tend to collapse. 
    Question 163
    CBSEENPH11018172

    Explain with an example that atmosphere exerts a huge pressure.

    Solution
    Atmosphere exerts a huge pressure.

    For example, let us consider a hollow sphere.

    i) Cut the sphere into two hemispheres.
    ii) Place the two hemispheres in contact with each other.
    iii) When there is air inside the hemispheres, two get easily separated. This is because there is air on both the sides of sphere at same pressure and hence net force on each hemisphere is zero and could be easily separated. 

     


    iv) Now remove the air by using vacuum pump from the two hemispheres by placing in contact with each other.
    v) When there is prefect vacuum inside, you will see that you are not able to separate them even by applying a large force.

    As shown in figure, if we have a sphere of surface area 1 m2, we need the weights more than 10 quintal to separate the two hemispheres. This experiment shows that air exerts a huge pressure.
    Question 164
    CBSEENPH11018173

    Why the passengers in aeroplane are advised to remove the ink from their fountain pens?

    Solution
    As altitude increases, the atmospheric pressure decreases in comparison to that on the surface of the earth. Ink on the fountain pens is filled at the atmospheric pressure on the surface of the earth. Therefore, as height increases ink will come out of the pen.
    Question 165
    CBSEENPH11018174

    To empty an oil tin, two holes are made diagonally. Why?


    Solution
    When hole A is made in the tin, the oil coming out will be less because the volume of the air above the oil increases and hence the pressure of the air decreases, obstructing the flow. When the pressure of air above oil plus pressure due to height column of oil becomes less than atmospheric pressure, the oil stops to ooze out. 

    If another hole B is made which is diagonal to the first hole, atmospheric air from this hole keeps on entering and hence pressure at A remains greater than atmospheric pressure and hence oil will continue to flow out from the tin.
    Question 166
    CBSEENPH11018175

    Straws are used to take soft drinks. Why?

    Solution
    Straws make use of the capillary action as a result of the difference in pressure.

    When a person suck though the straw, pressure inside the straw decreases and due to the difference in pressure outside and inside the straw, the liquid rises in the straw. 
    Question 167
    CBSEENPH11018176

    In a dropper, the liquid does not come out unless its bulb is pressed.

    Solution
    The pressure due to the height column of liquid inside the dropper is less than atmospheric pressure. That is the liquid is held inside the dropper against the atmospheric pressure. On pressing the dropper, the pressure inside is increased and when the pressure inside exceeds the atmospheric pressure, liquid comes out from the dropper. 
    Question 168
    CBSEENPH11018177

    Why does a siphon not work in vacuum?

    Solution
    The functioning of siphon is based on the atmospheric pressure. On sucking, the pressure of the air inside the siphon decreases and the atmospheric pressure on the liquid outside pushes the liquid up in the siphon. If there is vaccum, i.e., there is no atmospheric pressure, then there is no force on the liquid that can push the liquid into the siphon and siphon does not work. 
    Question 169
    CBSEENPH11018178

    Why it is easy to cut with a sharp knife than blunt one?

    Solution
    Sharp knife-edge has smaller area than a blunt one. When the force is applied on the knife, the sharp edged knife produces more pressure on the surface and gets penetrated into the surface.
    Thus, it is easy to cut with a sharp knife than blunt one.
    Question 170
    CBSEENPH11018179

    Why are paper pins and nails made to have pointed ends?

    Solution
    Pointed paper pins or nails have small area as compared to the blunt one. Lesser the area, greater is the force applied. When force is applied on the pins, they exert large pressure on the surface and hence easily pierce the surface and get penetrated into it. 
    Question 171
    CBSEENPH11018180

    Explain why water cannot be used in place of mercury in a barometer?

    Solution
    The atmospheric pressure at sea level = 76cm of Hg = 1.013x10Pascal. 

    If we fill the water in barometer tube, the height of water column required to balance the atmospheric pressure will be,

       h =Atmospheric PressureDensity of water×acceleration due to gravity   =1.013×1051000×9.81   =10.326m 

    That is, if water is used in barometer tube instead of mercury, the length of the tube must be greater than 10.326 cm. A greater height will be difficult to hold in a vertical position and needs a high tower to install the barometer. 

    So, we cannot replace mercury by water in the barometer. 

    Question 172
    CBSEENPH11018181

    Three vessels have same base area. These vessels are filled to same height by same liquid. In which vessel will the force on the base be maximum?

    Solution
    As all the vessels are filled with same liquid to the same height, therefore the pressure due to height column at the base will be the same. Also, since the base area of all the vessels is same, therefore the force acting on all the base of all vessels will be the same.
    Question 173
    CBSEENPH11018182

    What do you mean by absolute pressure and gauge pressure?

    Solution

    Absolute pressure is the pressure above that of vacuum. 

    Absolute pressure = Gauge pressure + One atmospheric pressure.

    Gauge pressure is the pressure above that of one atmospheric pressure. Since the pressure of vacuum is zero, it is referenced against an ambient air pressure.

    Note: Atmospheric pressure is the pressure at any point equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. 

    Question 174
    CBSEENPH11018183

    What is manometer?

    Solution
    Manometer is a U-tube open at both the ends having some mercury in it.


    It is used to measure the gauge pressure or pressure difference of gases on the two sides of manometer tube by measuring the height difference of mercury column in two limbs of manometer tube.  

    If, the difference in the height column of mercury in the two limbs of tube = h, then 

    Gauge pressure = ρgh 

    Absolute pressure = P
    a + ρgh.
    Question 175
    CBSEENPH11018184

    What height of water column produces the same pressure as 50 cm of Hg column?

    Solution

    The pressure due to height column, P = ρgh

    If the pressure due to height column of different liquids is same, then the relation between the height coulumn is given by,

                  ρ1gh1=ρ2gh2 h2=ρ1ρ2h1=13.61×50                         = 680cm                         =6.80m , 
    is the required height of the water column which will produce the same pressure as 50 cm of Hg column. 

    Question 176
    CBSEENPH11018185

    A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

    Solution

    Given, 
    Weight of the girl = 50 kg = 490 N
    Circular diameter of the heel = 1.0 cm = 10-2

    Area of the heel is given by, 

               A=14πD2=14π×10-4m2 

    Threfore, the pressure exerted by the heel on the ground is given by, 

           P = WA=49014π×10-4=6.2×106N/m2

    Question 177
    CBSEENPH11018186

    The atmospheric pressure at sea level is 760mm of mercury. At what height it will be 740mm of mercury? Take the density of air 1.293kg/m3 constant upto height about 1 km.

    Solution
    Given, atmospheric pressure, P1= 760 mm of Hg
    Atmospheric pressure, P2 = 740 mm of Hg
    Density of air, ρ = 1.293 kg/m3

    Let the height at which the atmospheric pressure is decreased be 'h'. 

    Now, Pressure due to h height of air = Pressure due to 2 cm or 0.02 m of mercury. 

    Threfore,

    1.293 x g x h = 13600 x g x 0.02 

                  h = 210 m, is the height at which the atmospheric pressure will decrease. 
    Question 178
    CBSEENPH11018187

    A force of 2x102 N is applied on one of the piston of area of cross section 100cm2of hydraulic lift to support a car placed on the second piston of lift of area of cross section 1960cm2. Find the mass of the car.

    Solution
    Given,
    Force, F = 2×102
    Area of cross-section of the first piston = 100 cm2 
    Area of cross-section of the second piston = 1960 cm

    Weight of the car = ?

    Let, M be the mass of car.

    The pressure on the piston on which force is applied is, 

                  P1=Fa=2×102100=2 N/cm   ...(1)

    The pressure on the second piston on which car is placed is, 

               P2=MgA=M×9.81960=M200N/cm    ...(2)

    Now, According to Pascal law we have

                           P=  P2    

     we get from equation (1) and (2),  

    Mass of the car, M=400  kg 
    Question 179
    CBSEENPH11018188

    A U-tube contains water and methylated spirit separated by mercury. The mercury in the two limbs of the tube is in level. If the water stands to a level 12.8cm above the mercury, then find the length of spirit in the U-tube. Take relative density of spirit as 0.8.

    Solution
    Given that, for both the limbs mercury is in level. Therefore, the pressure due to height column of water and spirit will be same. 

    Also, given water stands to a level of 12.8 cm above the mercury.

    Let, h be the length of the spirit in the U- tube. 

    So, Pressure due to 12.8cm  of water = Pressure due to h cm of spirit

    or       1 x g x 12.8 = 0.8 x g x h

    or                      h = 16 cm, is the length of the spirit in the U-tube. 
    Question 180
    CBSEENPH11018189

    Two limbs of hydraulic lift have area of cross-section 5 cm2 and 100 cm2. To lift the weight of 24 kg placed on 5 cm2 cross-section, what force must be applied on the piston of cross-section 100 cm2?

    Solution
    Given, two hydraulic lifts. 

    Area of cross-section of first limb = 5 cm2 

    Area of cross section of second limb = 100 cm

    Weight of the load = 24 

    Area of cross-section on which the load is kept = 5 cm2 

    Now, using Pascal's law, we have 

         F1a1 = F2a2  F2 = a2a1F  

    Here, a= 5 cm2 , F= 235 N
    a2 = 100 cm2 

    So, force F21005× 235 = 4700 N, is the force applied on a piston of cross-sectional area 100 cm2


    Question 181
    CBSEENPH11018190

    What is force of cohesion?

    Solution
    Force of cohesion is the intermolecular force between two same type of molecules.

    For example, solids have high cohesive properties, so they do not stick to the surface they come in contact with. 
    Question 182
    CBSEENPH11018191

    What is force of adhesion?

    Solution
    Force of adhesion is the intermolecular force between two different types of molecules. 

    E.g: The effect of capillary action is due to the adhesive force. The water molecules touching the surface of the container are at a higher level (meniscus) is also an example of adhesive force. 
    Question 183
    CBSEENPH11018192

    Define surface tension.

    Solution
    Surface tension is the property of liquid due to which the free surface of liquid at rest behaves like a stretched membrane and tries to acquire the minimum surface area.

    For e.g., oil and water do not mix, oil rises up in a cotton wick.  
    Question 184
    CBSEENPH11018193

    What are the units and dimensions of surface tension?

    Solution

    Unit of surface tension:

    SI system: N/m
    CGS system: dyne/cm

    Dimensional formula of surface tension is [M1L0T–2].

    Question 186
    CBSEENPH11018195

    Define surface energy.

    Solution
    Surface energy is defined as the amount of work required to increase the unit area of free surface under isothermal condition. 



    Question 187
    CBSEENPH11018196

    How surface tension is related to surface energy?

    Solution
    The surface energy is dimensionally and numerically equal to surface tension. 
    Question 188
    CBSEENPH11018197

    What is sphere of influence?

    Solution
    The vicinity around the iquid molecule within which it's influence can be experienced by another molecule is called as the sphere of influence.
    Question 189
    CBSEENPH11018198

    What is the order of the size of sphere of influence?

    Solution
    The sphere of influence is of the order of 10–7m. 
    Question 190
    CBSEENPH11018199

    What is the origin of surface tension?

    Solution
    Surface tension is originated as a result intermolecular forces. Surface tension is concerned with only liquid. 
    Question 191
    CBSEENPH11018200

    Why ends/edges of glass become round on heating?

    Solution
    Molten gas tries to acquire the minimum surface area and hence heats up due to surface tension. Rounding off the edges makes them acquire the minimum surface area. 
    Question 192
    CBSEENPH11018201

    Does the surface tension of a liquid depend on the area of the free surface of liquid?

    Solution
    Surface tension of the liquid is independent of the area of the free surface of liquid. 
    Question 193
    CBSEENPH11018202

    Why does the oil spread over the water?

    Solution
    Surface tension of oil is less than water, therefore, oil spreads over water and do not mix with each other. 
    Question 194
    CBSEENPH11018203

    Explain how the dental plate clings to the roof of mouth?

    Solution
    Dental plate clings to the roof of the mouth because of the presence of adhesive force between saliva and dental plate. 
    Question 195
    CBSEENPH11018204

    What is the thickness of surface film?

    Solution
    The thickness of surface film is equal to range of intermolecular forces of that liquid.
    Question 196
    CBSEENPH11018205

    Why do the two mercury drops coalesce when brought together?

    Solution
    Molecules of mercury coalesce due to strong cohesive force between the mercury molecules. 

    Cohesive force is the force of attraction between like molecules. 
    Question 197
    CBSEENPH11018206

    What is meniscus?

    Solution
    Meniscus is the curved upper surface area of the liquid near the walls of the container.
    Question 198
    CBSEENPH11018208

    What is the shape of meniscus of pure water in clean glass capillary tube?

    Solution
    The shape of meniscus of pure water in clean glass capillary tube is concave. 
    Question 199
    CBSEENPH11018209

    Can the shape of meniscus of water in capillary tube be convex?

    Solution
    Yes. The meniscus of water in capillary tube can be convex. 
    Question 200
    CBSEENPH11018210

    When is the shape of meniscus of liquid in the capillary tube convex upwards?

    Solution
    The shape of the meniscus of liquid in the capillary tube is convex that is, pointing upwards when the angle of contact of liquid is greater than 90°. 
    Question 201
    CBSEENPH11018211

    When is the shape of meniscus of liquid in the capillary tube concave upwards?

    Solution
    The shape of the meniscus liquid in the capillary tube concave upwards if the angle of contact of liquid is less than 90°.
    Question 202
    CBSEENPH11018212

    What happens to surface tension when impurity is mixed in liquid?

    Solution
    When impurities are mixed in liquid, surface tension of liquid decreases. The soluble substances when dissolved in water, decreases the surface tension of water. 
    Question 203
    CBSEENPH11018213

    How does the surface tension of liquid change with temperature?

    Solution
    With increase in temperature, the surface tension of liquid decreases. Surface tension of liquid is zero at boiling point and it vanishes at critical temperature.
    At critical temperature, liquid can expand without any restriction.

    Question 204
    CBSEENPH11018214

    Why does the wick of lamp is made of cotton?

    Solution
    Wick is made of cotton as the capillaries inside it enable the oil to rise in it. This makes the burning process more effective. 
    Question 205
    CBSEENPH11018215

    What is surface tension at critical temperature?

    Solution
    At critical temperature, surface tension reduces to zero. 
    Question 206
    CBSEENPH11018216

    A capillary is dipped in a liquid that does not stick with the walls of tube. How will the level of liquid change in the capillary tube?

    Solution
    The level of liquid in the capillary tube will fall when a capillary is dipped in a liquid that does not stick with the walls of the tube. 
    Question 207
    CBSEENPH11018217

    A capillary is dipped in a liquid that sticks with the walls of tube. How will the level of liquid change in the capillary tube?

    Solution
    When capillary is dipped in a liquid that sticks with the walls of the tube, the level of liquid in the capillary tube will rise. 
    Question 208
    CBSEENPH11018218

    On which side of the meniscus the pressure is more?

    Solution
    On the concave meniscus, pressure is more because of surface tension.

    Force exerted by the liquid is given by, F = Sl 

    Concave meniscus is formed as a result of water creeping up on the surface of the sides of the container. 
     
    Question 209
    CBSEENPH11018221

    How will the liquid rise in a capillary tube in state of weightlessness?

    Solution
    In the state of weightlessness, the liquid rises to brim of capillary tube.  
    Question 211
    CBSEENPH11018223

    Why do people prefer cotton dresses in summers?

    Solution
    Fine pores in the cotton fibres absorb the sweat as a result of capillary action. Therefore, people prefer wearing cotton clothes in summers.
    Question 212
    CBSEENPH11018224

    Air is blown in the soap bubble to increase its size. How will the air pressure inside the soap bubble change?

    Solution
    When air is blown in the soap bubble to increase it's size, the pressure inside the soap bubble decreases. 
    Question 213
    CBSEENPH11018226

    Write the expression for the height to which liquid rises or falls in the capillary tube when dipped in the liquid.

    Solution
    The height to which the liquid rises or falls in the capillary tube when dipped in the liquid is given by, 

                            
                           h=2T cosθrpg      
    Question 214
    CBSEENPH11018227

    What is the excess of pressure inside the soap bubble?

    Solution
    The excess pressure inside the soap bubble is given by, 

                              P = 4 Sr
    where, 
    S.A is the surface tension, and 
    r is the radius of the bubble. 
    Question 215
    CBSEENPH11018229

    A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the relative density of spirit?

    Solution
    Given,
    The level of mercury column with water in one arm, hw = 10cm
    Level of mercury column with sirit on the other arm, h
    s = 12.5cm

    Also, mercury columns in two arms are in level. 




    Therefore, pressure due to 10 cm of water = pressure due to 12.5 cm of spirit. 

    i.e.       10ρwg= 12.5ρsgor          ρsρw=10σ12.5=0.8 0.8 is the relative density.  
    Question 216
    CBSEENPH11018231

    A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. If 15.0 cm of water and spirit are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6).

    Solution
    The mercury column in two arms are in level with 10cm of water in one arm and 12.5cm of spirit in the second arm.

    Thus the density of spirit is,ρs = 1012.5×1 = 0.8 gm/cc 

    Let the difference in the level of mercury in two arms be x, on adding 15.0cm of water and 15.0cm of spirit in respective limbs. 

    Now, 

    Pressure due to 25 cm of water = Pressure due to x cm of mercury +  Pressure due to 27.5 cm of spirit

             ρwhwg = ρhg xg + ρshsg          ρwhw = ρhgx + ρshs        25×1 = (13.6)x + (0.8)(27.5)                x = 25 - (0.8)(27.5)13.6 = 0.221 cm 





     
    Question 217
    CBSEENPH11018233

    A manometer reads the pressure of a gas in an enclosure as shown in the figure. The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury. 
     
    Give the absolute and gauge pressure of the gas in the enclosure. 


    Solution
    Atmospheric pressure of mercury = 76 mm

    The difference in the Hg column in two arms of manometer as given in the figure is 20 cm.

    Therefore the gauge pressure is +20 cm of Hg.

    Positive sign indicates that the pressure of gas is greater than atmospheric pressure.

    So,
    Absolute pressure = Atmospheric pressure + gauge pressure 
                               = 76cm + 20cm

                               = 96cm of Hg. 
    Question 218
    CBSEENPH11018235

    Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg/m3. Determine the height of wine column for normal atmospheric pressure.

    Solution
    Density of wine = 984 kg/m3  

    Now, according to Pascal's law, we have

    Pressure due to height column of liquid is, 

                        P = ρgh         ... (1) 

    The normal atmospheric pressure is 760 mm og Hg. 

    If h' is the height column of wine corresponding to one atmospheric pressure, then 

                        P = σgh'        ... (2)

    where σ is the density of the wire. 

    Now from equations (1) ad (2), we get 

                ρgh = σgh'           h' = ρhσ                    = 13600×0.76984                     = 10.5 m , 

    is the height of wine column for normal atmospheric pressure. 



    Question 219
    CBSEENPH11018236

    A capillary tube is placed vertically in water and water rises to height h above the surface of water in the capillary tube. If the tube is inclined with vertical at angle θ, then it rises to height t. Find t.

    Solution
    Given, a capillary tube is placed in water and water rises to a height h above the surface of water in the capillary tube. 

    The vertical height to which the liquid rises in the capillary tube remains constant as shown.

     

    From the figure, we have 

    h/ t = cos θ 
    ⇒ t = h cos  [∵ cos θ < 1]

     ∴  t > h 



    Question 220
    CBSEENPH11018246

    State and prove Archimedes principle.

    Solution

    Archimedes principle: When a body is immersed partially or wholly in a liquid, its weight appears to be reduced and loss of weight is equal to the weight of liquid displaced by the body.
                      

    Proof: Consider a body of height h and area of cross-section A placed in liquid of density σ at a depth x below the free surface.
    Let ρ be density of the body.

    Now the pressure at the upper face of the body is,
                         straight P subscript 1 equals straight P subscript 0 plus xσg 
    where,  Pis atmospheric pressure.
    Pressure at the lower face of the body is, 
                      straight P subscript 2 equals straight P subscript 0 plus left parenthesis straight h plus straight x right parenthesis σg
    Therefore downward thrust on  upper face of the body is,
                      straight F subscript 1 equals straight P subscript 1 straight A
    and upward thrust on lower face of body is,
                      straight F subscript 2 equals straight P subscript 2 straight A
    Since P> P1  therefore F> F

    Body will experience the net force F- Fin upward direction known as upward thrust.
    
therefore space space space space space straight U equals straight F subscript 2 minus straight F subscript 1 equals left parenthesis straight P subscript 2 minus straight P subscript 1 right parenthesis straight A space

space space space space space space space space space space space equals straight A open square brackets open parentheses straight h plus straight x close parentheses σg minus xσg close square brackets equals Ahσg
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space
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#6 {main}</pre>
    where, V is volume of body
            = Weight of liquid displaced by body.

    Now, the  body experiences two forces: weight of the body itself and upward thrust.
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    i.e. Apparent weight is less than the actual weight by amount equal to weight of liquid displaced by body.
    Question 221
    CBSEENPH11018248

    Discuss how we measure the volume of an irregular body using Archimedes’ principle.

    Solution
    According to Archimedes principle, in order to measure the volume of a body, weigh it in air and in liquid.

    Let,
    W
    1 and W2 be the weight of body in air and liquid,
    V be the volume of body, and
    ρ be the density of liquid.

    Now, as per Archimedes’s principle, decrease in weight of body in liquid is equal to the weight of liquid displaced by body. 

    That is, 

              V ρ g = W1 - W2   V =W1 - W2ρg 



    Question 222
    CBSEENPH11018250

    Discuss using Archimedes' principle, how we can measure the specific gravity of a material?

    Solution
    The density of the body relative to water is known as the specific gravity of a body. 

    To measure the specific gravity of the body, the body is first weighed in air and then in water. 

    Let,
    W1 and W2 be the weights of the body in air and water respectively, 
    ρ - density of water, and 
    σ - density of body.

    According to Archimedes principle, loss of weight of body in liquid is given by, 


               W1-W2=Vρgi.e.,W1W1-W2=VσgVρg  W1W1-W2=σρ= σ                   ρ=1                   σ=W1W1-W2 
    Question 223
    CBSEENPH11018251

    Ships are made up of metals whose densities are more than density of water. Even then the ships float in water. Explain why.

    Solution
    The ships float in water even though they are made up of metals whose densities are more than the density of water. The ships are made to float in water by making these metals hollow. By making the metals hollow, the effective density of the ship becomes less than that of water. 
    Question 224
    CBSEENPH11018252

    Why does the ship rise as it enters the sea from river?

    Solution
    The density of sea water is greater than the river water. Therefore, to balance the weight of ship, the volume of sea water displaced by the ship will be less than that of the river water. Hence on entering the sea from river the ship rises a little. 
    Question 225
    CBSEENPH11018253

    A boat floating in a water tank is carrying a number of large stones. If the stones are unloaded into water, what will happen to the water level?

    Solution
    The level of water will fall when stones are unloaded into water. The density of stones is more than the density of water. So, when the stones are dropped into the water, the stone replaces the volume of water equal to the volume of stones. 
    Question 226
    CBSEENPH11018254

    The force required by a man to move his limbs immersed in water is smaller than the force for the same movement in air. Why? 

    Solution
    Upward thrust acting on the limbs is more in water than in air.
    Therefore, the force required to move the limbs in water is smaller than that in air.
    Question 227
    CBSEENPH11018255

    An ice piece floats in a glass of water. How does the level change when whole of ice melts?

    Solution
    When the whole ice melts, the water level will reamin unchanged. The floating ice piece will displace the volume of water equal to the weight of ice. When the ice melts the volume of water so formed is exactly equal to the volume of water displaced by ice. Hence, the water level will remain unchanged. 
    Question 228
    CBSEENPH11018256

    What is the fractional volume submerged of an ice cube in the pail of water placed in an enclosure which is falling freely under gravity?

    Solution
    A system is in a state of weightlessness, when the system is falling freely. So, weight of the ice cube as well as  upward thrust would be zero if the object is falling freely. So, ice cube floats with any volume submerged.


    Question 229
    CBSEENPH11018257

    A hollow piece of metal floats in water with 80% of its volume inside the water. Find the percentage of volume submerged when the same is taken on the moon.

    (The acceleration due to gravity on moon is one-sixth the acceleration due to gravity on the earth).

    Solution
    The fractional volume of body submerged in water depends only on the relative density of floating body and is independent of acceleration due to gravity.

    Therefore, the percentage of volume submerged is the same when taken on the moon. 

    Question 230
    CBSEENPH11018258

    Can cubic block be placed in tank of water in such a way that there is no upward thrust on the block?

    Solution
    Yes, this situation can be made possible if there is no contact between the block and the tank so that no water gets in between them. No water under the piece of bloack implies that there is no upward thrust on the block. 
    Question 231
    CBSEENPH11018259

    Two pieces of equal masses of lead and copper are immersed in liquid. Which will weigh more?

    Solution
    Density of copper is less than that of lead so, for the same mass copper occupies more volume than lead and hence, copper will experience more buoyant force than lead. Thus, copper will weigh less in liquid than lead. 
    Question 232
    CBSEENPH11018260

    Why hydrogen filled balloon rises up and a C02 filled balloon falls down?

    Solution
    Hydrogen gas is lighter than air therefore, the weight of air displaced by hydrogen balloon is greater than the weight of hydrogen. Therefore, the buoyant force on the hydrogen balloon is greater than its weight and hence rises up. But CO2 is heavier than air thus, the buoyant force on CO2 balloon is less than its weight and hence it falls down. 
    Question 233
    CBSEENPH11018261

    A balloon filled with gas is made to just float on the surface of water. What will happen to it if it is submerged a certain distance below the surface and released?

    Solution
    The weight of the balloon is just balanced by upward thrust on the surface. When it is pushed inside, due to additional pressure of water column, the gas in the balloon gets compressed as a result of which the upward thrust decreases. Now the weight of the balloon is greater than the upward thrust and the balloon will sink as the equilibrium is disturbed.
    Question 234
    CBSEENPH11018262

    An empty balloon weighs 100gm. How much will it weigh if 100gm of air is filled in the balloon at one atmospheric pressure?

    Solution
    Given that an empty balloon weighs 100 g. If this same amount of air is filled in the balloon at one atmospheric pressure, then the same volume of atmosphere will be displaced  as much it is inside the balloon. This increase in weight of the balloon due to weight of air is balanced by the upward thrust. So, the actual weight of the air filled balloon will remain the same, i.e., 100 g.



    Question 235
    CBSEENPH11018263

    A cork floating in water experiences an upward thrust of 6.3gf. If whole of system falls freely under gravity, then what will be the upward thrust on the cork?

    Solution
    When a body is falling freely it is in the state of weightlessness. Since the upward thrust is equal to weight of the water displaced by cork, which is zero due to free fall, thus upward thrust will be zero.
    Question 236
    CBSEENPH11018264

    A man is sitting in a boat floating in the pond. If the man drinks some water from the pond, then what will happen to the level of water in the pond?

    Solution
    When the man sitting on a boat drinks water, the level of the water tend to decrease as a result of decrease in the volume of water in the pond. As the man gulp down the water, his weight will also increase by mass m. Due to increase in the weight of the man, the boat will sink a little and displace the the same volume of water drank by the man. Hence, the level of water in the pond raised is equal to the level fallen when the man drank the water. 

    Thus, the level of water will remain unchanged. 
    Question 237
    CBSEENPH11018265

    Why do we make the bottom of a ship heavy? 

    Solution
    The bottom of a ship is made heavy to keep the centre of gravity of ship below the centre of buoyancy so that the ship is in stable equilibrium.

    If the ship gets disturbed from the equilibrium position due to any disturbance in the water medium, the torque produced by weight and buoyant force brings the ship back to equilibrium position. 


    Question 238
    CBSEENPH11018267

    A burning candle is floating in the water. What happens to candle as it burns?


    Solution
    Water exerts buoyant force to balance the weight of candle. Hence, the candle keep floating in the water. As the candle burns, the weight of the candle decreases and water has to exert less buoyant force. Thus the candle becomes less submerged and it rises. 
    Question 239
    CBSEENPH11018268

    What will happen if the lower portion of a ship is made light so that centre of buoyancy lies below the centre of gravity of the ship?

    Solution
    The ship will be in unstable equilibrium if the centre of buoyancy of the ship lies below the centre of gravity. If due to some sea disturbance, ship gets disturbed, the torque produced by weight of the ship and buoyant force will take the ship away from equilibrium position and ship will tumble down.
    Question 240
    CBSEENPH11018269

    What is meta-center?

    Solution
    A floating body which is in stable equilibrium, when disturbed from it's position it starts oscillating. The point about which the oscillations take place is called meta-center. Meta centre is the point where lines of action of force of buoyancy intersect. 
    Question 241
    CBSEENPH11018271

    A cork of density 0.18gm/cc floats in liquid of density 1.44 gm/cc with 20cc of volume inside the liquid. Find the mass and volume of the cork.

    Solution
    Given, 

    Density of the cork, σ = 0.18 gm/cc

    Density of liquid,ρ = 1.44 gm / cc

    Volume of liquid, V = 20 cc

    Now, according to the law of floatation,

    Weight of cork = Weight of liquid displaced by cork.

    Therefore mass of the cork is given by,


                  mg = Vρg                m =             m=20×1.44                     =28.8gm 

    Volume of the cork, Vc=mσ=28.80.18=160cc
    Question 242
    CBSEENPH11018272

    Wooden plank weighing 8kg floats in water with two-fifth of its volume floating in water. Find the maximum weight that can be placed over the plank so that it may float completely submerged.

    Solution
    Let V be the volume of the plank.
    Given that two-fifth volume of plank is submerged in water.
    Therefore by law of floatation the weight of wooden plank is equal to the weight of  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> volume of water.
    That is, 
    Weight of 2 over 5 v space of water = 8 kg
     Weight of V volume of water=20kg.
    Therefore when wooden plank is fully submerged in water, it displaces 20 kg of water.
    Hence a 20 kg weight can be made to float over the water.
    Thus additional weight that can be put over the plank is (20 – 8)kg = 12 kg.
    Question 243
    CBSEENPH11018275

    A weightless rubber balloon has 100 gm of liquid of density 2 gm/cc and is suspended by a string. What will be the tension in the string when balloon is fully immersed in water?

                          

    Solution

    Mass  of liquid  =  100 gm
    Density of liquid = 2 gm/cc

        Volume of ballon containing liquid = 50cc  

    When it is fully immeresed in water, it will displace 50cc of water.

    Therefore upward thrust on ballon is given by, 

                       U = 50 x 1
                          = 50 gmf 

    Tension in string is, 

                       T=100 - 50
                         = 50 gmf



    Question 244
    CBSEENPH11018279

    How does the fractional volume submerged of the floating body change with acceleration due to gravity?

    Solution
    Let a body of volume V and density σ float in a liquid of density ρ

    According to the law of floatation, 

    Weight of body = Weight f liquid displaced

    i.e.           Vσg = V' ρg

    where v' is the volume of liquid displaced.

    "
     
    Fractional volume submerged, 

                  V'V=σρ

    Since it is independent of g, therefore, the fractional volume submerged of the floating body does not change with acceleration due to gravity.
    Question 245
    CBSEENPH11018282

    The dimensions of the base of a boat are 3.3mx2.2m. When the man gets on to the boat, the boat sinks by 1 cm. What is the mass of the man?

    Solution
    The dimensions of the base of a boat are 3.3 m ×2.2 m

    The boat sinks by 1 cm, therefore volume of water displaced when man gets on the boat is,

                V = 3.3 x 2.2 x 0.01

                   = 72.6 x 10-3  m

    Mass of water displaced = 72.6 x 10-3 x 1000 = 72.6  kg

    The man is in a floating state int he boat.

    Therefore,

    Mass of man = Mass of water displaced
                        = 72.6 kg. 
    Question 246
    CBSEENPH11018283

    (a) Ice floats in water with about nine-tenth of its volume submerged. What is the fractional volume submerged for an iceberg floating on a fresh water lake of (hypothetical) a planet whose gravity is ten times that of the earth?

    (b) What is the fractional volume submerged of an ice-cube in a pail of water placed in an enclosure which is freely falling under gravity?

    Solution

    (a) Let a be the density of ice and ρ be the density of water.
    The ice floats in water.
    Therefore, weight of ice is equal to the weight of liquid displaced.

    Let the volume of ice be V and volume of water displaced be V'.
    
therefore space space space space space Vσ space straight g space equals space straight V apostrophe space ρg

rightwards double arrow space space space space fraction numerator straight V apostrophe over denominator straight V end fraction equals straight sigma over straight rho
    i.e. fractional volume displaced by ice is independent of g.
    Therefore, on the new hypothetical planet, fractional volume displaced by ice is also same as that on the earth i.e. nine-tenth.

    (b) In free falling state, ice-cube is in weightlessness state and also the thrust exerted by water is zero. Therefore, it can float with any volume submerged.

    Question 247
    CBSEENPH11018285

    Why do the hairs of a shaving brush cling together when taken out of water?

    Solution
    The hairs of a shaving brush cling together when taken out of water as a result of surface tension. The free surface of water tries to acquire a minimum surface area. 
    Question 248
    CBSEENPH11018286

    Why are rain drops spherical?

    Solution
    For any given volume, the sphere acquires a minimum surface area. Rain drops fall freely under the effect of gravity and is in a state of weightlessness. The only force on them is surface tension due to which the drops have tendency to acquire a minimum surface area. Therefore, the rain drops acquire spherical shapes.
    Question 249
    CBSEENPH11018287

    Why a piece of camphor placed on the surface of water begins to dance?

    Solution
    When impurities are dissolved in water, surface tension of water decreases. Hence, when camphor is dissolved in water, surface tension will decrease. Camphor being non-symmetrical, different amount of camphor is dissolved from different sides. Therefore, Surface tension will be different on different sides. Hence, unbalanced force sets the piece into motion and the piece of camphor begins to dance. 
    Question 250
    CBSEENPH11018288

    Why small mercury drops are spherical but large ones are flat?

    Solution
    A drop of mercury tries to acquire the minimum surface area due to surface area, for any given volume minimum area is for sphere. The effect of gravity is less for small droplets, hence the mercury drops are nearly spherical in shape. 

    As  the size of drop increases, force of gravity on the drop increases which tries to deform the shape. Therefore, bigger drops are flat.
    Question 251
    CBSEENPH11018289

    Why the nib of a pen is split?

    Solution
    The tip of nib of a writing pen is split to provide the capillary action.
    Due to capillarity, ink will rise in the split of nib and allows the pen to write continuously. 
    Question 252
    CBSEENPH11018290

    What is the difference between an ordinary stretched elastic membrane and that of the free surface of liquid?

    Solution
    Ordinary stretched elastic membrane obeys Hooke’s law i.e., tension increases with stretching while tension in the liquid does not increase on increasing the area. When the elastic membrane is stretched, the separation between the molecules increases.

    In case of liquid when surface area is increased the molecules move from bulk to surface film and the distance between the molecules remains constant. 
    Question 253
    CBSEENPH11018291

    When a sewing needle is gently placed on the surface of water, it floats. Why?

    Solution
    The water surface below the sewing needle gets slightly depressed when gently placed on the surface of water. The force due to surface tension on depressed curved surface is inclined upwards, whose resultant is vertically upward, which balances the weight of needle. therefore, the needle floats.
    Question 254
    CBSEENPH11018292

    The needle placed gently on the surface of clear water floats on it, what happens when some detergent is added to water?

    Solution
    Adding impurites to water decreases the surface tension of the liquid. So, when detergents are added into water, the surface tension decreases. Hence, vertical component of surface tension will not be able to support the needle and needle sinks.
    Question 255
    CBSEENPH11018293

    How does the surface tension depend on temperature? 

    Solution

    Surface tension of the liquid decreases with increase in temperature, reaching a value of 0 at the critical temperature. 

    This dependence of surface tension on temperature is given by the equation,

                           Tt = T0(1–αat)

    Question 256
    CBSEENPH11018295

    Define surface energy and show that surface energy is equal to surface tension.

    Solution
    The amount of mechanical work required to increase the unit area of a free surface under the isothermal condition is known as surface energy. 
    To show: Surface energy is equal to surface tension

    i) Take a wired rectangular frame whose three sides are rigid and one side AB is free to slide.
    ii) Dip the frame in a soap solution to form the film.
    iii) The surface tension force acts on all the sides of the frame.
    iv) Side AB is free to slide and the other three sides being fixed cannot displace. Due to surface tension force slides in left direction and free surface tries to contract.

    Let surface tension of soap solution be T and length of movable rod AB be l.
    There are two free surfaces of liquid which touch the movable wire AB.
    The surface tension of total inward force on the wire = 2Tl.
    If frame AB is displaced by distance x in right direction, then work done against surface tension to increase the surface area is,
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    Increase in surface area of film is
     
                        straight ell straight x plus straight ell straight x equals 2 straight ell straight x
    therefore space space space space space Surface space energy
               equals space fraction numerator Work space done space to space space increase space the space area over denominator Increase space in space area end fraction
        
space space space space space space space space space equals fraction numerator 2 straight T straight ell straight x over denominator 2 straight ell straight x end fraction equals straight T

    Question 257
    CBSEENPH11018296

    n identical drops of liquid of surface tension T each of radius r coalesce to form single drop. Find the energy released in the process.

    Solution
    Let R be the radius of the big drop.
    Since on coalesce of droplets, the volume of liquid remains constant, therefore
    Decrease space in space surface space area space during space coalesce space is comma space

increment straight A space equals space straight A subscript 1 space minus space straight A subscript 2 space

space space space space space space space space equals space 4 straight pi space left parenthesis nr squared space minus space straight R squared right parenthesis space equals space 4 πr squared left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

Energy space released space during space coalesce space is comma space

straight E space equals space increment straight A space cross times straight T space equals space 4 πTr squared space left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

rightwards double arrow 1 half straight rho 4 over 3 straight pi space straight R cubed straight v squared space equals space 4 πTr squared left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis

rightwards double arrow space 1 half straight rho 4 over 3 straight pi space nr cubed space straight v squared space equals space 4 πT space straight r squared space left parenthesis straight n minus straight n to the power of 2 divided by 3 end exponent right parenthesis space

rightwards double arrow space 1 over 6 space straight rho space nrv squared space equals space straight T space left parenthesis straight n minus straight n to the power of 2 divided by 3 end exponent right parenthesis space

rightwards double arrow space straight v space equals space square root of fraction numerator 6 straight T space left parenthesis straight n minus straight n to the power of 2 divided by 3 end exponent right parenthesis over denominator ρnr end fraction end root

    Question 258
    CBSEENPH11018298

    n identical drops (each of radius r) of liquid of surface tension T and density ρ coalesce to form single drop. The energy released in the process is converted into kinetic energy. Find the speed of drop.

    Solution

    Let R be the radius of big drop.

    When the drops coalesces, the volume of liquid remains constant.
    There are n identical drops of radius 'r'. 
    Therefore, 
    Decrease space in space surface space area space equals space straight A subscript 1 space minus space straight A subscript 2 space

increment straight A space equals space 4 straight pi space left parenthesis nr squared space minus space straight R squared right parenthesis space

space space space space space space space equals space 4 straight pi space left parenthesis thin space straight n space minus space straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space

Energy space which space is space released space during space coalesce space is comma space

straight E space equals space increment straight A space cross times space straight T space equals space 4 straight pi space Tr squared space left parenthesis space straight n minus straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space

The space enrgy space released space is space converted space into space Kinetic space energy. space

Therefore comma space

1 half mv squared space equals space 4 straight pi space Tr squared space left parenthesis space straight n minus straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space

rightwards double arrow space 1 half straight rho space 4 over 3 straight pi space straight R cubed straight v squared space equals space space 4 straight pi space Tr squared space left parenthesis space straight n minus straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space

rightwards double arrow space 1 half straight rho space 4 over 3 πnr cubed straight v squared space equals space space 4 straight pi space Tr squared space left parenthesis space straight n minus straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space

rightwards double arrow space 1 over 6 space ρnrv squared space equals space straight T space left parenthesis straight n space minus space straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis

rightwards double arrow space straight v space equals space square root of fraction numerator 6 straight T space left parenthesis straight n space minus space straight n to the power of bevelled 2 over 3 end exponent right parenthesis over denominator ρnr end fraction end root


    Question 259
    CBSEENPH11018301

    Given, n identical drops (each of radius r) of liquid of surface tension T and density ρ coalesce to form single drop. The energy released in the processes is converted into heat. Find the increase in the temperature of drop.

    Solution

    Given that 'n' drops coalesce to form a single drop.
    During the process, energy is released which is converted into heat. 
    Energy released is given by,
    straight E space equals space 4 πTr squared space left parenthesis straight n space minus space straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space
    This energy is converted to heat.

    Therefore, 
    space space space space space space space msJθ space equals space 4 πTr squared space left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

rightwards double arrow space straight rho space left parenthesis straight n 4 over 3 πr cubed right parenthesis sJθ space equals space 4 space straight pi space Tr squared space left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

straight i. straight e. comma space space space space straight theta space equals space fraction numerator 3 straight T over denominator rsρJ end fraction open parentheses fraction numerator straight n space minus space straight n to the power of 2 divided by 3 end exponent over denominator straight n end fraction close parentheses
    where,  θ is the increase in temperature drop.

    Question 260
    CBSEENPH11018302

    Explain that spraying results coldness.

    Solution

    The surface area of liquid increases, when the sprayed in small droplets. Work has to be done against the cohesive force to increase the surface area. This work is done at the cost of internal energy and hence the temperature of drop falls. 

    Question 261
    CBSEENPH11018303

    How does the surface tension of water change with impurities?

    Solution
    Adding impurities that are slightly soluble, decreases the surface tension of water. For e.g., addition of soap, camphor etc. decreases the surface tension of water.

    The surface tension of the water increases if the impurities are highly soluble. e.g. addition of sugar, common salt etc., increases the surface tension of water. 
    Question 262
    CBSEENPH11018304

    It is possible to produce a fairly vertical film of soap solution but not of pure water. Explain. 

    Solution

    When a vertical film of soap solution is formed, soap particles being heavy shift to lower side of film and hence surface tension on upper part is greater than the lower part which provides the necessary force to balance the weight of film. Hence, the liquid film is stable.




    But in vertical film of pure water surface tension forces are same at all the points i.e. on A' B' and C'D'. Therefore, a film of pure water will break in the absence of extra force required to balance the weight. 

    Question 263
    CBSEENPH11018306

    Work has to be done to increase the area of free surface of liquid. Explain it in accordance with the law of conservation of energy.

    Solution
    No net force is experienced by the molecules in the bulk of liquid and are free to move within. But, a net inward pull is experienced by the molecules in the surface film. When the area of the free surface is increased, the molecules from bulk move to surface film against the inward pull and hence work has to be done to increase the area of free surface. 
    This work is stored in the form of potential energy. Therefore, work done to increase the area of the surface of liquid is in accordance with the law of conservation of energy. 
    Question 264
    CBSEENPH11018308

    When large number of small droplets coalesces to from a single drop, it warms up. Explain why.

    Solution
    The surface area of the free surface decreases, when large number of small droplets coalesces to from a single drop. This results in decrease in the potential energy in the surface film which is converted into heat and increases the temperature of the drop. 
    Question 265
    CBSEENPH11018309

    Name two applications of surface tension in our daily life.

    Solution

    Applications of Surface tension in daily life are:

    (i) The surface tension of antiseptic ointments is low which make them to spread over the wound.

    (ii) Adding detergent in water decreases the surface tension of water and enhances the cleaning action of water and detergent both.

    Question 266
    CBSEENPH11018311

    What should be the ratio of cohesive force between liquid molecules and the adhesive force between liquid molecules and the tube molecules so that the shape of meniscus is: (i) concave (ii) convex? 

    Solution
    When we dip the capillary tube in liquid, at the contact, two types of forces are acting on liquid molecule as shown:



    (i) For the molecules which are perpendicular to the surface of the tube, adhesive force acts. 

    (ii) For liquid molecules at an angle of 45o, cohesive force will act on the molecules. 

    So,

        If          a>csin 45° =fc2,

    Resultant force or liquid molecule is outside the tube and meniscus is concave, and 


     If          a>csin 45° =fc2

    then, meniscus is convex.

    Question 267
    CBSEENPH11018312

    Derive an expression for excess of pressure inside a liquid drop.

    Solution
    Consider a liquid drop of radius R and surface tension T.
    Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.

    Therefore there is more pressure inside than outside.
    Let pi be the pressure inside the liquid drop and po be the pressure outside the drop.
    Therefore excess of pressure inside the liquid drop is,

                            p = p1– p0

    Due to excess of pressure inside the liquid drop the free surface of the drop will experience the net force in outward direction due to which the drop will expand.
    Let the free surface displace by dR under isothermal conditions.
    Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.
    The work done by excess of pressure in displacing the surface is, 
    dW  = Force x displacement
           = (Excess of pressure x surface area) x displacement of surface
           
           equals straight p cross times 4 πR squared xdR space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
    Increase in the  potential energy is,
    dU = surface tension x increase in area of the free surface
      
equals straight T open square brackets 4 straight pi open parentheses straight R plus dR close parentheses squared minus 4 πR squared close square brackets

equals straight T open square brackets 4 straight pi open curly brackets 2 RdR close curly brackets close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

From space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis

straight p cross times 4 πR squared cross times dR equals straight T left square bracket 4 straight pi open curly brackets 2 RdR close curly brackets right square bracket

rightwards double arrow space space space space space space space space space space space space space space space space space space space straight p equals fraction numerator 2 straight T over denominator straight R end fraction  
    The above expression gives us the pressure inside a liquid drop. 
         

    Question 268
    CBSEENPH11018315

    Derive an expression for excess of pressure inside a soap bubble.

    Solution

    Consider a soap bubble of radius R and surface tension T.
    There are two free surfaces of soap bubble. Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.
    Therefore there is more pressure inside than outside.
    Let pi be pressure inside the liquid drop and po the pressure outside the drop.
    Therefore excess of pressure inside the liquid drop is,

                         p =p1–Po


    Due to excess of pressure inside the liquid drop, the free surface of the drop will experience the net force in outward direction due to which the drop expands.
    Let the free surface be displaced by dR under isothermal conditions.

    Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.

    The work done by excess of pressure in displacing the surface is,
    dW= Force x displacement 
         = (excess of pressure x Surface area) x displacement of surface
        italic equals p italic cross times italic 4 pi R to the power of italic 2 cross times dR space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis 
    Increase in the potential energy is given by, 
    dU = surface tension x increase in area of the free surface
        
equals space straight T space open square brackets 2 open parentheses 4 straight pi open parentheses straight R plus dR squared close parentheses minus 4 πR squared close parentheses close square brackets
equals space straight T space open square brackets 2 open curly brackets 4 straight pi open parentheses 2 RdR close parentheses close curly brackets close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    From (1) and (2)
       straight p cross times 4 πR squared cross times dR equals space straight T left square bracket space 2 space left curly bracket 4 straight pi left parenthesis 2 RdR space right parenthesis space right curly bracket space right square bracket
rightwards double arrow space space space space space space space space space space space space space space space space straight p space equals space fraction numerator 4 straight T over denominator straight R end fraction
    p is the excess of pressure inside a soap bubble. 

    Question 269
    CBSEENPH11018316

    How does the surface tension help in skating?

    Solution
    We know that, with increase of pressure, melting point of ice decreases. Under the skater, ice melts and small tiny water droplets are formed. Due to the small size of water droplets, the excess pressure inside the droplets is sufficiently high and the drops remain spherical in shape. Thus, under the pressure of the skater, drops remain spherical in shape and acts as a ball bearing. Hence, the skater rolls over the drops. 



    Question 270
    CBSEENPH11018317

    Define angle of contact. Is it constant for liquid?

    Solution
    Angle of contact is the angle between the tangents drawn at the point of contact, one along the free surface and second along the solid surface inside the liquid. 
    Angle of contact is not fixed for a given liquid but depends upon the solid surface in contact with liquid and media which exists above the free surface of liquid. 

     
    Question 271
    CBSEENPH11018321

    Derive the ascent formula.

    Solution
    Consider a capillary tube of uniform bore be dipped vertically in a wet liquid. Since liquid is wet, therefore, the meniscus is concave.

    Let 'r' be the radius of capillary tube, 'R' be the radius of meniscus and 'θ' the angle of contact. 

     

    In figure (i),

    X is in atmosphere, Z is at the interface of mercury and atmosphere and Y is on convex side of meniscus.

    Therefore, pressure at X and Z is equal and is equal to the atmospheric pressure.

    But, pressure at Y is less than that at X and Z by 
    2TR.

    Therefore, the liquid is not in equilibrium.

    Now, inorder to attain the equilibrium, the liquid will rise in tube.

    Let at equilibrium, liquid rise to height h as shown in figure (ii). 

    Now in the equilibrium condition, 


    PY=PA=PZ                         ,,,,(1)PB=PA-2TR                       ...(2)PY=PB+ρgh                       ...(3)                     

    From (1), (2) and (3), we have

             PY=PB-2TR+ρgh 2TR=ρgh      h=2TRρg 

    But,      r = Rcos θ       R =rcosθ        h=2Tcosθrρg , 

    is the required ascent formula. 
    Question 272
    CBSEENPH11018322

    What happens if the capillary tube is of insufficient height?

    Solution
    The ascent formula is given by,

                          h=2TRpg  Rh=2Tρg=constant                   

    where,
    R is the radius of meniscus, and
    h is the height to which the liquid rises.

    If tube is of insufficient height, the liquid will rise to the top of the tube and spread over the brim. Thus, the radius of meniscus will adjust to a new value so that Rh remains constant. 
    Question 273
    CBSEENPH11018323

    How is the rise of liquid affected if the top of capillary tube is closed?

    Solution
    If the top of the capillary tube is closed, the liquid will rise to lesser height as compared to ordinary condition. This is because the air gets compressed and the pressure will increase. Thus, the top end of the tube being closed will affect the rise of a liquid. 
    Question 274
    CBSEENPH11018324

    Water rises higher in narrower tubes than in the broader tubes. Why? 

    Solution
    The height to which the liquid rises in capillary tube is inversely proportional to the radius of the tube. Therefore, water rises higher in narrower tubes than in the broader tubes. 
    Question 275
    CBSEENPH11018326

    Mercury barometer reads less or more than actual? Explain.

    Solution
    Mercury barometer reads less than the actual reading because mercury is dry liquid. Therefore, as a result of surface tension the level of the fluid falls in the barometer. Hence, reads less. 
    Question 276
    CBSEENPH11018327

    How does the ploughing of fields help in the preservation of moisture in the soil?

    Solution
    The formation of cappilaries take place in the field which are not ploughed for long. So, the moisture from inside of the soil reaches the surface and evaporates. But,  if the fields are ploughed, the capillaries will break and the water will not rise to surface and remains preserved in the soil.
    Question 277
    CBSEENPH11018328

    Where the molecules possess more energy-in the bulk or in surface film?

    Solution
    Molecules of surface film have additional potential energy due to net inward intermolecular forces. Therefore, the molecules of surface film posess more energy than on the bulk.
    Question 278
    CBSEENPH11018329

    A large bubble is formed at one end of capillary tube and a small one at the other end. Which one will grow at the expense of the other?

    Solution
    Given, a large bubble on one end of the capillary tube and a small one on the other end of the capillary tube. 



    We have,

    Excess of pressure inside the soap  bubble, p=4TR.

    This implies that pressure is inversely proportional to the radius of the bulb. 

    Therefore, pressure inside the bigger bubble will be less than that inside the smaller bubble. Therefore, the air will flow from smaller bubble to bigger bubble.
     
    Hence the bigger bubble will grow at the expense of smaller bubble.

    Question 279
    CBSEENPH11018330

    Why oil spreads on cold water and may remain as a drop on hot water?

    Solution
    Surface tension of oil is less than cold water and hence oil spreads over cold water. But, surface tension of oil is greater than that of hot water and hence resumes the shape of drop. 
    Question 280
    CBSEENPH11018332

    Why does coating of glycerine over the glass window prevent raindrops from sticking to it?

    Solution
    The angle of contact of water with clean glass is less than 90° i.e., acute. Therefore, the water sticks with glass window. By coating the layer of glycerine over glass window, the angle of contact becomes greater than 90° and makes the glass window non-sticky. 
    Question 281
    CBSEENPH11018333

    Why do the new earthen pots keep the water cooler than the old one?

    Solution
    There are fine pores on the surface of new earthen pots. The water rises in these pores due to capillary action and spreads over the surface. The water from the surface of pot, when evaporates, cools down the pot and keeps the water cold. With time, the dust particles  fill the capillaries of pot and hence the above phenomenon do not take place in old pots. 
    Question 282
    CBSEENPH11018334

    A capillary tube is dipped in water. The water rises to certain height. What will happen if we add some common salt in water?

    Solution
    On adding highly insoluble impurities in water, the surface tension decreases. And, also the height to which  the liquid rises into the capillary tube is directly proportional to the surface tension of the liquid. As salt is added in water, surface tension of water increases. Therefore, the height to which the water rises in the capillary tube increases on adding the salt. 
    Question 283
    CBSEENPH11018336

    Large force is required to draw apart two glass plates enclosing a thin layer of water film. Explain why.

    Solution
    A thin film is formed between the glass plates. And this thin film has a concave surface all around as shown in the figure below:




    The pressure on convex side is less than that on concave side because of surface tension. 

    i.e.,                         pi< p0

    Therefore, there is net inward force on the plates due to difference of pressure and require some force to separate them. 

    Question 284
    CBSEENPH11018337

    Water rises to height 1.2 cm in capillary tube on the surface of earth. How the height of water column will be affected when the apparatus is taken to the moon?

    Solution
    The height to which the liquid rises into the capillary tube is inversely proportional to the acceleration due to gravity. On the surface of the moon, acceleration due to gravity is 1/6th as that on the surface of the earth. Therefore, water will rise to a height 6 times than on the surface of the earth. 

    So, height of water on the surface of the moon = 7.2 cm
    Question 285
    CBSEENPH11018338

    Water spreads over a clean glass plate but mercury does not. Why?

    Solution
    Water spreads over a clean glass plate but mercury does not. In case of water, the surface tension force between glass plate and air is greater than the surface tension between liquid and solid. Therefore, angle of contact is less than 90o. Hemce, it will spread over the glass plate. 

    In case of mercury, the surface tension force between glass plate and mercury is less than the surface tension between liquid and solid. Therefore, angle of contact is greater than 90o. Hence, mercury will not spread over the glass plate. 



    Question 286
    CBSEENPH11018339

    Why a glass rod coated with wax does not become wet when dipped in water?

    Solution
    Adhesive force  between water and wax is very less as compared to the cohesive forces between water molecules. Therefore, a glass rod coated with wax does not become wet when dipped in water.
    Question 287
    CBSEENPH11018340

    Oil is poured to calm the sea waves. Explain why.

    Solution
    Oil when poured over water spreads over the surface of water because of surface tension. When the wind blows, the surface film of oil is carried in the forward direction, leaving behind the water of higher surface tension. This left behind water exerts a backward pull thus, opposing the motion of the surface by the wind. 
    Question 288
    CBSEENPH11018343

    Which produces the better cleaning action-clear water or soap solution?

    Solution
    Soap solution produces the better cleaning action than clear water. Surface tension of soap solution is less as compared to that of clean water.
    Therefore, the molecules of soap solution can penetrate more into the fine pores of the cloth, making the cleansing action more effective. 

    Question 289
    CBSEENPH11018344

    How the soldering paste makes the soldering process easy?

    Solution
    Adding of soldering paste to molten tin decreases the surface tension of the molten tin. Hence, it becomes easy to spread over the junction where soldering has to be done. 
    Question 290
    CBSEENPH11018345

    Why mercury does not cling to glass?

    Solution
    The force of cohesion between mercury molecules is much greater than force of adhesion between mercury and glass. Therefore, mercury does not cling to glass. 


    Question 291
    CBSEENPH11018349

    Water rises to a height 4.8cm in a certain capillary tube. When the same tube is dipped in some another liquid it falls by 12.5cm. Compare the surface tension of water and liquid. Density of liquid is 9.6 gm/cc and angle of contact of liquid is 120°.

    Solution
    Given, 

    Rise in water = 4.8 cm 
    Fall in water when tube is dipped in some other liquid = 12.5 cm
    Density of liquid = 9.6 gm/cc
    Angle of contact of liquid = 120o

    The height to which the liquid rises or falls in the capillary tube is given by the relation,

                            h=2T cosθrpg 

    The above relation can be simplified to 

                            T=rgph2 cos θ                    ....(1)

    Let T1 and T2 be the surface tension of water and liquid respectively.

    By using the equation (1), we have


    T1T2=ρ1h1ρ2h2cosθ2cosθ1

    Here, ρ1=1 gm/cc,        h1=4.8cm,        θ1=0°         ρ2=9.6gm/cc,      h2=-12.5cm,   θ2=120°    T1T2=1×4.89.6×(-1.25)cos 120°cos 0°=15        

     
     
    Question 292
    CBSEENPH11018351

    Calculate the force required to take away a flat circular plate of radius 5cm and mass 2gm from the surface of water. The surface tension of water is 72 dyne/cm.

    Solution

    Given, 
    Radius of the circular plate, r = 5 cm 
    Mass, m = 2 gm
    Surface tension of water, T = 72 dyne/cm

     Weight of plate is, W = mg=2 x 980 = 1960 dyne

    The circumference of circular plate is given by, =2πr=2π×5=31.4cm

    The down pull on plate due to surface tension is given by, 
     
    f=T=72×31.4=2260.8 dyne 

    Therefore, total force required to take plate away from water surface is,

    F=W+f 

     =1960+2260.8 

     =4220.8 dyne

     

    Question 293
    CBSEENPH11018352

    A big drop of water is sprayed into 1000 small droplets. Find the ratio of surface potential energy of drop to single droplet?

    Solution

    Let, 
    Radius of small droplet be 'r', 
    Radius of big drop be 'R'. 
    Volume of water will remain constant on splitting the drop into droplets. 
    Therefore, 
    R= nr3
    Here n = 1000, 
    Therefpre, R = 10 r
    Now, the surface of the potential energy of drop is, 
    straight E subscript 1 space equals space 4 πR squared straight T

and space the space surface space potential space energy space of space droplet space is comma space

straight E subscript 2 space equals space 4 πr squared straight T

Therefore comma space the space ratio space of space the space surface
space potential space energy space of space drop space to space single space drop space is comma space

straight E subscript 1 over straight E subscript 2 space equals space straight R squared over straight r squared space equals space fraction numerator left parenthesis 10 space straight r right parenthesis squared over denominator straight r squared end fraction space equals space 100 over 1

    Question 294
    CBSEENPH11018353

    A drop of mercury of radius 3mm splits into 27 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 N/m.)

    Solution

    Let r be the radius of small droplet.
    The volume of mercury will remain constant on splitting the drop into droplets.

    Therefore,
    straight R space equals space straight n to the power of begin inline style bevelled 1 third end style end exponent straight r space

Here comma space

straight R space equals space 3 space mm space and space straight n space equals space 27

therefore space straight r space equals space 1 space mm space

Surface space are space of space big space drop space is comma space

straight A subscript 1 space equals space 4 πR squared space equals space 4 straight pi space left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis squared space

space space space space space space equals space 1.13 space cross times 10 to the power of negative 4 end exponent space straight m squared space

Surface space area space of space 27 space droplets comma space

straight A subscript 2 space equals space 27 space cross times space 4 πr squared space

space space space space space space equals space 27 space cross times space 4 straight pi space left parenthesis 10 to the power of negative 3 end exponent right parenthesis squared space

space space space space space space equals space 3.39 space cross times space 10 to the power of negative 4 end exponent space straight m squared space space

Increase space in space surface space area comma space

increment straight A space equals space straight A subscript 2 space minus space straight A subscript 1 space equals space 2.26 space cross times 10 to the power of negative 4 end exponent space straight m squared space

Increase space in space potential space energy space of space surface space film comma space

increment straight E space equals space increment straight A cross times straight T space

space space space space space space space space space equals space 2.26 space cross times space 10 to the power of negative 4 end exponent space cross times space 0.465 space

space space space space space space space space space equals 10.51 space cross times space 10 to the power of negative 5 end exponent space straight J space
space space space space space space space space space 
    Question 295
    CBSEENPH11018354

    What are the limitations of Bernoulli's equation?

    Solution
    Limitations of Bernoulli's equations are as follows:

    (i) Bernoulli's equation is derived by assuming that there is no loss of mechanical energy i.e. sum of pressure energy, potential energy and kinetic energy is constant. But when fluid is in motion, a part of mechanical energy is converted into heat energy. 

    (ii) Bernoulli's equation is derived by assuming that liquid is non-viscous. But actually none of the liquids is non-viscous. All liquids have a tendency to flow, less or more. 

    (iii) Bernoulli's equation is derived by assuming that velocity of every fluid particle across any cross-section of pipe is same. But the particles of central layer have maximum velocity and velocity decreases towards the walls of pipe.
    Question 296
    CBSEENPH11018357

    Dust generally settles down in a closed room. Explain.

    Solution
    Dust particles being small spherical particles remain suspended in air during dusting.

    Terminal velocity is given by v ∝ r2. Since the dust particles are minutely small, these particles will acquire a small terminal velocity. Hence, the particles will settled down in a closed room after some time. 
    Question 297
    CBSEENPH11018358

    Two beakers containing water and honey are stirred rapidly and kept on a table. Which liquid will come to rest earlier?

    Solution

    The coefficient of viscosity of honey is very large as compared to that of water. Therefore, honey kept in the beaker after stirring will come to rest earlier than water. 

    Question 298
    CBSEENPH11018359

    Explain why a parachute is invariably used, while jumping from an aeroplane.

    Solution
    The parachute during descending, through air experiences very large viscous force due to large structure. The large structure of the parachute leads to a greater air resistance and hence a tendency to reach a slower terminal velocity. Therefore a person landing on ground will get very little hurt. 
    Question 299
    CBSEENPH11018364

    A soap bubble of radius 10mm is blown from soap solution of surface tension 0.06 N/m. Find the work done in blowing the bubble. What addition work will be done in further blowing to double the radius?

    Solution
    The soap bubble is formed from the soap solution.
    Therefore, increase in the surface area of soap bubble is equal to total surface area of soap bubble.
    Since the soap bubble has two free surfaces, therefore increase in area of free surface of bubble is,
    ΔA subscript 1 equals 2 cross times 4 πr squared
    We have,
    Radius of the soap bubble, r = 1mm=10-3 
    therefore

I n c r e a s e space i n space a r e a comma space increment straight A subscript 1 equals 2 cross times 4 straight pi left parenthesis 10 to the power of negative 3 end exponent right parenthesis squared

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.51 cross times 10 to the power of negative 6 end exponent straight m squared 
    Now the work done in blowing the bubble
    straight W subscript 1 equals capital delta A subscript 1 cross times straight T

space space space space space equals 2.51 cross times 10 to the power of negative 6 end exponent cross times 0.06

space space space space space equals 1.5 cross times 10 to the power of negative 7 end exponent straight J
space space space space space equals 1.5 space e r g s
    Additional work done in doubling the radius of bubble is,
    straight W space equals space ΔA cross times straight T

space space space space space equals 7.53 cross times 10 to the power of negative 6 end exponent cross times 0.06

space space space space space equals 4.5 cross times 10 to the power of negative 7 end exponent straight J equals 4.5 space ergs 

    Question 300
    CBSEENPH11018366

    A vessel, whose bottom has a hole of diameter 0.2mm, is filled with water. Find the maximum height to which the water can be filled without leaking it from the hole. (Surface tension of water is 70 dyne/cm)

    Solution
    Let the vessel be filled to a height of h.

    The force exerted by height column of water will be balanced by surface tension force on the free surface of water at the hole, at equilibrium.

    That is,

            ρgh×πr2=2πr×T                  h=2TrpgHere,   T= 70 dyne/cm              r=d/2=0.1mm=0.01 cm,             ρ=1 gm/cc,              g= 1000 cm/s2   h=2×700.01×1×1000=14cm 

    So, 'h' is the maximum height to which the water can be filled without leaking it from the hole. 
    Question 301
    CBSEENPH11018368

    A 10x1.5x0.2 cm3 glass plate weighs 8.6 gm in air. Now it is immersed half in water with longest side vertical. What will be its apparent weight? (surface tension of water is 70 dyne /cm.)

    Solution

    The different forces acting on the plate are:

    (i) Weight W vertically downward,
    (ii) Upward thrust U vertically upward,
    (iii) Surface tension force T in downward direction .

    These forces are given by, 

    Weight, W=8.6 x 980 = 8428 dyne

    Upward Thrust, U=weight of water displaced 

                              =Volume of water displaced x density of water x g
                               =12(10x1.5x0.2)x1 x 980=1470 dyne

    Let,

    T = Total length of the water in touch with plate x surface tension

       =2(1.5 + 0.2) x 70 = 238  dyne

    Now the apparent weight, 

    W+ W + T - U = 8428 + 238 - 1470

                             = 7196 dyne

                             = 7.434 gf

    Question 303
    CBSEENPH11018375

    Find the excess of pressure inside the soap bubble of diameter 1.2mm. The surface tension of soap solution is 18 dyne/cm.

    Solution
    Given, 
    Diameter of the soap bubble, d = 1.2 mm 

    Surface tension of soap, T = 18 dyne/cm

    So, excess of pressure inside the soap bubble is given by, 


                                                Δp=4TrHere, T=18 dyne/cm           D=1.2mm=0.12cm Excess pressure, p =4x180.06                                             =1.2×103 dyne/cm2                                              =1.2×102 N/m2 
    Question 304
    CBSEENPH11018377

    The excess of pressure inside the bubble of radius 2-4 cm is balanced by 0-12 cm height column of oil of density 0.8 gm/cc. Find the surface tension of liquid.

    Solution
    Given, the excess of pressure inside the bubble is, height column of oil of density 0.8 gm/cc. 

    Find: surface tension of liquid?

    The formula for excess pressure is given by, 

                       Δp=4Tr

    The pressure at height column of oil is given by, 

                       Pm = pgh

    Now, excess of pressure inside the bubble is balanced by height column of oil.

    Therefore, 

                          4Tr=ρgh                     T = ρrgh4Here       r=2.4 cm, ρ=0.8 gm/cc               h=0.12 cnm, g=980 cm/s2  Surface tension of the liquid, T = 0.8×2.4×980×0.124                                                             = 56.45 dyne/cm
    Question 305
    CBSEENPH11018380

    If an air bubble of 5.0 mm is formed at depth of 40.0 cm inside a container containing soap solution (of relative density 1.20), what would be the pressure inside the bubble?(1AP = 1.013 x105 Pa and surface tension of soap solution is 0-025 N/m)

    Solution
    Total pressure inside the air bubble in the soap solution is the sum of the atmospheric pressure, pressure due to height column of soap solution and excess of pressure due to surface tension.
    That is,
                      
straight P equals 1 Ap plus ρgh plus fraction numerator 2 straight T over denominator straight r end fraction 
    The relative density of soap solution is 1.2.
    Therefore,
    Relative density, straight rho=1.2 gm/cc = 1200 kg/m
    Therefore comma space

Pressure comma space ρgh equals 1200 cross times 9.8 cross times 0.4

space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4.7 cross times 10 cubed Pa equals 0.047 cross times 10 to the power of 5 space Pa

and space space space space fraction numerator 2 straight T over denominator straight r end fraction equals fraction numerator 2 cross times 2.5 cross times 10 to the power of negative 2 end exponent over denominator 5 cross times 10 to the power of negative 3 end exponent end fraction equals 10 Pa

Thus comma

Pressure space inside space the space bubble space is comma space

straight P subscript inside space equals 1.01 cross times 10 to the power of 5 plus 0.047 cross times 10 to the power of 5 plus 10

space space space space space space space space space space space space equals 1.05701 cross times 10 to the power of 5 Pa
    Question 306
    CBSEENPH11018386

    What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 0.465N/m. The atmospheric pressure is 1.01 x105 Pa. Also give the excess pressure inside the drop. 

    Solution
    Given, 

    Radius of drop of mercury, r = 3.0 mm 

    Surface tension of mercury = 0.465 N/m

    Atmospheric pressure, PA = 1.01 × 105 Pa

    Excess of pressure inside the liquid drop is given by,

                                        p=2TRHere,           T=0.465N/m         R=3mm=3×10-3mExcess pressure, p=2×4.65×10-13×10-3                                        =310 Pa                                      =0.0031 ×105 Pa  Pressure inside the drop =P0+p                                                  =1.01x105+0.0031×105 Pa                                                   = 1.0131×105Pa 
    Question 307
    CBSEENPH11018388

    A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and a light slider supports a weight of 1.5x10–2N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

    Solution
    Given, the thin soap film formed between the wire and slider has two free surfaces.

    Weight supported by the light slider = 1.5 × 10-2 N

    Length of the slider, L = 30 cm



    So, total length of free surface that touches with slider is 2L = 60cm = 0.60 m.

             Surface tension T=1.5×10-20.60                                            = 2.5×10-2N/m

    Question 308
    CBSEENPH11018389

    Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? (Surface tension of mercury at the temperature of the experiment is 0.465 N/m. Density of mercury = 13.6x10–3kgm–3.)

    Solution

    Here,

    Angle of contact, θ = 140°

    Radius of the narrow tube, r = 1.0mm=10-3m
    Surface tension, T = 0.465 N/m
    Density of mercury, ρ = 13600  kg/m3

    We know the height by which liquid rises in the tube is,

                 h=2T cosθrpg  =2×0.465 cos 14010-3×13600×9.8  = -0.00535m  = -5.35m 

    Negative sign means that mercury will dip down.

     Mercury will dip down by 5.35 mm.

    Question 309
    CBSEENPH11018390

    Two narrow bores of diameters 3-0 mm and 6.0 mm are joined together to form a U-tube open at both the ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? (Surface tension of water at the temperature of the experiment is 72 dyne/cm. Take angle of contact to be zero and density of water to be 1.0 g/cc.) 

    Solution
    We know water rises in the capillary tube.
    Height to which water rises in the capillary tube is,
                          
straight h equals fraction numerator 2 straight T space cos space straight theta over denominator rpg end fraction
    Therefore the height hof the water that rises in the limb of diameter 3 mm is given by, 
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    The height hof the water that will rise in the limb of diameter 6 mm is given by, 
    straight h subscript 2 equals fraction numerator 2 cross times 72 cross times 1 over denominator 0.3 cross times 1 cross times 980 end fraction equals 0.49 mm
    Hence difference in the levels of  two limbs is,
    Δh equals straight h subscript 1 minus straight h subscript 2 equals 0.98 minus 0.49 equals 0.49 mm
    Question 310
    CBSEENPH11018391

    Two soap bubbles of radii 'a' and 'b' of same liquid come together to form a double bubble. Find the radius of curvature of common interface of two bubbles.

    Take a < b and surface tension of soap liquid 'a' is T.

    Solution
    Given, two bubbles of radii 'a' and 'b'. 
    Let,
    p1be the excess of pressure inside the bubble of radius 'a', and
    p2 be the excess of pressure inside the bubble of radius 'b'.

    Therefore, we have
    P1=4Ta  and P2=4Tb
    The two bubbles come together to form a big bubble.
    Then let r be the radius of common interface.
    Now the difference of pressure on the two sides at interface is given by, 
    increment p space equals space p subscript 1 space minus space p subscript 2 space
space space space space space space space
space space space space space space space equals space 4 T space open parentheses 1 over a minus 1 over b close parentheses space

A l s o comma space increment p space equals space fraction numerator 4 T over denominator r end fraction

rightwards double arrow space fraction numerator 4 T over denominator r end fraction space equals space 4 T space left parenthesis fraction numerator b minus a over denominator a b end fraction right parenthesis

rightwards double arrow space r space equals space open parentheses fraction numerator b a over denominator b minus a end fraction close parentheses   
    'r' is the radius of curvature of common interface of two bubbles. 
    Question 311
    CBSEENPH11018392

    What is viscosity?

    Solution
    Viscosity is the property of the liquid by virtue of which it opposes the relative motion between its different layers.

    The resistance to the flow of liquid is also called viscosity.
    Question 312
    CBSEENPH11018393

    State Stoke’s law.

    Solution

    Stoke’s law states that the viscous force acting on a spherical body of radius 'r' moving with velocity 'v' in viscous media of viscosity 'η' is given by,

                                 F= – 6πηrv.

    Question 313
    CBSEENPH11018394

    What are practical applications of Stoke’s law?

    Solution
    Practical application of stoke's law are used in determining:

    1. the radius of liquid drop, and
    2. Coefficient of viscosity of liquids. 
    Question 314
    CBSEENPH11018395

    Is it necessary to apply some driving force to maintain the flow of liquid through pipe?

    Solution
    Yes, it is necessary to apply some driving force to maintain the flow of liquid through the pipe. The viscous force between the different layers of the liquid will retard the motion of the liquid and flow stops after certain distance. 
    Question 315
    CBSEENPH11018396

    In which liquid, the terminal velocity of an object will have lesser value-benzene or honey?

    Solution
    The terminal velocity of honey will be lesser because the coefficient of viscosity of honey is greater than that of benzene. 
    Question 316
    CBSEENPH11018397

    What are similarities between solid friction and viscosity?

    Solution
    Both solid friction and viscosity oppose the relative motion of the flow of liquid due to intermolecular forces.
    Question 317
    CBSEENPH11018398

    A body is dropped in a viscous liquid. How will velocity of the body change with time?

    Solution
    When a body is dropped in a viscous liquid, the velocity of the body first increases exponentially with time and then acquires constant value called terminal velocity. 
    Question 318
    CBSEENPH11018399

    Define coefficient of kinematic viscosity.

    Solution
    The coefficient of kinematic viscosity of a fluid is defined as the ratio of coefficient of viscosity to the density of fluid.
    Question 319
    CBSEENPH11018400

    Define fluidity.

    Solution
    Fluidity is measure of the ease with which the liquid can flow and is equal to reciprocal of coefficient of viscosity. 
    Question 320
    CBSEENPH11018401

    What are the cgs units of kinematic viscosity and fluidity?

    Solution
    CGS unit of kinematic viscosity is stokes (St), named after George Gabriel Stokes.

    CGS unit of fluidity is poise
    –1.  
    Question 321
    CBSEENPH11018402

    Two spherical balls of same radii but of density 4gm/cc and 2 gm/cc are dropped in water. What is the ratio of their terminal velocity?

    Solution
    Given, two spherical balls are of the same radii but of different density. 
    Density of first ball = 4gm/cc
    Density of second ball = 2 gm/cc
    Terminal velocity , v = straight r squared over straight eta left parenthesis straight rho space minus space straight rho subscript straight o right parenthesis straight g
    That is, terminal velocity is directly proportional to the difference in the density of body and liquid.
    Therefore, the ratio of the terminal velocity of the two balls will be 2:1. 
    Question 322
    CBSEENPH11018403

    Two equal raindrops are falling with terminal velocity v. State whether the terminal velocity will increase or decrease if the two-rain drops coalesce.

    Solution
    On coalesce the size of the raindrop will increase. Hence, the terminal velocity will also increase. 
    Question 323
    CBSEENPH11018404

    A ball is dropped from 20-storey building. The observers on 10th, 6 th and 1st floor measure the velocity of ball. Which observer measures the velocity maximum?

    Solution
    When a ball is dropped from a 20 storey building, all the observers will measure the velocity as same because ball is falling with it's terminal velocity. 
    Question 324
    CBSEENPH11018405

    What is the acceleration of a body falling freely through viscous liquid after it has acquired terminal velocity?

    Solution
    The acceleration of the body falling freely through viscous liquid after it has acquired terminal velocity is zero. 
    Question 325
    CBSEENPH11018406

    How does the velocity of a body falling freely through viscous liquid change with time before it acquires the terminal velocity?

    Solution
    The velocity of body increases exponentially with time before it acquires the terminal velocity.
    Question 326
    CBSEENPH11018407

    What is streamline flow?

    Solution
    Streamline flow is an orderly flow of fluid in which all the molecules of the fluid passing through a certain point have same velocity.
    Question 327
    CBSEENPH11018408

    What is turbulent flow?

    Solution
    An irregular flow in which different molecules passing through a given point possess different velocities is known as turbulent flow. 
    Question 328
    CBSEENPH11018409

    Define streamline.

    Solution
    Streamline is a line, straight or curved, tangent to which gives the direction of velocity of molecule of fluid at any point. 
    Question 329
    CBSEENPH11018410

    What is tube of flow?

    Solution
    A tube of flow is a bundle of streamlines having the same velocity of liquid elements over any cross-section perpendicular to the direction of flow. 
    Question 330
    CBSEENPH11018411

    What is laminar flow?

    Solution
    Laminar flow is a slow and steady flow of liquid in which different layers of liquid glide (move smoothly) over one another without intermixing.
    Question 331
    CBSEENPH11018412

    Define critical velocity.

    Solution
    Critical velocity is the velocity of flow at which there is a transition from streamline flow to turbulent flow. 
    The liquid transit from subcritical flow to supercritical flow. 
    Question 332
    CBSEENPH11018413

    What are the factors on which critical velocity depends?

    Solution
    The factors on which critical velocity depends on are:

    i) coefficient of viscosity of liquid,
    ii) density of liquid, and
    iii) diameter of pipe through which the liquid is flowing. 
    Question 333
    CBSEENPH11018414

    What are dimensions of Reynolds number?

    Solution
    Reynolds number is dimensionless quantity.
    Question 334
    CBSEENPH11018415

    Explain laminar flow and reynol'ds number.
    What is the range in the value of Reynold's number for the flow of viscous liquid to be laminar?

    Solution
    Laminar flow generally occurs when the fluid is moving very slowly or if the fluid is very viscous. 

    And Reynold's number is the ratio of how fast the fluid is moving to how viscous the fluid is. That is, reynold's number (NR) is directly proportional to velocity and inversely proportional to viscosity. 

    If Reynold's number lies between 0 to 2000 then flow of viscous liquid is laminar. 

    i.e.,                        NR < 2000
    Question 335
    CBSEENPH11018416

    What is the basic condition for the equation of continuity?

    Solution
    The basic condition for the equation of continuity to be satisfied is that there should not be any source or sink in the pipe through which liquid is flowing. 
    Question 336
    CBSEENPH11018417

     How does the velocity of efflux depend on height of free surface of liquid from an orifice?

    Solution
    Velocity of efflux is directly proportional to the square root of height.
    Question 337
    CBSEENPH11018419

    What is Poiseuille’s equation

    Solution
    Poiseuille studied the rate of flow of a liquid throught a horizontal capillary tube.
    He stated that the volume (V) of liquid flowing per second through a capillary tube varies, 
    i) directly as the pressure difference (p) across the two ends of the tube,
    ii) directly as the fourth power of the radius of the tube, 
    iii) inversely as the length of the tube,
    iv) inversely as the coefficient of viscosity of the liquid.
    That is,
    Error converting from MathML to accessible text.
    Question 338
    CBSEENPH11018420

    What is the pressure energy per unit mass stored in liquid of density ρ at pressure P?

    Solution

    Consider a tube of varying cross-section through which an ideal liquid is in streamline flow. 
    Let, 
    P1 = Pressure applied on the liquid at A, 
    P2 = Pressure at end B, against which the liquid is to move out
    a1, a2 = Area of cross section at the tube at A and B respectively. 
    h1, h2 = mean height of section A and B from ground or a reference level.
    v1, v2 = normal velocity of the liquid flow at section A and B, 
    straight rho = density of ideal liquid flowing though the tube.
    Liquid flows from A to B.
    That is,
    P1 > P2
    Mass m of the liquid crossing per second through any section of the tube is in accordance with the equation of continuity,  given by
    a1v1p = a2 v2 p = m 
    rightwards double arrow space straight a subscript 1 space straight v subscript 1 space equals space straight a subscript 2 space straight v subscript 2 space equals space straight m over straight rho space equals space straight V space

As comma space straight a subscript 1 space greater than thin space straight a subscript 2

therefore space straight v subscript 2 space greater than space straight v subscript 1 space

Force space on space liquid space at space section space straight A space equals space straight P subscript 1 straight a subscript 1

Force space on space liquid space at space section space straight B space equals space straight P subscript 2 straight a subscript 2 space

Work space done space per space second space on space the space liquid space
at space section space straight A space equals straight P subscript 1 straight a subscript 1 space cross times space straight v subscript 1 space equals space straight P subscript 1 straight V space

Work space done divided by second space by space the space liquid space
at space section space straight B space equals space straight P subscript 2 straight a subscript 2 space cross times space straight v subscript 2 space equals space straight P subscript 2 straight V

Net space work space done divided by second space on space the space liquid space by space the space
pressure space energy space in space moving space the space liquid space from space section space straight A space
to space section space straight B space is comma space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight P subscript 1 straight V space space minus space straight P subscript 2 straight V space

When space the space mass space straight m space of space the space liquid space flows space in space 1 space second space from space straight A space to space straight B comma space
it apostrophe straight s space height space increases space from space straight h subscript 1 space to space straight h subscript 2.

Therefore comma space

Increase space in space straight P. straight E space divided by sec space of space the space liquid space from space straight A space to space straight B space is comma space

mgh subscript 2 subscript space end subscript space minus space mgh subscript 1 space

Increase space in space straight K. straight E space divided by sec space of space the space liquid space from space straight A space to space straight B space is comma space

space space space space space space space space space space space space space space space space space space space space space 1 half mv subscript 2 squared space minus space 1 half mv subscript 1 squared

Now comma space ussing space work space energy space principle comma space

Work space done divided by sec space by space the space pressure space energy space equals
increase space in space straight P. straight E divided by sec space space plus space increase space in space straight K. straight E divided by sec space

straight P subscript 1 straight V space minus space straight P subscript 2 straight V space equals space left parenthesis mgh subscript 2 subscript space end subscript space minus space mgh subscript 1 space right parenthesis space plus space open parentheses space space space space space space 1 half mv subscript 2 squared space minus space 1 half mv subscript 1 squared close parentheses

That space is comma space

straight P subscript 1 straight V space plus space mgh subscript 1 space plus space 1 half mv subscript 1 squared space equals space space straight P subscript 2 straight V space plus mgh subscript 2 subscript space end subscript plus 1 half mv subscript 2 squared

Dividing space throughout space by space straight m comma space we space get

fraction numerator straight P subscript 1 straight V over denominator straight m end fraction space plus space gh subscript 1 space plus 1 half straight v subscript 1 squared space equals fraction numerator straight P subscript 2 straight V space over denominator straight m end fraction space plus gh subscript 2 subscript space end subscript space plus space 1 half straight v subscript 2 squared

Therefore comma space

straight P over straight rho plus space gh space plus space 1 half straight v squared space equals space straight a space constant

    Question 339
    CBSEENPH11018421

    How does the velocity of efflux depend on height of free surface of liquid from an orifice?

    Solution
    The velocity of efflux is directly proportional to the square root of height. 
    Question 340
    CBSEENPH11018430

    A tank containing water has an orifice at depth 4.9m below the free surface of water. Find the velocity of efflux? 

    Solution

    Given, depth of orifice, h = 4.9 m

    Therefore, the velocity of effulx is given by,

               ν=2gh  =2×9.8×4.9  =9.8 m/s

    Question 341
    CBSEENPH11018432

    The Bernoulli's theorem is a consequence of _________

    Solution
    Conservation of energy.
    Question 342
    CBSEENPH11018445

    Define viscosity. State the factors on which the viscous force depends.

    Solution
    Viscosity is the property of fluid by virtue of which an internal resistance comes into play when fluid is in motion and disturbs its motion.

                  
    Let a liquid is flowing on a horizontal surface. 
    Consider the two layers of liquid each of area A having a velocity gradient dv over dx as shown in the figure. 
    According to Newton. the viscous force is,
    i) proportional to area of contact of two layers. 
    That is, 
                                straight F space proportional to space straight A

    ii) proportional to the velocity gradient between two layers. 
    i.e., straight F space proportional to space dv over dx space

Now comma space combining space both space the space factors comma space we space have

space space space space space space space space straight F space proportional to space straight A space space dv over dx space

rightwards double arrow space straight F space equals space straight eta space straight A space dv over dx space

where comma space straight eta space is space called space coefficient space of space viscosity. space 
    Question 343
    CBSEENPH11018446

    Define coefficient of viscosity. What are its units and dimensions?

    Solution
    The viscous force acting between the two layers of fluid having relative velocity is given by,

                         F=-ηAdvdxIf A = 1 unit , dvdx=1 unit,then,             η=F 

    Coefficient of viscosity of liquid is numerically equal to viscous force per unit area, which maintains a unit velocity gradient between two parallel layers.

    SI unit of viscosity is deca-poise and CGS unit of viscosity is poise.

    Dimensional formula of viscosity is [M1L–1T–1]. 

    Question 344
    CBSEENPH11018450

    Derive an expression for terminal velocity acquired by a spherical body falling in viscous media under gravity.

    Solution
    Consider, a spherical body of radius 'r' and density 'ρ' is falling under gravity in viscous media of coefficient of viscosity 'η' and density 'a'.
                                

    Let during fall, v be the velocity of body at any instant.
    The different forces acting on the body at that instant are:

    (i) Weight W, acting downward,
    (ii) Upward thrust U, acting upward,
    (iii) Viscous force F, acting upward.
    Net force on body is,  
          space space space space space space straight f equals straight W minus straight U minus straight F

rightwards double arrow space space straight f equals 4 over 3 πr cubed ρg minus 4 over 3 πr cubed σg minus 6 πηrν
rightwards double arrow space space space stack stack straight f equals 4 over 3 πr cubed left parenthesis straight rho minus straight sigma right parenthesis straight g with underbrace below with straight I below minus stack stack 6 πηrν with underbrace below with II below
    We can see that the body falls under gravity.
    Therefore, its velocity will increase due to gravity.
    But as the velocity increases, the viscous force (factor II) acting on the body also increase.
    Since factor I is constant, therefore the net force on the body decreases with time.
    At some velocity v = vo, the resultant of all the forces acting on body reduces to zero and body acquires the constant velocity known as terminal velocity.
    Therefore, 
    straight i. straight e. space space space space space space space space straight f equals 4 over 3 pi r cubed rho g minus 4 over 3 pi r cubed sigma g minus 6 pi eta r nu
rightwards double arrow space space space space space space space 4 over 3 pi r cubed open parentheses straight rho minus straight sigma close parentheses cubed sigma g minus 6 pi eta r nu
rightwards double arrow space space space space space space space space space straight v equals 2 over 9 fraction numerator straight r squared left parenthesis straight rho minus straight sigma right parenthesis straight g over denominator straight eta end fraction
    The above expression represents the terminal velocity acquired by a spherical body falling in viscous media under gravity.



    Question 345
    CBSEENPH11018452

    Gale blows over the roof of a house. In which direction the roof experiences the force?

    Solution
    Since the roof is blown over, the roof experiences force in upward direction.  
    Question 346
    CBSEENPH11018453

    Water is flowing through a pipe. The Reynolds number of the flow is 1800. What type of water flow is there in the pipe?

    Solution
    Since Reynold's number is less than 2000, the water flow is laminar or streamline. 

    Question 347
    CBSEENPH11018454

    Water is flowing through a pipe of non-uniform cross section. The ratio of the radii of tube at entrance and exit is 3:5. What is the ratio of the velocity of liquid entering and leaving the tube?

    Solution
    Given, ratio of the velocity of liquid entering and leaving tube is 3:5.
    Now, using equation of continuity, 
    Ratio of the velocity of liquid entering and leaving the tube = 5:3. 
    Question 348
    CBSEENPH11018455

    What is the pressure energy stored in liquid per unit volume of liquid?

    Solution
    The amount of pressure energy stored per unit volume of liquid is equal to hydrostatic pressure of liquid.
    Question 349
    CBSEENPH11018457

    What does the sudden fall in barometric height indicate?

    Solution
    Sudden fall in barometric height indicates the strong wind or storm.
    Question 350
    CBSEENPH11018458

    What are pressure head and velocity head?

    Solution

    The internal energy of a fluid due to pressure exerted on it's container is called pressure head. 
    straight P over ρg is called the pressure head. 
    The static pressure required to produce the velocity of fluid is known as velocity head. 
    The factor fraction numerator straight v squared over denominator 2 straight g end fraction is called velocity head. 

    Question 351
    CBSEENPH11018459

    What does Venturimeter measure?

    Solution
    Venturimeter measures the rate of flow of fluid when flow of fluid is steady. 

    Venturimeter is typically a gauge.  
    Question 352
    CBSEENPH11018460

    A spinning cricket ball in air does not follow a parabolic trajectory. Why?

    Solution
    A spinning cricket ball in air does not follow a parabolic trajectory due to Magnus effect. 

    Note: Magnus effect is responsible for the swerving of balls when hit or thrown with a spin. Magnus effect is a particular manifestation of Bernoulli's Theorem.
    Question 353
    CBSEENPH11018461

    Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.

    Solution
    Bernoulli's equation is only applicable to streamline flow or laminar flow. Therefore, Bernoulli's equation cannot be used to describe the flow of water rapidly through a river. 
    Question 354
    CBSEENPH11018462

    How does the pressure change with the velocity of flow?

    Solution
    Pressure is inversely proportional to the velocity of flow. So, pressure decreases with the increase of velocity of flow. 
    Question 355
    CBSEENPH11018464

    What is Torecelli’s theorem?

    Solution
    According to Toricelli's theorem, velocity of efflux, i.e., the velocity with which the liquid flows out on an orifice ( a nozzle) is equal to that which a freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid. 
    Question 356
    CBSEENPH11018465

    What is principle on which the atomizer works?

    Solution
    The atomizer works on the principle of Bernoulli's theorem.

    The difference between the reduced pressure at the top of the tube and the higher atmospheric pressure inside the bottle pushes the liquid from the reservoir up the tube and into the moving stream of air where it is broken up into small droplets, and carried away with the stream of air. 

    Question 357
    CBSEENPH11018466

    What is the shape of the wings of an aircraft?

    Solution
    The wings of the aircraft have upper side convex outside and lower surface concave inside. 
    This causes the air that flows over the top of the airfoil to move faster than the air that flows beneath it. 
    Question 358
    CBSEENPH11018468

    What are the differences between solid friction and viscosity?

    Solution
    Difference between solid friction and viscosity is as follows: 

    Viscosity is proportional to the surface area whereas solid friction is independent of area of solid surfaces in contact. 
    Viscous force on the body depends upon its velocity in the viscous media. But, friction does not depend on the velocity of the body. 




    Question 359
    CBSEENPH11018469

    Why machines are sometimes jammed in winter?

    Solution
    The machines are lubricated with oils to reduce the force of friction. In winter, due to fall in temperature the viscosity of oil increases and hence machines get jammed.
    Question 360
    CBSEENPH11018470

    How does the raindrop attain constant velocity after some time?

    Solution
    The stoke's law states that  the viscous force acting on the body is directly proportional to its velocity. The velocity of the falling raindrop increases due to gravity acting on it. So, the viscous force also increases.
    After a certain amount of velocity is attained, the weight W becomes equal to the upward forces. That is, the viscous force and upward thrust of the raindrop acquires a constant velocity. 

    [Note: According to stoke's law, the backward dragging force opposes the motion of the body. And, backward dragging force increases with increase in velocity of the moving body.]
    Question 361
    CBSEENPH11018471

    Hotter liquids move faster than colder ones. Why?

    Solution
    With the increase in temperature the coefficient of viscosity decreases. So, the velocity of the flowing liquid increases. Therefore, hotter liquids move faster than the colder ones. 
    Question 362
    CBSEENPH11018472

    Small raindrops fall slower than bigger raindrops through air. Why?

    Solution
    We know, that terminal velocity is directly proportional to the square of radius of drop.
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#6 {main}</pre>
    Smaller the radius, smaller is the terminal velocity and vice-versa.
    Hence small raindrops fall slower than bigger raindrops through air.
    Question 363
    CBSEENPH11018476

    Why do clouds seem floating in the sky?

    Solution

    Tiny droplets of water are contained in clouds. We kniow that terminal velocity acquired by a spherical body falling through a viscous media is directly proportional to the square of the radius of a body. 
    So, very small drops falling in air under gravity have very small terminal velocity and remain in sky. Hence, clouds appear to be floating in the sky.

    Question 364
    CBSEENPH11018477

    Dust generally settles down in a closed room. Explain.

    Solution
    Dust particle are very small spherical particles. Now, using the formula of terminal velocity, 

                                       v ∝ r

    The small dust particles acquire a very small terminal velocity and remain suspended in air because of their small size. Therefore, they will settle down in a closed room after some time.
    Question 365
    CBSEENPH11018479

    Two beakers containing water and honey are stirred rapidly and kept on a table. Which liquid will come to rest earlier?

    Solution
    Coefficient of viscosity of honey is greater than that of water. Therefore honey will come to rest earlier than water.
    Question 366
    CBSEENPH11018480

    Explain why a parachute is invariably used, while jumping from an aeroplane.

    Solution
    A parachute while descending through air experiences large viscous force as a result of it's large structure. As the body is falling down, drag force is created opposing the motion. This balances out the gravitational force acting downwards. The terminal velocity of the falling object decreases as a result. Hence, a person landing would not get hurt.
    Question 367
    CBSEENPH11018481

    Why does an object entering the earth’s atmosphere at high speed catch fire?

    Solution
    When an object moving at a high speed and enters the Earth's atmosphere, kinetic energy of body is converted into heat. The viscosity of ar is high in te Earth's atmosphere. Hence, a high apeed object catches fire. 
    Question 368
    CBSEENPH11018482

    If a body is thrown with velocity u in a viscous media free from any field, then what will be terminal velocity of the body?

    Solution
    When a body is thrown with velocity 'u' in a viscous media free from any field, then there is no force that can keep the velocity constant against the viscous force. Hence the velocity of the body decreases to zero.
    Question 369
    CBSEENPH11018483

    A body is projected downward with velocity twice the terminal velocity from a very tall building. Find the velocity with which it will reach the ground.

    Solution
    When body is thrown with velocity greater than terminal velocity, the viscous force on the body is greater than the weight of the body. Therefore the body starts to retard and velocity of the body decreases. When the velocity of body decreases to terminal velocity, then net force on the body becomes zero and falls with constant velocity. Thus, the body reaches the ground with terminal velocity.
    Question 370
    CBSEENPH11018485

    The velocity of water at the surface is 10 m/s in a 5m deep river. Find the sheer stress between the horizontal layers of water. Take η = 0.01 poise.

    Solution
    The viscous force between the layers of water is given by,
    straight F space equals space straight eta space straight A space dv over dx

straight F over straight A space equals space straight eta space dv over dx space

straight eta space equals space 0.01 space poise

dv over dx space equals space 10 over 5 space equals space 2 space straight s to the power of negative 1 end exponent space

therefore space straight F over straight A space equals space 0.01 space cross times space 2 space
space space space space space space space space space space space space equals space 0.02 space dyne space divided by cm squared
                       
    Question 371
    CBSEENPH11018487

    A plate of area 20x10 cm2 is placed over the honey poured on the table. The layer of honey between plate and the table is 2.5mm thick. To move the plate with velocity 3 cm/s a horizontal force of 30720 dyne is required. Find the coefficient of viscosity of honey.

    Solution

    The formula for viscous force is given by,
    straight F space equals space straight eta space straight A dv over dx

rightwards double arrow space straight eta space equals space straight F over straight A dx over dv space space space space... thin space left parenthesis 1 right parenthesis thin space
    Here given that,
    Horizontal force, F = 30720 dyne    
    Area of the plate, A = 200 cm
    Velocity with which the plate is moved = 3 cm/sec
    Thickness of the layer of honey between the plate and table = 2.5 mm 
    Putting the values in equation (1), we get
    dv over dx space equals space fraction numerator 3 over denominator 0.25 end fraction space equals space 12 space straight s to the power of negative 1 end exponent space

Therefore comma space

straight eta space equals space 30720 over 200 cross times 1 over 12 space equals space 12.8 space poise
    is the required coefficient of viscosity of honey. 

    Question 372
    CBSEENPH11018488

    The coefficient of viscosity of water is 0.01 poise. What is fluidity and coefficient of kinematic viscosity?

    Solution

    Given that,
    Coefficient of viscosity,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> =0.01 poise
     p = 1 gm/cc
    therefore space space space space space space Fluidity space equals space 1 over straight eta equals 100 space poise to the power of negative 1 end exponent]
    Coefficient of kinematic viscosity eta divided by rho =0.01 stokes

    Question 373
    CBSEENPH11018491

    Distinguish between streamline and turbulent flow of a liquid.

    Solution

    Difference between streamline motion and turbulent motion.

    i) It is an orderly type of motion in which the liquid flows in parallel layers while, turbulent motion is disorderly type of motion.

    ii) Every particle of the liquid follows the path of it's preceeding particle and travel with the same velocity in magnitude and direction whereas, the motion of particles of the liquids becomes different at different points in turbulent flow.

    iii) The velocity of streamline flow is less than critical velocity but, in turbulent motion the liquid moves with a velocity greater than the critical velocity of the liquid.

    iv) Streamline flow is laminar whereas, turbulent flow is non-laminar. 

    Question 374
    CBSEENPH11018493

    Two streamlines cannot cross each other. Why?

    Solution
    Streamline is a line, straight or curved tangent to which at any point gives the direction of flow of liquid at that point. If two streamlines cross each other, then at that point there will be two tangents and hence two directions of flow of liquid, which is impossible.
    Therefore, two streamlines cannot cross each other.
    Question 375
    CBSEENPH11018494

    Why is energy loss in a streamlined flow minimum?

    Solution
    In streamline flow, the different layers glide over one another without intermixing. Therefore there is no collision between the molecules of different layers. Hence the energy loses is minimum in a streamline flow.
    Question 376
    CBSEENPH11018495

    What is critical velocity?

    Solution
    Critical velocity is that velocity of liquid flow, upto which it's flow is streamlined and above which it's flow becomes turbulent. 
    Question 377
    CBSEENPH11018649

    What is wave?

    Solution
    Wave is the disturbance which is transported from one particle to the other due to repeated periodic vibrations of particles of media about mean position, without any net transport of medium.
    Question 378
    CBSEENPH11018650

    Is it logical to say that wave motion is a process of propagation of mass?

    Solution
    In wave motion, transportation of matter does not take place in the process. 
    Question 379
    CBSEENPH11018651

    What is common property of all types of waves?

    Solution
    The medium in which the wave travels does not travel along with the disturbance. 
    Question 380
    CBSEENPH11018652

    What are different types of wave motions?

    Solution
    There are two types of wave motions:
    (i) Transverse wave motion, and
    (ii) Longitudinal wave motion.
    Question 381
    CBSEENPH11018653

    Define transverse wave motion.

    Solution
    A transverse wave motion is that in which particles of the medium vibrate about their mean position perpendicular to the direction of propagation of wave. 
    Question 382
    CBSEENPH11018654

    Define longitudinal wave motion.

    Solution
    Longitudnal motion is the wave motion in which the individual particles of the medium vibrate about their mean position in the direction of propagation of wave.
    Question 383
    CBSEENPH11018655

    Which of the two waves - elastic waves or electromagnetic waves does need any material media for its propagation?

    Solution
    Electromagnetic waves do not need any material media for it's propagation. 
    Question 384
    CBSEENPH11018656

    What is sound?

    Solution
    Sound is a form of energy that produces the sensation of hearing. Sound is produced due to the vibration of body. 
    Question 385
    CBSEENPH11018657

    How sound is produced?

    Solution
    Sound is produced as a result of vibrating bodies. 
    Question 386
    CBSEENPH11018658

    Can sound travel in vacuum?

    Solution
    Sound cannot travel in vacuum. For the proapagation of sound waves a material medium is required.  
    Question 387
    CBSEENPH11018659

    Can two persons hear each other on the moon? Explain.

    Solution
    Atmosphere is absent on the moon. And sound waves require medium for propagation. Therefore, two persons on the moon cannot hear each other. 
    Question 388
    CBSEENPH11018660

    Can electromagnetic wave propagate in vacuum?

    Solution
    Yes, electromagnetic waves can travel through vacuum. It does not require any material medium for propagation. 
    Question 389
    CBSEENPH11018661

    Can we listen in vacuum?

    Solution
    Sound wave cannot travel in vacuum. Therefore we cannot listen in vacuum. 
    Question 390
    CBSEENPH11018662

    Which of the two moves forward during wave motion - particle or disturbance?

    Solution
    In wave motion it is the disturbance that moves forward and not the particles of the medium. 
    Question 391
    CBSEENPH11018663

    What type of wave is sound wave when it travels in air?

    Solution
    Sound waves are longitudinal in nature. 
    Question 392
    CBSEENPH11018664

    What type of wave is sound wave when it travels in solids?

    Solution
    Sound waves can be either longitudinal or transversal wave, when it travels in a solid medium. 
    Question 393
    CBSEENPH11018665

    Which of the two - sound or light is longitudinal in nature?

    Solution
    Sound waves are longitudinal in nature.
    Question 394
    CBSEENPH11018666

    In what form the sound travels in air?

    Solution
    The sound travels in air in form of compression and rarefaction.
    Question 395
    CBSEENPH11018667

    Define frequency.

    Solution
    Frequency of wave is the number of complete wavelengths traversed by the wave in one second. 
    Question 396
    CBSEENPH11018668

    Define time period.

    Solution
    Time period is the time taken by the vibrating body to complete one vibration. 
    Time taken by the wave to travel a distance equal to one wavelength is called as the time period. 
    Question 397
    CBSEENPH11018669

    What is wavelength?

    Solution

    It is the distance between two nearestive particles vibrating in same phase.
                                             or 
    It is distance travelled by a wave in the direction of propagation in the duration equal to time period of vibrating body that produces the wave.

    Question 398
    CBSEENPH11018670

    What is the distance between compression and its nearest rarefaction in a sound wave?

    Solution
    Distance between the compression and rarefaction is λ/2 
    Question 399
    CBSEENPH11018671

    Define wave number.

    Solution
    Wave number is defined as the number of waves in one meter and is equal to reciprocal of wavelength.
    Question 400
    CBSEENPH11018672

    What is the unit of frequency?

    Solution
    Unit of frequency is hertz (Hz).
    Question 401
    CBSEENPH11018673

    How is wavelength of wave related with velocity of air?

    Solution
    Relation between wavelength and velocity of air is given by, 
                               v = straight nuλ
    Question 402
    CBSEENPH11018674

    What is the unit of wavelength?

    Solution
    In SI system, unit of wavelength is metre.
    In CGS system the unit of wavelength is cm. 
    Question 403
    CBSEENPH11018675

    What is amplitude?

    Solution
    Amplitude is the maximum displacement of the vibrating particle from its mean position. 
    Question 404
    CBSEENPH11018676

    What is the significance of phase?

    Solution
    The phase of the particle at any instant gives the position and direction of motion of particle at that instant. This is the significance of phase. 
    Question 405
    CBSEENPH11018677

    Can transverse mechanical waves be propagated in air?

    Solution
    No, transverse mechanical waves cannot propagate through air. They require a medium for propagation. 
    Question 406
    CBSEENPH11018678

    What is wave front?

    Solution
    The locus of all the points which are in the same phase is called as wavefront. 
    Question 407
    CBSEENPH11018679

    What is the shape of wave front due to a point source?

    Solution
    The wave front due to a point source is a spherical in shape. 
    Question 408
    CBSEENPH11018680

    What is the shape of wave front due to a line source?

    Solution
    The wave front due to a line source is a cylindrical in shape. 
    Question 409
    CBSEENPH11018681

    Is it necessary that amplitude be constant over a given wave front?

    Solution
    No, if and only if the media is homogeneous, amplitude at every point on the wave front is constant. 
    Question 410
    CBSEENPH11018682

    Is there any transport of media during wave motion?

    Solution
    There is no net transport of media during wave propagation. The particles of medium only vibrate about their mean position. 
    Question 411
    CBSEENPH11018683

    What are the required characteristics of media to propagate the wave?

    Solution
    The required characteristics of media to propagate the wave are, the medium must possess property of elasticity and inertia. The medium also should have  least frictional resistance. 
    Question 412
    CBSEENPH11018684

    Which travels fast - sound of bomb explosion or sound produced by a humming bee?

    Solution
    The velocity of the wave is independent of loudness, quality or pitch. Therefore, both the sounds will travel at the same speed. 
    Question 413
    CBSEENPH11018685

    Does the speed of sound depend upon wavelength of sound?

    Solution
    The velocity of sound is independent of the wavelength of the wave. 
    Question 414
    CBSEENPH11018686

    In which gas - CO2 or N2, will sound travel with grater speed under same physical condition?

    Solution
    In N2 gas, the sound will travel with a greater speed.
    Speed of sound in N2 gas is 349 m/s and speed of sound in CO2 is 267 m/s. 
    Question 415
    CBSEENPH11018687

    Name the gas in which the sound travels fastest under similar physical conditions.

    Solution
    The sound travels fastest under similar physical condition in H2 gas.
    Speed of sound in hydrogen gas is 1270 m/s. 
    Question 416
    CBSEENPH11018688

    Will a vibrating source always produce sound?

    Solution
    Vibrating source produces sound only if its frequency of vibration lies in audio region i.e. 20Hz to 20,000Hz. 
    Question 417
    CBSEENPH11018689

    How does a bee buzz?

    Solution
    The buzzing of a bee is due to the vibration of its wings.
    The frequency of vibration lies in the audible range.
    Question 418
    CBSEENPH11018690

    What propagates faster - radio signal or sound in air?

    Solution
    Radio signal travels faster than sound in air.
    Velocity of radio signal is 3 x 108 m/s, while the velocity of sound in air is 330 m/s. 
    Question 419
    CBSEENPH11018691

    What is supersonic body?

    Solution
    Any body which is moving with a velocity greater than the velocity of sound is called supersonic body.
    Question 420
    CBSEENPH11018692

    Why there is always a time interval between observing flash and hearing a thunder in the sky?

    Solution
    The flash of light is always seen before we hear the sound of the thunder. This is because, light travels much faster than sound. 
    This is the reason that we see lightning faster than we hear the sound of thunder. 
    Question 421
    CBSEENPH11018693

    In which media sound travels faster - solids or gases?

    Solution
    Sound travels faster in solids than in gases because young's modulus of elasticity of solids is large as compared to bulk modulus of gases. 
    Question 422
    CBSEENPH11018694

    What is the range of audible sound?

    Solution
    The audible range of sound is from 20 Hz to 20,000 Hz.
    Question 423
    CBSEENPH11018695

    Out of the following, name the medium which can transmit mechanical wave:

    air, steel, water, vacuum. 

    Solution
    Mechanical waves can be transmitted through air, steel and water.
    Mechanical waves cannot be transmitted through vacuum as they require some material medium for propagation.  
    Question 424
    CBSEENPH11018696

    What are water waves?

    Solution
    The waves on the surface of water are neither purely longitudinal nor purely transverse. The waves that do not belong to both categories are called water waves. 
    Question 425
    CBSEENPH11018697

    Why a vibrating pendulum does not produce sound? Explain.

    Solution
    The frequency of vibrating pendulum is less than audible range.
    Hence, no sound is heard from the vibrating pendulum. 
    Question 426
    CBSEENPH11018698

    What are ultrasonic waves?

    Solution
    Ultrasonic waves are waves having frequency more than the upper audible limit of 20,000 Hz. 
    Question 427
    CBSEENPH11018699

    What are subsonic waves?

    Solution
    Subsonic waves are waves having frequency less than the lower audible limit of 20 Hz. 
    Question 428
    CBSEENPH11018700

    What are shock waves?

    Solution
    When a supersonic body moves in air, it produces disturbance, which moves in backward direction and diverges in the form of cone. Such a disturbance is called shock wave. 
    Question 429
    CBSEENPH11018701

    What is Newton's formula for the velocity of sound in medium?

    Solution
    Newton's formula for velocity of sound in a medium is given by, 
                              straight v equals square root of straight E over straight rho end root 
    where,
    E is volume elasticity, and
    ρ the density of media.
    Question 430
    CBSEENPH11018702

    How does the velocity of sound vary with temperature?

    Solution
    The velocity of sound is directly proportional to the square root of absolute temperature.
    That is, 
                                    v ∝ √T  
    Question 431
    CBSEENPH11018703

    What is the Laplace's formula for the velocity of sound in medium?

    Solution
    The Laplace's formula for velocity of sound in a medium is given by, 
                                  straight v equals square root of γE over straight rho end root 
    where γ ratio of the specific heats of air. 
    Question 432
    CBSEENPH11018704

    How does the velocity of sound in gas vary with density of gas?

    Solution
    The velocity of sound in a gas is inversely to the square root of the density of gas. 
    Question 433
    CBSEENPH11018705

    Why does the velocity of sound in air increases after rain?

    Solution
    When it rains, there is increase in humidity. Therefore, the density of air decrease and hence the velocity of sound increases.
    Question 434
    CBSEENPH11018706

    Is there any affect of pressure variation at constant temperature on the velocity of sound in gases?

    Solution
    No, the velocity of sound in gas at constant temperature is independent of pressure of gas.
    Question 435
    CBSEENPH11018707

    The velocity of sound in still air is 340m/s. The wind starts blowing opposite to the direction of sound with velocity 25m/s. What is the velocity of sound in blowing wind?

    Solution
    Velocity of sound in still air = 340 m/s
    Velocity of the wind = 25 m/s
    The velocity of sound in wind is blowing opposite to the direction of sound is,
    V = v – u
       = 340 – 25
       = 315 m/s 
    Question 436
    CBSEENPH11018708

    In which state of medium is the velocity of sound maximum?

    Solution
    Velocity of sound is maximum in solids, less in liquids and minimum in gases. 
    More denser is the state of matter, more faster is the velocity of sound. 
    Question 437
    CBSEENPH11018709

    What is temperature coefficient?

    Solution
    The change in velocity of sound per degree change in temperature is called as temperature coefficient. 
    Question 438
    CBSEENPH11018710

    What is wave function?

    Solution
    Wave function is a periodic function of position and time.
    Question 439
    CBSEENPH11018711

    What is a harmonic wave function?

    Solution
    Harmonic wave function is a periodic wave function which is sinusoidal in shape. 
    Question 440
    CBSEENPH11018712

    Express the harmonic wave in terms of wavelength and time period.

    Solution

    The harmonic waves in terms of wavelength and time period is given as, 
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#6 {main}</pre>
    where,
    straight lambda space is space the space wavelength comma space and space

straight T space is space the space time space period space of space the space oscillation. space 

    Question 441
    CBSEENPH11018713
    Question 442
    CBSEENPH11018714

    Express the harmonic wave in terms of its wavelength and velocity.

    Solution

    Harmonic wave in terms of wavelength and velocity is given as, 
                       space straight y equals rsin open square brackets fraction numerator 2 straight pi over denominator straight lambda end fraction left parenthesis vt minus straight x right parenthesis close square brackets
    where, 
    straight lambda space is space the space wavelength comma space and

straight v space is space the space velocity space of space the space wave. space 

    Question 443
    CBSEENPH11018715

    Write the expression for the velocity of particle of the media located at x in a simple harmonic wave at any instant t.

    Solution

    The velocity of particle of the media located at x in a simple harmonic wve at any instant t is given by, 
                   straight v subscript straight p equals straight r fraction numerator 2 straight pi over denominator straight lambda end fraction straight v space cos open square brackets fraction numerator 2 straight pi over denominator straight lambda end fraction left parenthesis vt minus straight x right parenthesis close square brackets

    Question 444
    CBSEENPH11018716

    What is the ratio of maximum particle velocity to the wave velocity in a harmonic wave?

    Solution

    In a harmonic wave, the ratio of maximum particle velocity to the wave velocity in a harmonic wave is, 

                     fraction numerator left parenthesis straight v subscript straight p right parenthesis subscript max over denominator straight v end fraction equals straight r fraction numerator 2 straight pi over denominator straight lambda end fraction equals rk

    Question 445
    CBSEENPH11018717

    Which is constant an a medium - particle velocity or wave velocity?

    Solution
    Wave velocity is constant. Velocity of wave is dependent on the medium in which it travels. Therefore, as far as the wave is travelling in the same medium, wave velocity does not change. 
    Question 446
    CBSEENPH11018718
    Question 447
    CBSEENPH11018719

    What a wave transmits - energy or momentum?

    Solution
    Both energy and momentum is transmitted by the wave. 
    Question 448
    CBSEENPH11018720

    Write the differential form of one -dimensional wave.

    Solution

    One-dimensional wave in the differential form is written as,
                          fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 over straight v squared fraction numerator straight d squared straight y over denominator dt squared end fraction

    Question 449
    CBSEENPH11018721

    Can we concentrate sound waves at a particular point?

    Solution
    Yes, by using converging lens we can concentrate sound at a particular point. 
    Question 450
    CBSEENPH11018722

    What is wave motion? What are the essential properties of propagating a wave in media? What is propagated in wave motion?

    Solution
    Wave motion is a means of propagating energy and momentum from one point to another without any actual transportation of matter between these points. 
    To propagate a wave in media, it must possess the property of inertia and elasticity.
    In a wave motion, energy and momentum is propagated.
    Question 451
    CBSEENPH11018723

    State the properties of wave motion in brief.

    Solution

    The properties of wave motion are:
    (i) Wave motion is a disturbance propagated in media due to periodic motion of the particles of media.
    (ii) In wave motion, it is the momentum and energy which is propagated, not the mass.
    (iii) In wave motion, the particles of media vibrate about mean position.
    (iv) The velocity of vibrating particle is different at different position.
    (v) The velocity of wave is constant.
    (vi) Different particles of media vibrate with different phase and phase difference between two consecutive particles is constant.

    Question 452
    CBSEENPH11018724

    Distinguish between transverse wave and longitudinal wave.

    Solution

    Transverse wave:
    Transverse wave motion is the disturbance in which particles of medium vibrate in the direction perpendicular to the direction of propagation of wave. Transverse wave is propagated through media in the form of crests or troughs.
    Longitudinal wave:
    Longitudnal wave motion is the disturbance in which particles of media vibrate in the direction of propagation of wave. These waves propagate in the form of compressions and rarefactions.

    Question 453
    CBSEENPH11018725

    Discuss the propagation of transversal wave in brief.

    Solution

    Propagation of transverse waves in a media.
    Consider nine particles of media on a reference line AB.

    i) Let the particles vibrate perpendicular to line AB with amplitude ‘a’ and the wave propagates along AB from left to right.
    ii) When the wave propagates, the different particles of media vibrate in different phase because it takes some time to transfer the disturbance (momentum and energy) from one particle to next particle.
    iii) Consider that the disturbance takes T/8 seconds to travel from one particle to next.
    iv) At t =0, all the particles are at mean position.
    v) At t = T/8 sec, particle 1 gets displaced by 0.707a distance in the upward direction, while the disturbance reaches the particle 2.
    vi) At t = 2T/8 sec, particle 1 reaches positive extreme position, the particle 2 gets displaced by 0.101a distance in the upward direction and the disturbance reaches the particle 3. 
    vii) At t = 3T/8 sec, after completing three eighth of vibration, particle 1 comes  back to 0.707 a; the particle 2 reaches positive extreme position; particle 3 undergoes the displacement of 0.707a and the disturbance just reaches the particle 4. 
    viii) In this way the disturbance continues and the position of different particles at 4778, 5778, 6778 and 7778 seconds is as shown in the figure above. 
    viii) After T seconds, the particle 1 completes one vibration and particle 9 is just at the point to start its first vibration.
    Thus, the particle 1 leads the particle 9 in phase by an angle of space space 2 straight pi.

    Question 454
    CBSEENPH11018726

    Discuss the propagation of longitudinal wave in a media. Show that the longitudinal wave propagates in the form of compression and rarefaction.

    Solution

    Propagation of longitudinal waves in a media:
    i) Consider nine particles of media on a reference line AB.
    ii) Let the particles vibrate perpendicular to line AB with amplitude ‘a’ and the wave propagates perpendicular to AB from left to right.
    iii) When the wave propagates, the different particles of media vibrate in different phase because it takes some time to transfer the disturbance (momentum and energy) from one particle to the next particle.
    iv) Let disturbance take 778 seconds to travel from one particle to next.
    v) At t = 0, all the particles are at mean position. 
    vi) At t = T/8 sec, particle 1 gets displaced by 0.707a distance in the right direction, while the disturbance reaches the particle 2. 
    vii) At t = 2T/8 sec, particle 1 reaches positive extreme position, the particle 2 gets displaced by 0 707a distance in the right direction and the disturbance reaches the particle 3. 
    viii) At f = 3T/8 sec, particle 1 after completing three eighth of vibration, comes back to 0.707a, the particle 2 reaches positive extreme position, particle 3 undergoes the displacement of 0.707a and the disturbance reaches the particle 4. 
    The disturbance of the particles is as shown below: 

    viii) In this way, the disturbance continues and the position of different particles at 4778, 5778, 6778 and 7778 seconds is as shown in fig. above. 
    ix) After T seconds, the particle 1 completes one vibration and particle 9 is just at the point to start its first vibration. Thus the particle 1 leads the particle 9 in phase by angle space space 2 straight pi. 
    The longitudinal wave propagates in the form of compression and rarefaction. Let us see how:
    i) Draw the instantaneous position of different particles and their relative displacement from their mean positions after the particle 1 has completed on vibration.
    The illustration of the disturbance is as shown below: 

    ii) After one complete vibration, the particles 1,5,9 are at the mean position; particles 2,3 and 4 move close towards particle 1 and particles 6,7 and 8 move towards particle 9. That is the particles crowd nearby 1 and 9. Thus the positions of particles 1 and 9 are the positions of compression.
    iii) On the other hand, the particles 2,3 and 4 move away towards left from particle 5 and particles 6,7 and 8 move away towards right from particle 5. Thus the position of particle 5 is the position of rarefaction. 
    iv) Thus, there is an alternate formation of compression and rarefaction in media and hence the wave propagates in the form of compression and rarefaction.
    v) In the longitudinal wave, the distance between the two consecutive position of maximum compression or rarefaction is equal to the wavelength of wave.

    Question 455
    CBSEENPH11018727

    What do you mean by crest or trough?

    Solution
    Crest: The maximum displacement of particle from mean position in positive direction is called crest.
    Trough: It is the maximum displacement in negative direction.
    Question 456
    CBSEENPH11018728

    What do you mean by compression and rarefaction?

    Solution
    Compression:
    The regions where the density of media is greater than normal are called compression. 
    Rarefaction:
    The regions where the density of media is less than the normal are called rarefaction. 
    Question 457
    CBSEENPH11018729

    Find the velocity of wave in a string of linear mass density m when stretched to tension T?

    Solution
    Let velocity of wave in the string vary with ath power of m and bth power of T.
    That is, 
                     straight v proportional to straight m to the power of straight a

space space space proportional to straight T to the power of straight b
space space space 
    ∴            straight v proportional to straight m to the power of straight a straight T to the power of straight b 
    rightwards double arrow      space space space space space straight v equals straight k space straight m to the power of straight a space straight T to the power of straight b          ...(1)
    where,
    k is constant of proportionality.
    If this relation is true then dimensions of each term on both sides will be same.
    Substituting the dimension formula of each quantity in equation (1),
              left square bracket straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent right square bracket space equals space left square bracket straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of 0 right square bracket to the power of straight a space open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets to the power of straight b
                              equals straight M to the power of straight a plus straight b end exponent straight L to the power of negative straight a plus straight b end exponent straight T to the power of negative 2 straight b end exponent
    Comparing the dimensions of M,L and T on both sides, we get
                             space space straight a plus straight b equals 0 comma space

minus straight a plus straight b equals 1
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#6 {main}</pre> 
    On solving we get, 
                 straight a equals negative 1 half space space and space space straight b space equals space 1 half 
    Thus,   space space space space straight v equals KT to the power of 1 divided by 2 end exponent straight m to the power of negative 1 divided by 2 end exponent 
    rightwards double arrow             straight v equals square root of straight T over straight m end root 
    Question 458
    CBSEENPH11018730

    Derive Newton's formula for the velocity of sound in air.

    Solution
    Velocity of longitudinal wave in elastic media is, 
                                straight v equals square root of straight E over straight rho end root 
    where E is elasticity and ρ is density of media.
    Newton assumed that sound wave travels in air under isothermal condition.
    Therefore,
                           space space straight v equals square root of straight E subscript iso over straight rho end root 
    For isothermal process, 
                          PV equals Constant 
    On differentiating, we get 
                       PdV plus VdP equals 0 
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#6 {main}</pre>
    Now volume elasticity is,
                      space space space straight E equals negative straight V dP over dV
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#6 {main}</pre> 
    Thus,                straight v equals square root of straight P over straight rho end root 
    At NTP, the density of air is 1.293 kg/m3
    ∴  
    Velocity of sound in air is given by, 
    space space space space straight v equals square root of fraction numerator 1.01 cross times 10 to the power of 5 over denominator 1.293 end fraction end root equals 280 space straight m divided by straight s        
     
    This value is approximately 16% less than the experimental value 332 m/s.
    Question 459
    CBSEENPH11018731

    What correction was applied by Laplace to remove the discrepancy given by Newton's formula and why?

    Solution
    Laplace correction:
    Laplace suggested that sound waves travel through air under adiabatic conditions and not under isothermal conditions because air is a bad conductor of heat. Also, compression and rarefaction of air takes place rapidly.
    Therefore according to Laplace, the velocity of sound in air is,
                         straight v equals square root of straight E subscript adi over straight rho end root 
    For adiabatic changes, 
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#6 {main}</pre> 
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#6 {main}</pre> 
    Thus,                 space straight v equals square root of γP over straight rho end root 
    Since air is diatomic, therefore γ = 1.4.  
    Substituting the values of γ, P and ρ, we get 
    straight v equals square root of fraction numerator 1.4 cross times 1.01 cross times 10 to the power of 5 over denominator 1.293 end fraction end root space equals space 331 space straight m divided by straight s 
    This result agrees with the experimental value.
    Question 460
    CBSEENPH11018732

    Derive the relation between velocity of sound V in gas and rms velocity v of gas 

    Solution
    According to kinetic theory of gases, 
                           straight P equals 1 third ρv squared 
    ∴                       straight v equals square root of fraction numerator 3 straight P over denominator straight rho end fraction end root           ...(1) 
    Also the velocity of sound in gas is given by, 
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#6 {main}</pre>           ...(2)
    From (1) and (2), 
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#6 {main}</pre> 
    This expression gives us the relationship between the velocity of sound in gas and rms velocity of gas. 
    Question 461
    CBSEENPH11018733

    What would be the effect of pressure on the velocity of sound? Explain.

    Solution
    If pressure is increases at a constant temperature, the density will also increase such that P/ ρ remains constant. 
    That is,
    straight v subscript straight s equals square root of γP over straight rho end root
    At constant temperature P/ρ is constant.
    Therefore, the velocity of sound, vis also constant.
    Thus at constant temperature velocity of sound is independent of pressure. 
    Question 462
    CBSEENPH11018734

    How does the velocity of sound change with temperature?

    Solution

    For one mole of gas, ideal gas equation is given by, 
                     PV equals RT 
    rightwards double arrow          PM over straight rho equals RT 
    rightwards double arrow             straight P over straight rho equals RT over straight M proportional to straight T 
    Now the velocity of sound in gas at temperature T is, 
                                   space space straight v equals square root of γP over straight rho end root
    Since, straight P over straight rho proportional to straight T 
    ∴                             straight v proportional to square root of straight T 
    Hence the velocity of sound in gases varies directly to the square root of temperature in Kelvin. 

    Question 463
    CBSEENPH11018735

    Define temperature coefficient of velocity of sound and find its value.

    Solution
    The increase in velocity per unit rise in temperature is called the temperature coefficient of velocity. 
    Let v0 be the velocity of sound at zero degree centigrade and vt be the velocity of sound at t°C. 
    Since,   
                                 straight v proportional to √ straight T 
    ∴        space space straight v subscript straight t over straight v subscript straight o equals square root of fraction numerator straight t plus 273 over denominator 273 end fraction end root equals square root of 1 plus straight t over 273 end root 
    Using binomial expansion, we get 
                        straight v subscript straight t over straight v subscript straight o equals straight i plus straight t over 546 
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#6 {main}</pre> 
    rightwards double arrow             fraction numerator straight v subscript straight t minus straight v subscript straight o over denominator straight v subscript straight o end fraction equals straight t over 546 
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#6 {main}</pre>
                              equals 0.61 space straight m divided by straight s to the power of straight o straight C     
    Therefore the temperature coefficient of velocity of sound is 0-61 m/s°C. 
    Question 464
    CBSEENPH11018736

    How does the velocity of sound change with humidity? 

    Solution

    Velocity of the sound wave is given by,
                           space space straight v equals square root of γP over straight rho end root 
    Humid air contains large proportion of water vapours.
    Therefore, the density of humid air is less than dry air.
    Since velocity of sound is inversely proportional to the square root of density, therefore, sound travels faster in moist air than dry air. 

    Question 465
    CBSEENPH11018737

    If you set your watch by sound of distant siren, will it go slow or fast?

    Solution
    The speed of sound in air has a finite value. The sound of the distant siren will take a finite time to reach us. Hence, the watch set by the sound of a distant siren will go a little slower. 
    Question 466
    CBSEENPH11018738

    An approaching train can be felt by applying ears onto rails but not easily through air. Why?

    Solution
    The density of a medium affects the loudness of sound. More the medium is denser, more is the intensity and hence more is the loudness of the sound. 
    Therefore, sound of distant train travelling through rails is louder than air. So, approaching train can be felt by applying ears onto rails but not easily through air. 
    Question 467
    CBSEENPH11018739

    If an observer places his ear at the end of a long pipe, he can hear two distinct sounds when pipe is hammered at the other end. Why?

    Solution
    Sound propagates both through solid pipe as well as in air, when the pipe is hammered.
    The velocity of sound in solids is greater than air. Therefore, sound through solid pipe will reach earlier than sound through air. Hence a person hears two distinct sounds. 
    Question 468
    CBSEENPH11018740

    Why is sound heard with more intensity through CO2 than air?

    Solution
    Intensity of sound depends upon density. More is the density, greater is intensity of sound. Since CO2 is heavier than air, therefore, sound travelling through CO2 is louder than air. 
    Question 469
    CBSEENPH11018741

    What is the most fundamental property of wave and why?

    Solution
    The most fundamental property of the wave is frequency because it is the only physical quantity that is independent of the nature of media through which the wave is propagating.
    Question 470
    CBSEENPH11018742

    A spherical balloon is filled with: (i) CO2 and (ii) air. What type of lens will it behave for sound?

    Solution
    The velocity of sound in CO2 is less than in air. Therefore, spherical balloon filled with CO2 placed in air will act like a converging lens. While balloon filled with air placed in air will not act like any lens. 
    Question 471
    CBSEENPH11018743

    Deaf people can be made to dance to music. Explain how?

    Solution
    Deaf people can listen to music, if there is nothing wrong with internal ear. This can be possible by placing one end of a rod on musical instrument and other end of rod gripped in teeth. The music would reach the auditory nerve via rod and bones of head.
    Question 472
    CBSEENPH11018744

    Why are sound waves called pressure waves?

    Solution
    Sound wave travels in the media in form of compressions and rarefactions.
    These compressions and rarefactions change the volume and hence pressure of media.
    That is why sound waves are called pressure waves.
    Question 473
    CBSEENPH11018745

    Why a diver under water is unable to hear the sound produced in air?

    Solution
    When the sound produced in air, falls on the surface of water, reflects about 99.9% of incident intensity back into air and refracts only 0.1% of incident intensity.
    Hence, the diver in water cannot hear the sound produced in air. 
    Question 474
    CBSEENPH11018746

    Why does an empty vessel produce more sound than the filled vessel?

    Solution
    The intensity of sound is directly proportional to the square of amplitude of vibration.
    When empty vessel is struck, the air molecules are set in vibration and when filled vessel is struck the liquid molecules are set in vibration.
    Since, the amplitude of vibration of air molecules is greater than liquid molecules, therefore empty vessel produces louder sound than the filled vessel.
    Question 475
    CBSEENPH11018747

    While watching the science fiction movie, you see and hear a distant explosion in space simultaneously. Is it possible? If not, then what is wrong?

    Solution
    Explosions happen in the universe often. Even scientists suggest that universe has been created from an exlosion which is called as the Big Bang theory.
    But what is wrong in the movie is, we see and hear a distant explosion in space simultaneously. 
    Since, velocity of sound and light is different, therefore we cannot see and hear the explosion simultaneously.
    Moreover there is vacuum in the space and sound cannot travel in vacuum, therefore we cannot hear the explosion. 
    Question 476
    CBSEENPH11018748

    Intensity of sound of public address system is found to be little higher, immediately after rain. Explain.

    Solution
    After rain, the air becomes humid and hence the density of air decreases.
    This results in increase in the velocity of sound in air. Since intensity of sound is directly proportional to the velocity of wave, hence the intensity of wave increases a little after rain.
    Question 477
    CBSEENPH11018749

    Give a simple experiment to show that there is transfer of energy by a wave.

    Solution
    An experiment to show there is a transfer of energy by wave.
    1. Place a cork in the pond of water.
    2. Drop a stone from some height at some distance from the cork.
    3. You will see when the waves reach the cork, the waves will take the cork along with them.
    4. This shows that there is transfer of energy by a wave.
    A disturbance is created and you will see the waves travelling. 
    Question 478
    CBSEENPH11018750

    The audible range of human car is 20Hz - 20KHz. What is the corresponding wavelength range? (Speed of sound in air at ordinary temperature is 340 m/s)

    Solution

    Wavelength of the wave is given by,
                            space space lambda equals nu over straight v
    Wavelength corresponding to frequency 20 Hz is, 
                            straight lambda subscript max equals 340 over 20 equals 17 straight m 
    and
    Wavelength corresponding to frequency 20KHz is, 
                         straight lambda subscript min equals fraction numerator 340 over denominator 20 comma 000 end fraction equals 0.017 straight m 
    Wave length range of audible sound is 0.017m to 17m.

    Question 479
    CBSEENPH11018751

    The velocity of longitudinal wave in water is nearly 1450 m/s at 20°C. What is the adiabatic compressibility of water?

    Solution

    Given,
    Velocity of longitudnal wave, v = 1450 m/s
    And the formula for velocity of a longitudanal wave is,
     
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    ∴          straight C subscript adi equals fraction numerator 1 over denominator straight v squared straight rho end fraction 
    The given values are,
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#6 {main}</pre> 
    ∴            straight C subscript adi equals fraction numerator 1 over denominator left parenthesis 1450 right parenthesis squared cross times 1000 end fraction 
                       equals 5 cross times 10 to the power of negative 10 end exponent straight m squared divided by straight N , is the adiabatic compressibiltiy of water. 

    Question 480
    CBSEENPH11018752

    A string of mass 2.50 kg is under the tension of 200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

    Solution

    Velocity of wave in stretched string is given by, 
                               straight v equals square root of straight T over straight m end root 
    where,
    T is the tension, and 
    m is the mass per unit length.
    If Error converting from MathML to accessible text. is length of string then time taken by wave to reach the other end of string is, 
    straight t equals ell over straight v equals ell square root of straight m over straight T end root 
    We have,
    ell equals 20 straight m comma space space space space space space space space space space space space space space straight T equals 200 straight N 
     straight m over ell equals fraction numerator 2.5 over denominator 20 end fraction equals 1 over 8 kg divided by straight m 
    Therefore,
    Time taken to reach the other end of the string is,
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

    Question 481
    CBSEENPH11018753

    How far does the sound travel in air when a tuning fork of frequency 288 makes 36 vibrations? The velocity of sound is 340m/s.

    Solution
    Velocity of sound in air, v = 340m/s
    Time taken to make 36 vibrations by tuning fork of frequency 288, 
                             straight t equals 36 over 288 equals 1 over 8 straight s 
    The distance travelled by sound in <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  is, 
      space space straight S equals νt equals 340 cross times 1 over 8 equals 42.5 straight m
    Question 482
    CBSEENPH11018754

    A stone dropped from top of the tower 300 m high splashes into water of pond near the base of tower. When is the splash heard at the top? Velocity of sound in air = 340m/s, g = 9.8 m/s2.

    Solution
    Let t be the time taken by stone to reach the surface of water in pond.
    When the stone strikes the surface of water, splash is produced.
    Let, t2 be the time taken by the splash to reach top of tower.
    The total time after which the splash is heard after dropping the stone is, 
                           t = t1 + t
    The stone is dropped from height 300m.
     The time taken by the stone to reach the surface of water can be calculated by using the thrid equation of motion,  
                           straight h equals ut subscript 1 plus 1 half at subscript 1 squared 
                     space space space 300 equals 1 half cross times 9.8 cross times straight t subscript 1 superscript 2
                           straight t subscript 1 equals square root of fraction numerator 2 cross times 300 over denominator 9.8 end fraction end root space equals space 7.825 straight s 
    The time taken by splash to reach the top of tower is,
                            space space space space straight t subscript 2 equals fraction numerator Distance space travelled space by space sound over denominator Velocity space of space sound end fraction 
         equals 300 over 340 equals 0.882 straight s  
    Therefore, total time taken to hear the splash at the top is, 
                              
    Total time straight t equals straight t subscript 1 plus straight t subscript 2 equals 7.825 plus 0.882 equals 8.707 straight s
    Question 483
    CBSEENPH11018755

    A steel wire has a length of 12.0m and mass 2.10kg. What should be the tension in wire so that the speed of transverse wave on the wire equals to the speed of sound in dry air at 20°C i.e. 343 m/s?

    Solution
    Velocity of wave in stretched string is given by, 
                       straight v equals square root of straight T over straight m end root equals square root of TL over straight M end root
    rightwards double arrow               straight T equals fraction numerator straight v squared straight M over denominator straight L end fraction 
    where,
     M is the mass and L is the length of wire 
    We have,
    Velocity of sound, straight v equals 343 space straight m divided by straight s comma space
    Mass of steel wire, M = 2.1kg 
    Length of the string, L = 12m 
    Therefore,
    Tension in the wire is,
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Question 484
    CBSEENPH11018756

    A bat emits ultrasonic sound of frequency 100 KHz in air. If this sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? [Speed of sound in air = 340 m/s Speed of sound in water = 1486 m/s]

    Solution

    (a) Frequency of wave, 
           straight v equals 100 space KHZ space equals space 10 to the power of 5 HZ 
    Velocity of reflected wave space space straight v equals 340 space straight m divided by straight s 
    ∴  Wavelength, straight lambda equals straight u over straight v equals 340 over 10 to the power of 5 
                              equals 3.4 cross times 10 to the power of negative 3 end exponent straight m 
                              equals 0.34 cm 
    (b) Velocity of wave in water  = 1486 m/s 
    Since the frequency of wave does not change on refraction, therefore wavelength of transmitted wave is,
                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                   space space space space equals 1.486 space cm 

    Question 485
    CBSEENPH11018757

    The velocity of sound in hydrogen gas at temperature 0°C is 1320m/s. Find the velocity of sound in a mixture consisting of two parts by volume of hydrogen and one part by volume of oxygen at same temperature.

    Solution

    Let ρ1 be the density of hydrogen and ρ2 be the density of oxygen at 0°C.
    The molecular weight of oxygen is 16 times as that of hydrogen.
    Therefore,
                        ρ2= 16 ρ
    Density of the mixture consisting of two parts by volume of hydrogen and one part by volume of oxygen is. 
    straight rho equals fraction numerator 2 straight rho subscript 1 plus 1 straight rho subscript 2 over denominator 3 end fraction equals fraction numerator 2 straight rho subscript 1 plus 16 straight rho subscript 1 over denominator 3 end fraction equals 6 straight rho subscript 1 
    As, velocity of longitudnal wave is given by,
                            straight v equals square root of 1 over straight rho end root 
    ∴                   straight v subscript straight m over straight v subscript straight h equals square root of straight rho subscript 1 over straight rho end root 
    rightwards double arrow                   straight v subscript straight m equals straight v subscript straight h square root of straight rho subscript 1 over straight rho end root equals 1320 square root of 1 over 6 end root 
                                equals 538.77 straight m divided by straight s 

    Question 486
    CBSEENPH11018758

    A wave of frequency 400Hz has a velocity 340m/s. What is the distance between two nearest points which are out of phase by 120°?

    Solution

    Given,
    Frequency of the wave, v = 400HZ
    Velocity of the wave, v = 340m/s 
    ∴ Distance between two nearest points which are out by phase 120o is given by the wavelength of the wave, 
                       lambda equals v over straight v equals 340 over 400 equals 0.85 straight m

    Question 487
    CBSEENPH11018759

    A wave vibrates up and down 16 times a second and travels a distance 4m in one second. Find the wavelength of wave?

    Solution
    Distance travelled by the wave in 1 sec = 4 m
    Therefore,
    Velocity of wave = 4m/s.
    Also the wave vibrates up and down 16 times a second, therefore the frequency of wave is 16Hz. 
    Now using the formula,
                             v equals vλ 
    ∴                        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Question 488
    CBSEENPH11018760

    A tuning fork of frequency 275 Hz produces sound of wavelength 1.2m in air at NTP. Calculate the increase in wavelength when temperature of air is 27.3°C.

    Solution

    Velocity of a wave is given by,
                        v equals vλ 
    ∴             space space v subscript 27.3 end subscript over v subscript o equals straight lambda subscript 27.3 end subscript over straight lambda subscript straight o                        ...(1) 
    Also,  
                                 v proportional to square root of straight T 
    ∴     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>        ...(2) 
    From (1) and (2), 
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  
    rightwards double arrow               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                               equals 1.259 straight m 
    Increase in wavelength is, 
               straight lambda subscript 27.3 end subscript minus straight lambda subscript straight o equals 1.259 minus 1.2 space equals space 0.059 straight m

    Question 489
    CBSEENPH11018761

    Compressional wave impulses are sent to the bottom of sea from the ship and the echo is heard after 1.8s. Find the depth of sea. Assume that the density of sea water is 1000kg/m3 and bulk modulus of water is 2.25x109 N/m2.

    Solution

    Given,
    Density comma space straight rho equals 1000 space k g divided by straight m cubed comma

E n e r g y comma space straight E space equals space 2.25 cross times 10 to the power of 9 space straight N divided by straight m squared 
    Therefore velocity of sound in sea water, 
    straight v equals square root of straight E over straight rho end root equals square root of fraction numerator 2.25 cross times 10 to the power of 9 over denominator 1000 end fraction end root equals 1500 space straight m divided by straight s 
    Let d be the depth of the sea.
    As the echo is heard after 1.8s, therefore the wave travels 2d distance in water in time 1.8s. 
    i.e.   2 straight d equals vt 
    rightwards double arrow     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Question 490
    CBSEENPH11018762

    Derive an analytical expression for a simple harmonic plane progressive wave.

    Solution
    Let a plane wave originate from O and progress in all the directions.
    All the particles of media will execute simple harmonic motion of same amplitude and time period about mean position when wave propagates in media.
    Let A be the amplitude, co be the frequency and λ be the wavelength of the wave. 
                          

    The displacement of particle O at any instant is given by y(0,t) = A sin ωt
    The disturbance is handed over from one particle to next and will reach point P at a distance x from O, a bit later.
    Therefore, phase of particle at P lags behind the phase of particle at O by, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    where,              fraction numerator 2 straight pi over denominator straight lambda end fraction equals straight k 
    Therefore the displacement of particle at P is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                            equals straight A space sin left parenthesis ωt minus kx right parenthesis 

    Therefore simple harmonic progressive wave propagating towards positive direction of x-axis is, 
    y(x,t) = A sin (ωt – kx).
    If wave propagates towards left direction then replacing x by -x, we get
    y(x,t) = A sin (ωt + kx).



    Question 491
    CBSEENPH11018763

    What is the difference between wave velocity and particle velocity?

    Solution

    Wave velocity:
    Wave velocity is the velocity with which progressive wave front travels forward. It is constant throughout the media if the media is homogenous and given by,
                          v = v λ.
    Particle velocity:
    Particle velocity is the velocity with which particles vibrate during the wave propagation. The particle velocity is changing and given by, 
                   v = rω cos(ωt – kx).

    Question 492
    CBSEENPH11018764

    Define particle and wave velocity and find the relation between them.

    Solution
    Particle velocity is the velocity of particle of media at any instant as a progressive wave travels through media.
    Wave velocity is the velocity with which a progressive wave travels in media.
    Displacement of particle at any instant t is, 
                      straight y equals Asin left parenthesis ωt minus kx right parenthesis                     ...(1) 

    ∴          straight v subscript straight p equals dy over dt equals Aω space cos left parenthesis ωt minus kx right parenthesis             ...(2) 
    and       space space dy over dx equals negative Akcos left parenthesis ωt minus kx right parenthesis                ...(3) 
    From (2) and (3), 
                        straight v subscript straight p equals negative straight omega over straight k dy over dx 
    But straight omega over straight k is wave velocity. 
    So, the relation between particle and wave velocity is given by,
                            straight v subscript straight p equals negative straight v subscript straight omega dy over dx
    Question 493
    CBSEENPH11018765

    Derive the differential equation of travelling wave equation.

    Solution
    The general equation of simple harmonic progressive wave is, 
               straight y left parenthesis straight x comma space straight t right parenthesis space equals space straight A space sin left parenthesis ωt minus kx right parenthesis            ...(1) 
    Differentiating (1) with respect to time, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>         ...(2) 
    Differentiating (2) with respect to time, 
               space space fraction numerator straight d squared straight y over denominator dt squared end fraction equals negative Aω squared space sin left parenthesis ωt minus kx right parenthesis 
                       equals negative straight omega squared straight y                       ...(3) 
    Differentiating (1) with respect to x, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>     ...(4) 
    Differentiating (4) with respect to x, 
                  fraction numerator straight d squared straight y over denominator dx squared end fraction equals negative Ak squared sin left parenthesis ωt minus kx right parenthesis space space space space space space space space space 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>                       ...(5) 
    From (3) and (4), we get  
                    fraction numerator straight d squared straight y over denominator dt squared end fraction equals straight omega squared over straight k squared fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight v squared fraction numerator straight d squared straight y over denominator dx squared end fraction  
    rightwards double arrow              fraction numerator straight d squared straight y over denominator dt squared end fraction equals straight v squared fraction numerator straight d squared straight y over denominator dx squared end fraction                  ...(6) 
    This is the required differential equation of a travelling wave. 
    Question 494
    CBSEENPH11018766

    A transverse harmonic wave on a string is described by
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
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#6 {main}</pre>
    where x, y are in cm and t in s. The positive direction of x is from left to right.
    (a) Is this a travelling or a stationary wave? If this is travelling, then what is the speed and direction of its propagation?
    (b) What is its amplitude and frequency?
    (c) What is the initial phase at the origin?
    (d) What is the least distance between two successive crests in the wave?

    Solution
    The harmonic wave equation is given by, 
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#6 {main}</pre>       ...(1) 
    The general harmonic wave equation is, 
           straight y left parenthesis straight x comma space straight t right parenthesis space equals space straight A space sin left parenthesis ωt minus kx plus straight ϕ right parenthesis                    ...(2) 
    Comparing (1) and (2), we get 
    Amplitude space of space the space wave comma space straight A equals 3.0 cm comma space

Angular space frequency comma space straight omega equals 36 space rad divided by straight s

Wave space vector comma space straight k space equals space 0.018 space cm to the power of negative 1 end exponent 
    (a) The given wave is a travelling wave.
    The speed of wave is, 
        space space space straight v equals straight omega over straight k equals fraction numerator 36 over denominator 0.018 end fraction equals 20000 cm divided by straight s equals 20 straight m divided by straight s 
    (b) Amplitude, A = 3.0cm 
    Frequency, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre> 
    (c) Initial phase, straight ϕ equals straight pi divided by 4 
    (d) The least distance between two successive crests in the wave is equal to wavelength. 
               straight lambda equals fraction numerator 2 straight pi over denominator straight k end fraction equals fraction numerator 22 over denominator 7 cross times 0.018 end fraction cm almost equal to 3.5 straight m
    Question 495
    CBSEENPH11018767

    The equation of a travelling plane sound wave is y = 60 sin (456t – 1.2x), where y is in mm and x is in meter. Find the ratio of displacement amplitude with which the particles of media vibrate to the wavelength. Also verify that the maximum particle velocity of vibrating particle of media is equal to velocity of wave times the maximum deformation of media.

    Solution

    The given travelling wave is, 
                   straight y equals 55 space sin space left parenthesis 456 straight t minus 1.2 straight x right parenthesis 
    On comparing with general wave equation 
    space straight y equals straight A space sin left parenthesis ωt minus kx right parenthesis comma 
    We get,
     Amplitude comma space straight A equals 55 cm equals 0.55 straight m comma space

Angular space velocity comma space straight omega equals 456 space straight s to the power of negative 1 end exponent

Wave space vector comma space straight k space equals space 1.2 space straight m to the power of negative 1 end exponent
    Now,  
        straight A over straight lambda equals fraction numerator AK over denominator 2 straight pi end fraction equals fraction numerator 0.55 cross times 1.2 over denominator 2 straight pi end fraction equals 0.105 
    Maximum particle velocity of media is, 
                   left parenthesis straight v subscript straight p right parenthesis subscript max equals Aω equals 0.55 cross times 456 space

space space space space space space space space space space space space space equals space 250.8 space straight m divided by straight s            ...(1)
    The velocity of wave is, 
                         straight v subscript straight w equals straight omega over straight k equals fraction numerator 456 over denominator 1.2 end fraction equals 380 space straight m divided by straight s    ...(2) 
    Maximum deformation in media, 
                   open parentheses dy over dx close parentheses subscript max equals Ak equals 0.55 cross times 1.2 space equals space 0.66 
    Now,    
     straight v subscript straight w open parentheses dy over dx close parentheses subscript max equals 380 cross times 0.66 equals 250.8 straight m divided by straight s       ...(3)
    From (1) and (3),
                   space space left parenthesis straight v subscript straight p right parenthesis subscript max equals straight v subscript straight w open parentheses dy over dx close parentheses subscript max

space space space space space space space space space space space space space space equals 250.8 straight m divided by straight s 

    Question 496
    CBSEENPH11018768

    For a travelling harmonic wave y(x,t) = 2.0 cos [2straight pi (10t – 0.008x + 0.35)] what is the phase difference between oscillatory motion at two points separated by a distance of
    (i) 4m (ii) 0.5m (iii) λ/2?
    Here x and y are in cm and t in seconds?

    Solution
    The given travelling wave is, 
          space straight y left parenthesis straight x comma space straight t right parenthesis space equals space 2.0 space cos open square brackets 2 straight pi left parenthesis 10 straight t minus 0.008 straight x plus 0.35 right parenthesis close square brackets space

space space space space space space space space space space space space equals 2.0 space cos open square brackets 20 πt minus 0.016 πx plus 0.7 straight pi close square brackets  
    On comparing with general wave equation, 
    straight y left parenthesis straight x comma space straight t right parenthesis space space equals space straight A space sin left parenthesis ωt minus kx plus straight ϕ right parenthesis comma

we space get comma 
            straight k equals 0.016 straight pi space cm to the power of negative 1 end exponent space equals space 1.6 πm to the power of negative 1 end exponent 
    We know the phase different between two points separated by distance space Δx is, 
                 increment straight ϕ equals straight k increment straight x 
    (i) Distance of seperation, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
                  increment straight ϕ space equals space left parenthesis 1.6 straight pi right parenthesis 4 space equals space 6.4 straight pi space rad 
    (ii) Distance of seperation, increment straight x space equals space 0.5 straight m         

                  increment straight x equals straight lambda divided by 2 
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#6 {main}</pre> 
    Question 497
    CBSEENPH11018769

    A sound wave of frequency 512hz and amplitude 2.4 x 10–3m is going through air of density 1.293kg/m3. Find the intensity of wave. Velocity of sound in air is 330m/s.

    Solution

    Given that,
    Frequency of the sound wave = 512 Hz
    Amplitude of the sound wave = 2.4 cross times 10-3
    Density of air = 1.293 kg/m3 
    Intensity of the wave is, 
                     Error converting from MathML to accessible text.
    nu italic space italic equals italic space italic 512 italic space H z
rho italic space italic equals italic space italic 1 italic. italic 293 italic space k g italic divided by m to the power of italic 3 italic space
v italic space italic equals italic space italic 330 italic space m italic divided by s
A italic space italic equals italic space italic 2 italic. italic 4 italic space italic cross times italic 10 to the power of italic minus italic 3 end exponent m 
    Therefore, 
    Intensity, I = 2 space straight pi squared space cross times space 1.293 space cross times space 330 space cross times space left parenthesis 512 right parenthesis squared space cross times space left parenthesis 2.4 space cross times space 10 to the power of negative space 3 end exponent right parenthesis squared space 
    I = 1.2728 space cross times space 10 to the power of 4 space straight J divided by straight m squared space straight s 

    Question 498
    CBSEENPH11018770

    The displacement of particle in wave equation is represented as <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> . If this wave is made to incident on the rigid wall and its amplitude falls to 40% on reflection from wall, then what is the equation of reflected wave?

    Solution
    It is given that amplitude of reflected wave reduces to 40%.
    Therefore, amplitude of reflected wave is, 
    straight A subscript ref equals fraction numerator 1.2 cross times 40 over denominator 100 end fraction equals 0.48 
    We know that on reflection, the direction of propagation of wave reverses and on reflection from rigid wall, the phase changes by space straight pi. 
    Therefore equation of reflected wave is, 
    straight y equals 0.48 space sin open square brackets 120 πt minus 0.20 left parenthesis negative straight x right parenthesis plus straight pi close square brackets 
      equals 0.48 space sin open square brackets 120 πt plus 0.20 straight x plus straight pi close square brackets 
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#6 {main}</pre>
    Question 499
    CBSEENPH11018771

    The transverse displacement of a string (clamped at its two ends) is given by
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#6 {main}</pre>
    where x,y are in meter and t is in second. The length of the string is 1.5 m and its mass is 3.0 x 10–2kg. Answer the following: 

    (a) What are the wavelength, frequency and speed of propagation of each wave? 

    (b) Determine tension in the string.

    Solution
    (a) The transverse displacement of a string is given by, 
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#6 {main}</pre>       ...(1) 
    The general wave equation of wave in the string clamped between two points is given by, 
    space space straight y left parenthesis straight x comma space straight t right parenthesis space equals space 2 rsin left parenthesis kx right parenthesis space cos left parenthesis ωt right parenthesis                     ...(2) 
    where,
    r is the amplitude of each wave superimposed. 
    On comparing (1) and (2), we get, 
    space space space space space space space space space space space space space 2 straight r equals 0.06 space space

rightwards double arrow space space space space space space space space space space straight r space equals space 0.03

Therefore comma space
Wave space vector comma space straight k equals fraction numerator 2 straight pi over denominator straight lambda end fraction equals fraction numerator 2 straight pi over denominator 3 end fraction space

space rightwards double arrow space space space space space space space space space space space space space space space space space space space straight lambda equals 3 straight m 
    Angular frequency, straight omega equals 120 straight pi space rad divided by straight s space space space space
     
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#6 {main}</pre> 
    We know, v = space v straight lambda 
    Therefore,

    Speed of wave, straight v equals 60 cross times 3 equals 180 space straight m divided by straight s
    (b) We know speed of wave in the stretched string is given by, 
                               straight v equals square root of straight T over straight m end root 
     
    ∴                         straight T equals mv squared 
    where,
     m is mass per unit length of wire. 
    It is given that mass of string is 3.0 x 10–2kg and length of string is 1.5 m. 
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#6 {main}</pre> 
    So,
    Tension in the string is,
    straight T equals mv squared equals 2 cross times 10 to the power of negative 2 end exponent cross times left parenthesis 180 right parenthesis squared equals 648 straight N 
    Question 500
    CBSEENPH11018772

    What is a string?

    Solution
    String is a perfectly flexible chord having uniform area of cross section whose length is much larger than its diameter.
    Question 501
    CBSEENPH11018773

    When the two waves <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> superpose, then what is the resultant wave?

    Solution

    According to the principle of superposition of waves, 
                                stack straight y subscript 1 with rightwards arrow on top equals stack straight y subscript 1 with rightwards arrow on top plus stack straight y subscript 2 with rightwards arrow on top

    Question 502
    CBSEENPH11018774

    What are stationary waves?

    Solution
    When two waves of same nature, either longitudnal or transverse, having the same amplitude and frequency, travelling in opposite direction with the same speed are superimposed, then stationary waves are produced.
    Question 503
    CBSEENPH11018775

    Why stationary wave is so called?

    Solution
    In stationary waves there is no propagation of disturbance from one place to another.
    That is, the rate of propagation of energy is zero. 
    Hence, these waves are known as stationary waves. 
    Question 504
    CBSEENPH11018776
    Question 506
    CBSEENPH11018778

    A string stretched between two points is plucked. What is the fundamental frequency of vibration of string?

    Solution

    The fundamental frequency of the vibration of string is, 
                        straight v subscript straight o equals fraction numerator straight v over denominator 2 straight L end fraction equals fraction numerator 1 over denominator 2 straight L end fraction square root of straight T over straight m end root

    Question 507
    CBSEENPH11018779

    Can the string stretched between two rigid supports be made to vibrate in even harmonic?

    Solution
    Yes. When the strings are stretched, there is a transfer of energy from the string, the strings vibrate in even harmonic. 
    Question 508
    CBSEENPH11018780

    What is the frequency of vibrations of a string stretched between two points vibrating in nth mode?

    Solution

    Frequency of vibrations of a string between two points vibrating in the mode, 
                         straight v equals fraction numerator straight n over denominator 2 straight L end fraction square root of straight T over straight m end root

    Question 509
    CBSEENPH11018781

    A string stretched between two points when plucked vibrates in nth overtone. What is the frequency of vibration of string?

    Solution

    The frequency of the vibration of the string is given by, 
                        straight v equals fraction numerator straight n plus 1 over denominator 2 straight L end fraction square root of straight T over straight m end root

    Question 510
    CBSEENPH11018782

    A string stretched between two points is plucked at a distance L/6 from one end. What is the frequency of vibration of string? What is the overtone of vibration?

    Solution

    The frequency of the vibration of string is given by, 
                       straight v equals fraction numerator 3 over denominator 2 straight L end fraction square root of straight T over straight m end root semicolon 
    The overtone of vibration is second. 

    Question 511
    CBSEENPH11018783

    A string is vibrating in nth mode. What is its overtone?

    Solution
    Overtone for a string vibrating in the nth mode is,
                                (n–1)th.
    Question 512
    CBSEENPH11018784
    Question 513
    CBSEENPH11018785

    What are nodes?

    Solution
    The points in the stationary waves where the particles do not vibrate and strain is maximum are called nodes. 
    Question 514
    CBSEENPH11018786

    What are antinodes?

    Solution
    The points where the amplitude of vibration is maximum and strain is minimum are called as antinodes. 
    Strain in the points of antinodes is zero. 
    Question 515
    CBSEENPH11018787
    Question 516
    CBSEENPH11018788

    What is the distance between two consecutive nodes or antinodes?

    Solution
    The distance between two consecutive nodes and anti-nodes is X/2.
    Question 517
    CBSEENPH11018789

    What is the distance between node and next antinode?

    Solution
    Distance between node and next anti-node is λ/4.
    Question 518
    CBSEENPH11018790

    Is the frequency of vibration of string same for different modes of vibrations?

    Solution
    No, for different modes of vibrations, the frequency of vibration of string is different. 
    Question 519
    CBSEENPH11018791

    What is the distance between two consecutive nodes?

    Solution
    The distance between two consecutive nodes is given by, 
    straight L over straight n equals straight lambda over 2 
    Question 520
    CBSEENPH11018792

    What is organ pipe?

    Solution
    Organ pipe is a hollow pipe in which sound is produced by setting up stationary waves in its air column.
    Question 521
    CBSEENPH11018793

    What is the fundamental frequency of stationary waves produced in one end closed organ pipe?

    Solution

    The fundamental frequency of stationary waves produced in one end closed organ pipe is given by, 
                    space space space straight v subscript straight o equals left parenthesis 2 straight n minus 1 right parenthesis fraction numerator straight v over denominator 4 straight L end fraction
space space space space space space equals left parenthesis 2 straight n minus 1 right parenthesis fraction numerator 1 over denominator 4 straight L end fraction square root of γP over straight rho end root

    Question 522
    CBSEENPH11018794

    What is the fundamental frequency of stationary waves produced in both end open organ pipe?

    Solution

    The fundamental frequency of stationary waves produced in both end open organic pi[e is given by, 
    space space space space space space space space space space straight v subscript straight o equals straight n fraction numerator straight v over denominator 2 straight L end fraction

where comma space

straight L space is space the space length space of space the space organ space pipe comma space
straight nu subscript straight o space end subscript is space the space fundamental space frequency.

    Question 523
    CBSEENPH11018795

    What is the position of node and antinode in one end closed pipe in its fundamental mode of vibrations?

    Solution
    Node is formed at the closed end of pipe and antinode at the open end of pipe.
    Question 524
    CBSEENPH11018796

    How does the pressure change at nodes and anti-nodes?

    Solution
    At nodes, the change in pressure is maximum and that at antinodes, the pressure is zero.
    Question 525
    CBSEENPH11018797

    Why are the strings of different thickness and materials used in a sitar or violin?

    Solution
    The strings of different thickness and materials are used in a sitar or violin to produce different fundamental frequencies. This is done so as to distinguish between different tones and pitch. 
    Question 526
    CBSEENPH11018798

    What types of waves are produced in the sonometer wire?

    Solution
    Stationary transverse waves are produced in the sonometer wire.
    Question 527
    CBSEENPH11018799

    What is the frequency of stationary transverse waves produced in the sonometer wire?

    Solution

    Frequency of stationary transverse waves produced in the sonometer wire is given by,  
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    Question 528
    CBSEENPH11018800

    Does the frequency of sonometer string remain the same for different mode of vibrations?

    Solution
    With increase in mode of vibrations, the frequency of sonometer string increases. 
    Question 529
    CBSEENPH11018801

    If the length of the wire stretched between two rigid supports is doubled, then how will the fundamental frequency change?

    Solution
    Ig the length of the wire stretched between two rigid support is doubled, the fundamental frequency will decrease to half. 
    Question 530
    CBSEENPH11018802

    If we want to keep the frequency of the string same by doubling the length of the wire, what should we do?

    Solution
    If the material of the wire is the same, then increase the tension to four times. In this case the frequency will remain same even when the length of the wire is doubled. 
    Question 531
    CBSEENPH11018803

    A string fixed at rigid supports is vibrating in its fundamental mode. How many nodes and antinodes are formed in the string?

    Solution
    When a string which is fixed at rigid support is vibrating in it's fundamental mode, two nodes and one antinode is formed in the string.
    Question 532
    CBSEENPH11018804

    Where are the nodes and antinodes formed in the vibrating wire?

    Solution
    The nodes are formed at the positions occupied by the wedges while antinode is formed in the middle.
    Question 533
    CBSEENPH11018805

    What is a fundamental note?

    Solution
    Fundamental note is the note of lowest frequency. 
    Question 534
    CBSEENPH11018806

    What are overtones?

    Solution
    The frequency of vibration string which becomes higher than the fundamental frequency or fundamnetal note are called overtones. 
    Question 535
    CBSEENPH11018807

    Can a standing wave be produced on a string by superposing two waves with same frequency but different amplitude travelling in opposite direction?

    Solution
    Yes, by superposing two waves of same frequency with differnt amplitude, it is possible to produce standing waves. 
    Question 536
    CBSEENPH11018808

    What is resonance?

    Solution
    Resonance is the phenomenon of vibratory motion in which a vibrator is set into motion with its natural frequency by another body vibrating with the same frequency.
    Question 537
    CBSEENPH11018809

    What is the amplitude of vibration at resonance?

    Solution
    At resonance, the amplitude of vibration is maximum. 
    Question 538
    CBSEENPH11018810

    Why does the rider fall from the string when sonometer wire is in resonance with tuning fork?

    Solution
    At resonance, the amplitude of vibration of string is maximum and rider placed over it loses its contact from string and falls down. 
    Question 539
    CBSEENPH11018811

    What type of vibrations are produced in resonance tube?

    Solution
    Stationary longitudinal waves are produced in resonance tube. 
    Question 540
    CBSEENPH11018812

    What is the frequency of vibration of air column in resonance pipe?

    Solution

    The frequency of vibration of air column in resonance pipe is, 
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    Question 541
    CBSEENPH11018813

    Is the antinode situated exactly at the open end?

    Solution
    No, antinode is situated slightly above the open end. 
    Question 542
    CBSEENPH11018814

    What is the relation between end correction and diameter of pipe?

    Solution
    The end correction is 0.3 times the internal diameter of pipe.
    Question 543
    CBSEENPH11018815

    What is the ratio of the frequency of vibration of air column at first resonance length to second resonance length?

    Solution
    The ratio of the frequency of vibration of air column at first resonance length to second resonance length is one.
    Question 545
    CBSEENPH11018817

    Is first resonance length equal to λ /4, where straight lambda is the wavelength of stationary longitudinal wave produced in the air column? 

    Solution
    Given, 
    Resonance length = straight lambda over 4 
    where, straight lambda is the wavelength of the stationary longitudnal wave. 
    So, first resonance length produced in the air column is less than λ/4. 
    Question 546
    CBSEENPH11018818

    What is the difference in second and first resonance length?

    Solution
    The difference in second and first resonance length is λ/2.
    Question 547
    CBSEENPH11018819

    If in resonance tube, water is replaced by movable piston then how does the frequency change?

    Solution
    The frequency for same length of air column does not depend on the material of closing end of the tube. So, the frequency does not change. 
    Question 548
    CBSEENPH11018820

    What is a tuning fork?

    Solution
    Tuning fork is a U-shaped bar of metal with a stem at the middle.
    Tuning fork is a simple source of sound.
    Question 549
    CBSEENPH11018821

    Why tuning fork is so called?

    Solution
    Tuning fork produces musical sound when striked and possesses the shape of a fork. 
    Therefore, it has been named as tuning fork. 
    Question 550
    CBSEENPH11018822

    On what factors does the frequency of tuning fork depend?

    Solution
    The frequency of tuning fork depends upon:
    i) length,
    ii) thickness, and
    iii) material of fork.
    Question 551
    CBSEENPH11018823

    On what factors the velocity of sound in material of tuning fork depend?

    Solution
    The velocity of sound in material of tuning fork depends upon:
    i) Density, and
    ii) Young's modulus of elasticity of the material of the fork. 

    Question 552
    CBSEENPH11018824

    How the frequency of tuning fork can be changed?

    Solution
    The frequency of tuning fork can be increased by filing the ends of the prongs and can be decreased by loading the ends of prongs. 
    Question 553
    CBSEENPH11018825

    Why do the vibrations of tuning fork die away after some time?

    Solution
    The vibrations of tuning fork die away after some time due to loss of energy in overcoming the frictional forces and radiating sound.
    Question 554
    CBSEENPH11018826

    What happens when a vibrating tuning fork is held near the car and rotated about its stem?

    Solution
    When a vibrating tuning fork is held near the car, the intensity of sound varies due to the interference of sound from prongs. 
    Question 555
    CBSEENPH11018827

    Does the frequency of tuning fork change with temperature?

    Solution
    When temperature of the tuning fork increases, the length of prongs increase and Young's modulus decreases.
    This results in a slight decrease in frequency. 
    Question 556
    CBSEENPH11018828

    Why do we strike the tuning fork gently on rubber pad?

    Solution
    On striking the tuning fork gently, it produces only fundamental tone and when it is struck hard, higher overtones are produced. 
    So, we strike the tuning fork gently on a rubber pad. 
    Question 557
    CBSEENPH11018829

    Why have tuning fork two prongs?

    Solution
    If there is only one prong, then the vibration will die out as soon as the stem is touched.
    Hence, tuning fork has two prongs.  
    Question 558
    CBSEENPH11018830

    What is the function of a tuning fork?

    Solution
    The purpose of tuning fork is to produce monochromatic wave. 
    Question 559
    CBSEENPH11018831

    What is frequency of a tuning fork?

    Solution
    The number of vibrations made by either of the prong in one second is called as the frequency of a tuning fork. 
    Question 560
    CBSEENPH11018832

    Which part of the tuning fork is set into vibrations when its prong is struck against the pad-prongs or stem?

    Solution
    Both prongs as well as its stem are set into vibrations. 
    Question 561
    CBSEENPH11018833

    What types of vibrations are set up in the prongs of tuning fork?

    Solution
    Transverse vibrations are set up in the prongs of tuning fork.
    Question 562
    CBSEENPH11018834

    What types of vibrations are set up in the stem of tuning fork?

    Solution
    In the stem of tuning fork, longitudnal vibrations are set up. 
    Question 563
    CBSEENPH11018835

    Are the frequencies of the vibrations of stem and prongs different?

    Solution
    No, the frequencies of longitudinal vibrations in stem and transverse vibration in prongs are same.
    Question 564
    CBSEENPH11018836

    Is there any effect of temperature on the frequency of tuning fork?

    Solution
    Yes, the frequency of fork decreases, with increase in temperature. 
    Question 565
    CBSEENPH11018837

    How does the frequency of vibration of the prong of a tuning fork change when it is loaded with wax?

    Solution
    The frequency of vibration decreases on loading with wax.
    Question 566
    CBSEENPH11018838

    If the prong of a tuning fork is filled, will its frequency change?

    Solution
    Yes, if the prong of a tuning fork is filled, its frequency will increase.
    Question 567
    CBSEENPH11018839

    How do both the prongs vibrate, though we strike only one?

    Solution
    The energy is transmitted from one prong to another through the material of the fork. 
    Therefore, both prongs begin to vibrate. 
    Question 568
    CBSEENPH11018840

    Locate the position of nodes and antinodes in a vibrating tuning fork.

    Solution
    The free ends and the middle of bend are antinodes.
    Nodes lie very close to bend on either side of it.
                                
    In the given fig.,
    A, B and C are antinodes, and
    D and E are nodes. 
    Question 569
    CBSEENPH11018841

    Why is sound so loud when we keep the stem of the fork pressed against the board?

    Solution
    When the stem is pressed against the board, then the particles of the board are set into forced vibrations.
    So, the area of vibrating surface increases. Hence, the intensity of sound increases. 
    Question 570
    CBSEENPH11018842

    When two notes of frequencies v1 and v2 are sounded together, what are the number of beats produced in one second?

    Solution
    Number of beats produced in one second is given by,
                              | v1 – v2|
    Question 571
    CBSEENPH11018843

    Is it necessary for beat production that amplitude of two waves should be exactly same?

    Solution
    For production of beats, it is not necessary that amplitude of two waves should be equal.
    Question 572
    CBSEENPH11018844

    What can be the maximum frequency difference of two nodes to hear the beats?

    Solution
    The maximum frequency difference of two nodes to hear the beats distincyly is 10 Hz.
    Question 573
    CBSEENPH11018845

    What is the frequency of beats produced by sounding the two tuning forks of frequencies 204 and 208.5 Hz?

    Solution
    Beat frequency is given by, 
    b = 208.5 – 204
       = 4.5 Hz. 
    Question 574
    CBSEENPH11018846

    A note produces 6 beats per second with a tuning fork of frequency 256Hz. Find the frequency of the tuning fork. What may be the frequency of note?

    Solution
    Let the frequency of the tuning fork = n
    It produces 6 beats per second with tuning fork of frequency 265 Hz
    n = 256 plus-or-minus  6 = 250 or 262 Hz
    250 or 262 Hz is the frequency of the note. 
    Question 577
    CBSEENPH11018849

    What is interference?

    Solution
    Interference is the superposition of two waves of same frequency travelling in same direction with same speed. 
    Question 578
    CBSEENPH11018850

    Is the phenomenon of interference characteristic of all wave motion?

    Solution
    Yes, both transversal and longitudinal waves can undergo interference. 
    Question 579
    CBSEENPH11018851

    Does interference phenomenon reveal the nature of wave?

    Solution
    No, interference does not reveal the true nature of waves, because interference can be obtained from both transversal as well as longitudinal waves.
    Question 580
    CBSEENPH11018852

    Does the interference violate the law of conservation of energy?

    Solution
    No. Interference is in accordance with the law of conservation of energy. In interference, energy that disappears at position of destructive interference appears at the position of constructive interference.
    Question 581
    CBSEENPH11018853

    In interference, what happens to the energy that disappears at position of destructive interference?

    Solution
    During the process of interference, energy that disappears at position of destructive interference appears at the position of constructive interference.
    Question 582
    CBSEENPH11018854

    State the most essential condition for observing interference.

    Solution
    The most essential conditions for observing interference are:
    i) Two sources of sound must be coherent, and
    ii) The sources should emit monochromatic waves. 
    Question 583
    CBSEENPH11018855

    When the two sources of sound are said to be coherent sources?

    Solution
    When the initial phase difference between two sounds coming from two sources is constant, two sources of sound are said to be coherent. 
    Question 584
    CBSEENPH11018856

    Why it is not possible to have interference between the waves produced by two violins?

    Solution
    Two violins are not coherent sources of sound. That is why the process of interference is not possible in violins. 
    Question 585
    CBSEENPH11018857

    When two waves interfere, does one wave alter the progress of the other?

    Solution
    In interference, no wave alters the progress of other wave.
    Question 586
    CBSEENPH11018858

    Two waves of amplitude a and b are superposed with phase difference ϕ. What is the amplitude of resultant wave?

    Solution
    The amplitude of resultant wave is given by, 
    space space square root of straight a squared plus straight b squared plus 2 abcosϕ end root 
    where, 
    a and b are amplitudes of two waves, and 
    straight phi spaceis the phase differnce. 
    Question 587
    CBSEENPH11018859

    Two waves, each of amplitude a, are superposed with phase difference 90°. What is the amplitude of resultant wave?

    Solution

    Amplitude of two waves = a
    Phase difference, straight phi space equals space 90 to the power of 0

    Therefore, 
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    Question 588
    CBSEENPH11018860

    Two waves of intensity I1, and I2 are superposed with phase difference ϕ. What is the intensity of resultant wave?

    Solution

    Resultant Intensity of the wave, I = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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    where, 
    straight phi is the phase differnce. 

    Question 589
    CBSEENPH11018861

    Two waves of intensity I1 and I2superpose in phase. What is the intensity of resultant wave?

    Solution

    Resultant Intensity of the waves which are in phase is given by, 
    I = straight I subscript 1 plus straight I subscript 2 plus 2 square root of straight I subscript 1 straight I subscript 2 end root 

    Question 590
    CBSEENPH11018862

    What is the relation between path difference and wavelength for the constructive interference between two waves?

    Solution
    Relation between path differnce and wavelength for constructive interference is given by,
                                    x = n λ 
    where, 
    x = path diffrence. 
    Question 591
    CBSEENPH11018863

    What is the phase difference between two waves for constructive interference?

    Solution

    Phase differnce between two waves for constructive interference isspace 2 nπ

    Question 592
    CBSEENPH11018864

    What is the relation between path difference and wavelength for the destructive interference between two waves?

    Solution

    The relation between path difference and wavelength for the destructive interference between two waves is,
                               straight x equals fraction numerator 2 straight n plus 1 over denominator 2 end fraction straight lambda 
    where, 
    x is the path differnce, and 
    straight lambda is the wavelength. 

    Question 593
    CBSEENPH11018865

    What is the phase difference between two waves for destructive interference?

    Solution

    The phase difference between two waves for destructive interference is given by, 
                               space ϕ equals space left parenthesis 2 straight n plus 1 right parenthesis straight pi

    Question 594
    CBSEENPH11018866

    Crest of one wave of amplitude a falls on the trough of second wave of amplitude b. What is the amplitude of resultant wave?

    Solution
    Amplitude of the crest of the wave = a
    Amplitude of trough of second wave = b
    Therefore, the amplitude of the resultant wave = a – b
    Question 595
    CBSEENPH11018867

    Crest of one wave of amplitude a falls on the crest of second wave of amplitude b. What is the amplitude of resultant wave?

    Solution

    Amplitude of the crest = a
    Amplitude of trough = b
    Amplitude of the resultant wave = a + b

    Question 596
    CBSEENPH11018868

    Two coherent waves of amplitude 3mm and 4mm superimpose with a phase difference of 90°. What is the amplitude of resultant wave?

    Solution
    Amplitude of first wave = 3mm
    Amplitude of second wave = 4 mm
    Resultant amplitude is given by, R = square root of straight a squared plus straight b squared plus 2 ab space cosφ end root space
    That is, 
    R = square root of 3 squared plus 4 squared end root space equals space square root of 25 space equals space 5 space m m space
    Question 597
    CBSEENPH11018869

    Two coherent waves of intensity I1 and I2 interfere. What is the range of variation of intensity on screen?

    Solution

    The range of variation of intensity on the screen is, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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    Question 598
    CBSEENPH11018870

    Two coherent waves of intensities I1 and I2 interfere. What is the ratio of maximum to minimum intensity in interference pattern?

    Solution

    Ratio of maximum to minimum intensity in interference pattern is given by, 
    straight I subscript max over straight I subscript min space equals space open parentheses square root of straight I subscript 1 end root plus square root of straight I subscript 2 end root close parentheses squared over open parentheses square root of straight I subscript 1 end root minus square root of straight I subscript 2 end root close parentheses squared 

    Question 599
    CBSEENPH11018871

    What do you mean by superposition?

    Solution
    Superposition of waves is the process in which two or more different waves travelling through a media simultaneously overlap one another without losing their individual nature and shape. 
    Question 600
    CBSEENPH11018872

    State principle of superposition. What are different phenomena guided by superposition of waves?

    Solution
    Principle of superposition: When two or more waves of the same nature travel past a point simultaneously, the resultant displacement at the point is vector sum of the instantaneous displacements due to individual wave. 
    If stack straight y subscript 1 with rightwards arrow on top comma space stack straight y subscript 2 with rightwards arrow on top space..... are the displacement vectors due to individual waves at a point at a certain instant, then the resultant displacements due to individual wave is given by, 
                       straight y with rightwards arrow on top equals stack straight y subscript 1 with rightwards arrow on top plus stack straight y subscript 2 with rightwards arrow on top...... 
    Some important cases of superposition of waves are:
    (i) Stationary waves,
    (ii) Beats,
    (iii) Interference
    Question 601
    CBSEENPH11018873

    Discuss analytically, the different modes of vibration in the stretched string tied between two rigid supports.

    Solution
    Consider a string of length L which is stretched between two points A and B.
    Let T be the tension in the string and 'v' be the velocity of wave in the string when plucked. 
                     
    When the string is plucked, a progressive transverse wave be sent along the string toward right.
    On reflection from B, it will travel along the string toward left.
    Both incident and reflected waves superimpose in the string and give rise to stationary wave. 
    Incident and reflected wave are represented by,
    space space space space space straight y subscript 1 equals rsin left parenthesis ωt minus kx right parenthesis              ...(1) 
        straight y subscript 2 equals negative rsin left parenthesis ωt plus kx right parenthesis          ...(2) 
    On superposition, 
                  straight y equals straight y subscript 1 plus straight y subscript 2 
                    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow      straight y equals negative 2 rcosωt space sinkx    ...(3) 
    which is the resultant stationary wave. 
    The string is rigidly tied at A and B, therefore the boundary conditions are, 
    At A:            at x = 0;        y = 0 
    At B:           at x = L;         y = 0 '
    From equation (3), it follows that 
               straight y equals negative 2 rcosωt space sink left parenthesis 0 right parenthesis space equals space 0 
    The solution is trivial. 
                     straight y equals negative 2 rcosωt space sinkL space equals space 0 
    rightwards double arrow              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow           kL space equals space nπ, when n is an integer. 
    rightwards double arrow           fraction numerator 2 straight pi over denominator straight lambda end fraction equals nπ over straight L         
    rightwards double arrow               straight lambda equals fraction numerator 2 straight L over denominator straight n end fraction 
    ∴           space space space straight v equals v over lambda equals straight n fraction numerator v over denominator 2 straight L end fraction 
    Therefore,
    1st mode of vibration i.e. n = 1 

                straight lambda equals 2 straight L    and        straight v equals fraction numerator v over denominator 2 straight L end fraction 
    This is known as fundamental mode of vibration. 
                       

    Here, the point of minimum amplitude of vibration i.e., A and B are known as nodes. Point C is the point of maximum amplitude of vibration and is known as anti-node.
    In first mode of vibration, the string vibrates in one segment.
    2nd mode of vibration i.e. n = 2 
        straight lambda equals straight L       and         space space space straight v italic equals italic 2 fraction numerator v over denominator italic 2 L end fraction 
    Frequency of second mode of vibration is two times the fundamental frequency, and hence is known as second harmonic.
    In second harmonic, the string vibrates in two segments. 
    Similarly, in nth mode of vibration, the string vibrates in n segments with frequency n times the frequency of fundamental mode.
    In nth mode of vibration the number of nodes are n+1, and number of antinodes are n. 
    Question 602
    CBSEENPH11018874

    What are nodes and antinodes? Find the position of nodes and antinodes in a stretched string between two points vibrating in nth harmonic.

    Solution
    The position or points in the string where the amplitude of vibration is maximum are called antinodes. 
    Nodes are the points in the string where the particles in the string do not vibrate. 
    The stationary wave produced in the string is given by,
                   straight y equals Asinkx equals Asin fraction numerator 2 straight pi over denominator straight lambda end fraction straight x
    Position of nodes:
    These are the positions where y = 0. 
    i.e.                Asin fraction numerator 2 straight pi over denominator straight lambda end fraction straight x equals 0 
    rightwards double arrow                    space space space fraction numerator 2 straight pi over denominator straight lambda end fraction straight x equals mπ
    i.e.,                       straight x equals straight m straight lambda over 2 
    where m = 1, 2, 3...n
                         
    Let L be the length of the string.
    The wavelength of wave when string vibrates in nth modes is,
                           straight lambda equals fraction numerator 2 straight L over denominator straight n end fraction 
    ∴                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Thus the  position of nodes is, 
                space space space straight L over straight n comma space 2 straight L over straight n comma space 3 straight L over straight n comma space.......... straight n straight L over straight n

    Position of antinodes:
     
    These are the positions where straight y equals plus-or-minus straight A 
    i.e.    Asin fraction numerator 2 straight pi over denominator straight lambda end fraction straight x equals plus-or-minus straight A
    rightwards double arrow          fraction numerator 2 straight pi over denominator straight lambda end fraction straight x equals fraction numerator 2 straight m minus 1 over denominator 2 end fraction straight pi
    rightwards double arrow       straight x equals fraction numerator 2 straight m minus 1 over denominator 4 end fraction straight lambda equals fraction numerator 2 straight m minus 1 over denominator 2 end fraction open parentheses straight L over straight n close parentheses
    where m = 1, 2, 3...n
    Position of nodes is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
     

         
     
    Question 603
    CBSEENPH11018875

    Derive an expression for the fundamental frequency of one end closed pipe. Discuss the various overtones emitted by the pipe.

    Solution
    Consider, a hollow pipe of length L closed at one end.
    A progressive longitudinal wave be sent in the pipe towards left.
    The wave gets reflected from closed end.
    Both incident and reflected waves superimpose in the pipe and give rise to stationary wave.
    Incident and reflected waves are represented by, 
              space space space space space space straight y subscript 1 equals rsin left parenthesis ωt plus kx right parenthesis                      ...(1) 
                   straight y subscript 2 equals negative rsin left parenthesis ωt minus kx right parenthesis                  ...(2) 
    On superposition, 
                   straight y equals straight y subscript 1 plus straight y subscript 2 
                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                   space space equals 2 rcosωt space sinkx 
                    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>                                  ...(3) 
    which is the resultant stationary wave.
    Pipe is open at end B and closed at end A.
    Since the air is free to vibrate at open end, therefore open end of the pipe is antinode.
    At closed end, the amplitude of vibration is zero and hence closed end is node.
    Thus the boundary conditions are: 
    At A,  x = 0;       y = 0 
    At B,   x = L;      space space straight y equals plus-or-minus straight A 
    From equation (3), it follows that 
                         straight y equals straight A space sink left parenthesis 0 right parenthesis space equals space 0 
    The solution is trivial. 
     
             space space straight y equals AsinkL space equals space plus-or-minus straight A 
     rightwards double arrow          space space sin space kL space equals space plus-or-minus 1 
    rightwards double arrow               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  where n is an integer. 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  
    rightwards double arrow                space space straight lambda equals fraction numerator 1 over denominator left parenthesis 2 straight n minus 1 right parenthesis end fraction 4 straight L 
    ∴             straight v equals v over straight lambda equals left parenthesis 2 straight n minus 1 right parenthesis fraction numerator v over denominator 4 straight L end fraction 
    where, v is the velocity of wave in the air column.
    Fundamental frequency or 1st mode of vibration i.e. n = 1
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    This mode is the fundamental mode of vibration. 
     
    First overtone or 2nd mode of vibration i.e. n = 2. 
    straight lambda equals fraction numerator 4 straight L over denominator 3 end fraction space space and space space space space space space space space space space straight v equals 3 fraction numerator v over denominator 4 straight L end fraction equals 3 fraction numerator v over denominator 4 straight L end fraction 
    The Frequency of second mode of vibration is three times the fundamental frequency.
    Hence, it is known as the third harmonic.
    3rd mode of vibration i.e. n = 3 
    straight lambda equals fraction numerator 4 straight L over denominator 5 end fraction space space and space straight v equals 5 fraction numerator v over denominator 4 straight L end fraction equals 5 fraction numerator v over denominator 4 straight L end fraction 
    Frequency of this mode of vibration is five times the fundamental frequency, hence known as fifth harmonic. 
    Similarly, in nth mode of vibration, the frequency of vibration is,
     straight v subscript straight n equals left parenthesis 2 straight n minus 1 right parenthesis fraction numerator v over denominator 4 straight L end fraction
    overtone emitted is (n–1)th.
    Since (2n–1) is an odd integer, therefore the overtones emitted by one end closed pipe have only odd harmonics. 
    Question 604
    CBSEENPH11018876

    Write an expression for the fundamental frequency of a both end open pipe. What is the difference in the overtones emitted by one end closed pipe and both end open pipe?

    Solution
    For longitudnal vibrations in open end pipe, the fundamental frequency is given by, 
                      straight v equals straight n fraction numerator straight v over denominator 2 straight L end fraction equals fraction numerator straight n over denominator 2 straight L end fraction square root of γP over straight rho end root 
    The difference in the overtones emitted by two pipes is that one end closed pipe emits only odd harmonics while both end open pipe emits both even and odd harmonics. 
    Question 605
    CBSEENPH11018877

    Compare travelling and stationary waves.

    Solution

    Travelling waves

    Stationary waves

     

    1. Each particle of media transfers disturbance to next particle

    2. Amplitude of vibration of each particle is constant.



    3. Every particle passing through mean position has same velocity.


    4. There is a gradual phase difference between successive particles.

    5. None particle is permanently at rest.

    6. There is a flow of energy along the length of propagation of wave.

    1. Disturbance is not
    communicated from
    one particle to another.

    2. Amplitude of vibration
    of different particle is different.
    The amplitude of
    vibration increases 
    gradually from zero to maximum from node
    to antinode and then decrease
    gradually to from antinode to node in a loop.

    3. Every particle in a loop passing through mean position has different 
    velocity. The particle
    at antinode has maximum velocity at mean position.

    4. All the particles between two consecutive nodes vibrate in same phase.

    5. The particles at nodes are permanently at rest.

    6. There is no flow of energy along the length of propagation of wave

     
     

    Question 606
    CBSEENPH11018878

    Write some characteristic of the stationary wave.

    Solution

    Characteristics of stationary waves:

    (i) The disturbance is confined to a particular region between the starting point and the reflecting point of the wave. 


    (ii) There is no onward motion of the disturbance from one particle to the adjoining particle and so on beyond this particular region.
    (iii) The total energy assosciated with a stationary wave is twice the energy of each of the incident and the reflected wave. There is no flow of energy along the stationary waves.

    (iv) In a stationary wave, the medium splits up into a numvber of segments. Each segment is vibrating up and down as a whole. All the particles in one particular segment vibrate in the same phase. 
    (v) There are certain points in the medium of standing wave, which is permanently at rest. These are called nodes. And certain other points in the medium where the amplitude is maximumwhich are called antinodes. 

    Question 607
    CBSEENPH11018879

    In case of standing waves no energy can be transferred past a node. Why?

    Solution
    In case of standing wave, energy is returned back through reflected wave and hence the net energy transferred past a node is zero. 
    Question 608
    CBSEENPH11018880

    Where will a man hear the loud sound - at node or antinode? Explain.

    Solution
    Sound is produced due to variation of pressure and it is louder where pressure variation is maximum.
    The strain is maximum at nodes and hence the pressure, therefore the sound is louder at nodes. 
    Question 609
    CBSEENPH11018881

    What is sonometer? What type of waves are produced in sonometer wire?

    Solution

    i) Sonometer consists of a hollow wooden box.
    The arrangement for producing sonometer waves is shown in the figure.

    ii) A string is stretched over the sonometer box with its one end fixed rigidly at one end of the sonometer box.
    iii) The other end is connected to a hanger. The string is passed over the frictionless pulley fixed at the other end of rigid support.
    iv) Two wedges are placed below the string. The two wedges behave as two rigid supports for the string.
    v) Transverse stationary waves are produced in the string between two rigid supports.

    Question 610
    CBSEENPH11018882

    A vibrating string is heated to higher temperature. How will the frequency of note produce change?

    Solution
    When the string is heated to higher temperature, its length increases and Young's modulus decreases. Therefore, the frequency of vibration of the string will decrease.
    Question 611
    CBSEENPH11018883

    What type of organ pipe will you choose for making a flute and why?

    Solution
    Open organ pipe will be chosen because all even and odd harmonics are present in the open end pipe.
    While in one end closed pipe only odd harmonics are present. Larger the number of harmonics, better is the quality of sound.
    Question 612
    CBSEENPH11018884

    What is end correction?

    Solution
    When the vibrations excite in the pipe, the antinode is not exactly at the open end but at some distance above the open end. The distance by which the antinode is above the open end is called end correction.
    Question 613
    CBSEENPH11018885

    Why does the rider placed on the string fly off when the string and fork act in unison?

    Solution
    Sonometer vibrations are forced vibrations.
    When the string and the fork are in unison i.e. frequency of tuning fork is equal to the frequency of string, resonance occurs and string vibrates with a larger amplitude and hence the rider flies off.
    Question 614
    CBSEENPH11018886

    When we start filling an empty bucket with water from the tap, the pitch of the sound produced goes on changing. Why?

    Solution
    The pitch of the sound produced is determined by the length of the air column in the bucket.
    The fundamental frequency of note emitted by one end closed pipe is given by,
                               straight v equals fraction numerator v over denominator 2 straight L end fraction 
    As the water fills the bucket, the length of the air column in the bucket goes on decreasing and as a result the pitch goes on increasing. 
    Question 615
    CBSEENPH11018887

    A wire of length L is made of material of density ρ and Young's modulus Y. It is suspended from a rigid support. On loading at the free end, it elongates by x. When the string is plucked, a stationary transversal wave is produced in the wire. Find the fundamental frequency of vibration of string, assuming the loaded end of string is a node.

    Solution
    Let r be the radius of the wire.
    The tension in the wire is, 
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#6 {main}</pre> 
    Mass per unit length of wire is, 
                     straight m equals ρA equals ρπr squared 
    Now the frequency of fundamental note of vibration is,
                     straight v space equals fraction numerator 1 over denominator 2 straight L end fraction square root of straight T over straight m end root

space space equals fraction numerator 1 over denominator 2 straight l end fraction square root of fraction numerator Yπr squared straight x over denominator Lρπr squared end fraction end root 
                     space space equals fraction numerator 1 over denominator 2 straight L end fraction square root of Yx over Lρ end root
    Question 616
    CBSEENPH11018888

    A tuning fork produces resonance in a closed pipe. But the same tuning fork is unable to produce resonance in open pipe of equal length. Why? What should be the length of open pipe to produce the resonance with same tuning fork?

    Solution
    Let L be the length of closed pipe and V be the velocity of sound in air.
    Since the tuning fork is in resonance with this pipe, therefore, frequency of tuning fork is,
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#6 {main}</pre> 
    The fundamental frequency of open pipe is, 
                   straight v equals fraction numerator v over denominator 2 straight L end fraction 
    Therefore, tuning fork is not in resonance with open pipe.
    To produce resonance, let the length of tube required be l.
    The resonance condition is, 
                      fraction numerator v over denominator 2 ell end fraction equals straight v equals fraction numerator v over denominator 4 straight L end fraction 
    ∴                   space space space ell space equals space 2 straight L 
    Thus, to produce the resonance with same tuning fork, the length of open end tube should be double the length of one end closed pipe. 
    Question 617
    CBSEENPH11018889

    The fundamental frequency of a sonometer wire increases by 7Hz on increasing the tension by 10-25%. Find the fundamental frequency of sonometer wire.

    Solution
    We know that fundamental frequency of sonometer wire is directly proportional to the square root of tension in the string.
    straight i. straight e. comma space space space space space straight nu space proportional to space square root of straight T space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

When space the space tension space is space increased space by space 10.25 space percent sign comma
straight i. straight e. comma space by space 0.1025 space straight T comma space the space fundamnetal space frequency space
becomes space straight nu space plus space 7.

Therefore comma space

straight nu space plus space 7 space proportional to space square root of 1.1025 space straight T end root space space space space space space space space... space left parenthesis 2 right parenthesis space

Now comma space dividing space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis comma space we space get

fraction numerator straight nu space plus space 7 over denominator space space straight nu end fraction space equals space square root of 1.1025 space straight T end root space equals space 1.05 space

rightwards double arrow space straight nu space plus space 7 space equals space 1.05 space space space straight nu space

rightwards double arrow space 0.05 space straight nu space space equals space 7 space

rightwards double arrow space straight nu space space equals space 140 space Hz
    140 Hz is the fundamental frequency of sonometer wire. 
    Question 618
    CBSEENPH11018890

    When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers. Why?

    Solution
    According to the equation of continuity,

                           av = constant

    where, 'a' is the area of cross-section, and
    'v' is the velocity of the liquid.

    'a' and 'v' are inversely proportional t each other. 

    Therefore, when we close the water tap with our fingers, the area at that point, from where water flows out, decreases and hence velocity of water increases. 
    Question 619
    CBSEENPH11018891

    How the rate of flow of fluid through capillary tube will change when the diameter of the capillary tube is reduced to half?

    Solution

    According to Poiseuille’s equation, the rate of flow of fluid through capillary tube is directly proportional to fourth power of diameter.

    i.e.,                      V ∝ d4

    If d is changed to d/2, then rate of flow will decrease to V/16. 

    Question 620
    CBSEENPH11018892

    What are the assumptions of Bernoulli's theorem?

    Solution
    The assumptions of Bernoulli's theorem are:

    (i) The liquid is non-viscous,
    (ii) The liquid is incompressible, 
    (iii) The flow is steady, and
    (iv) The velocity of the flow of liquid is less than critical velocity. 
    Question 621
    CBSEENPH11018893

    Why do the two sheets of paper come closer towards each other when air is blown between the two sheets?

    Solution
    When air is blown between two sheets of paper, the pressure between the two sheets decreases. So the atmospheric pressure exerts more force from outside than the force exerted by pressure inside. 
    Hence, two sheets come closer with each other.


    Question 622
    CBSEENPH11018894

     If a person is standing near a fast moving train there is a danger that he will fall towards it. Explain why.

    Solution
    A person is standing near a fast moving train will tend to fall towards it because of Bernoulli's principle.
    The molecules of air between the person and the train is high due to the dragging effect. And, the velocity of the air molecules behind the person is constant. Therefore, as per Bernoulli’s principle, a pressure difference is created. So, the air behind the person pushes them towards the train and hence he falls towards the train.
    Question 623
    CBSEENPH11018895

    To keep a piece of paper horizontal you should blow over, not under it. Why?

    Solution
    When we blow the air above a piece of paper, velocity of air above paper increases and hence the pressure above the piece of paper decreases.
    Therefore, the air below the paper exerts more force than the air above and hence does not fall.
    Question 624
    CBSEENPH11018896

    The accumulation of snow on an aeroplane wings may reduce the lift. Explain.

    Solution
    Due to accumulation of snow on an aeroplane wings the structure of wings no longer remains as that of airfoil. This results in the decrease of lift. And as a result of Bernoulli's principle, more pressure difference is created than required. 
    Question 625
    CBSEENPH11018897

    Why does a flag flutter, when strong winds are blowing on a certain day? Explain.

    Solution
    When strong winds blow over the flag, the velocity of wind at the top is more than that at the bottom. According to Bernoulli's theorem, there exists pressure difference above and below the flag and hence it flutters. 
    Question 626
    CBSEENPH11018898

    Does it matter if one uses gauge pressure instead of absolute pressure in applying Bernoulli's equation? Explain.

    Solution

    Yes, it does matter if one uses gauge pressure instead of absolute pressure if atmospheric pressure at the two points. Bernoulli's equation applied is different at different points.

    If the atmospheric pressure is same at all the points then it does not matter whether we use absolute or gauge pressure. 

    Question 627
    CBSEENPH11018899

    What are the limitations of Bernoulli's equation?

    Solution

    Limitations of Bernoulli's equation are:

    (i) The Bernoulli's equation is derived by assuming that there is no loss of mechanical energy i.e. sum of pressure energy, potential energy and kinetic energy is constant. But when fluid is in motion, a part of mechanical energy is converted into heat energy.

    (ii) The Bernoulli's equation is derived by assuming that liquid is non-viscous. But actually none of the liquids is non-viscous.

    (iii)The Bernoulli's equation is derived by assuming that velocity of every fluid particle across any cross-section of pipe is same. But the particles of central layer have maximum velocity and velocity decreases towards the walls of pipe.

    Question 628
    CBSEENPH11018900

    Two similar vessels contain water and mercury upto same level on a table. A hole is made on the side wall of vessels at equal depth from the free surface of liquid. In which case the liquid will seep (leak) out with greater velocity? In which case the stream of flow has greater horizontal range?

    Solution
    The velocity of efflux from an orifice is independent of density of liquid but depends on the height of free surface of liquid from orifice. Since the height of free surface of water and mercury is same from the holes made in the two vessels, so the velocity of efflux will be same. Also the height of orifices from the ground is same, therefore the horizontal range will also be same. 
    Question 629
    CBSEENPH11018901

    Derive equation of continuity for steady flow of incompressible liquid.

    Solution
    Consider that a liquid is flowing through a pipe of varying cross-section as shown in figure.



    Let the liquid enter at A with velocity 'v1' whose area of cross-section is 'a1' and exit from end B with velocity 'v2' whose area of cross-section is 'a2'.

    Let 'ρ1' and 'ρ2' be the densities of liquid at ends A and B respectively.

    Now the volume of liquid that enters in one second at end A is given by,

                            V1 = a1 v1

    Mass of liquid entering per second at end A is, 

                           m1 = a1v1ρ1

    Similarly the mass of liquid leaving per second at end B is, 
                           m2 = a2v2ρ2

    If there is no source or sink of liquid, then mass of liquid that enters at end A in one second is equal to the mass of liquid that leave in one second.

    i.e.               a1 v1 ρ= a2 v2 ρ2        ... (1)

    If the liquid is incompressible, then
                               ρ1 = ρ2.

    Therefore equation (1) reduces to

                          a1v1= a2v2                   ....(2)

    Equation (2) is called the equation of continuity.

    Question 630
    CBSEENPH11018902

    If an addition of 1250 g tension of string raises its pitch to interval fifth, what is the original tension?

    Solution
    Interval fifth corresponds to the frequency ratio 3:2.
    Let T be the original tension in the string. 
    Relation between velocity and tension is given by, 
                                  straight v proportional to square root of straight T 
    ∴                           straight v subscript 2 over straight v subscript 1 equals square root of straight T subscript 2 over straight T subscript 1 end root 
    rightwards double arrow                   3 over 2 equals square root of fraction numerator straight T plus 1250 over denominator straight T end fraction end root 
    rightwards double arrow                   9 over 4 equals fraction numerator straight T plus 1250 over denominator straight T end fraction 
    rightwards double arrow                  space 9 straight T equals 4 straight T plus 5000 
    i.e.,                    straight T equals 1000 gm, is the original tension of the string. 
    Question 631
    CBSEENPH11018903

    Why does deep water run slow?

    Solution
    Consider, water be flowing in a river of width b.

    Let d1 and d2
     (d1 < d2) be the depth of river at two different places in the river.

    The velocity of water at these two places be respectively v
    1 and v2.

    Therefore, according to the equation of continuity, 



    That is why depth is more; speed is less and vice versa.

    Hence deep water runs slow and shallow water runs fast.

    Question 633
    CBSEENPH11018905

    A stone hangs in the air from a wire stretched over a sonometer. The wire is in unison with a tuning fork for distance l1 between the bridges. When the stone is dipped in water, the length between the bridges must be altered to l2 to bring it again in unison. Find the relative density of stone.

    Solution
    Let a be the density, V be the volume of stone and v be the frequency of tuning fork.
    When the stone is in air, the tension in the string is Vag.
    The wire is in unison for distance /, between the bridges, therefore
    straight v equals fraction numerator 1 over denominator 2 straight l subscript 1 end fraction square root of Vσg over straight m end root                    ...(1) 
    When the stone is put in water, due to upward thrust on the stone, the tension in the string decreases to F(σ-1)g.
    Now the wire is in unison for distance l2 between the bridges.
    Thus,

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#6 {main}</pre>              ...(2) 
    From (1) and (2), 
                fraction numerator straight sigma minus 1 over denominator straight sigma end fraction equals straight l subscript 2 squared over straight l subscript 1 squared 
    rightwards double arrow      straight sigma equals fraction numerator straight l subscript 1 squared over denominator straight l subscript 1 squared minus straight l subscript 2 squared end fraction, is the relative density of the stone. 

    Question 635
    CBSEENPH11018907

    Two wires of length 120cm and 90cm and radii 0-8mm and 0.9mm respectively are made from same material. Find the interval between their fundamental frequencies when two wires are stretched by equal forces.

    Solution
    The fundamental frequency of vibration of stretched string is given by their fundamental frequencies when two wires are stretched by equal forces.
    That is, 
    straight v equals fraction numerator 1 over denominator 2 straight l end fraction square root of straight T over straight m end root equals fraction numerator 1 over denominator 2 straight l end fraction square root of straight T over ρπr squared end root equals fraction numerator 1 over denominator 2 lr end fraction square root of straight T over ρπ end root 
    The fundamental frequency of vibration of first wire is,
                  straight v subscript 1 equals fraction numerator 1 over denominator 2 straight l subscript 1 straight r subscript 1 end fraction square root of straight T over πρ end root 
    The fundamental frequency of vibration of second wire is, 
                  straight v subscript 2 equals fraction numerator 1 over denominator 2 straight l subscript 2 straight r subscript 2 end fraction square root of straight T over πρ end root 
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#6 {main}</pre> 

    We have,
     space space space straight l subscript 1 equals 120 cm comma space space space space space space space space space space straight r subscript 1 equals 0.8 mm
space space space straight l subscript 2 equals 90 cm comma space space space space space space space space space space space space space straight r subscript 2 equals 0.9 mm 
    ∴     straight v subscript 1 over straight v subscript 2 equals fraction numerator straight l subscript 2 straight r subscript 2 over denominator straight l subscript 1 straight r subscript 1 end fraction equals fraction numerator 90 cross times 0.9 over denominator 120 cross times 0.8 end fraction equals 27 over 32 
    When two wires are stretched by equal forces, the fundamental frequencies lie between 27 and 32. 
    Question 636
    CBSEENPH11018908

    Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave.

    State which of these represent

    (i) a travelling wave

    (ii) a stationary wave  or

    (iii) none at all: 

    (a) y = 2 cos 3x sin 10t 

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#6 {main}</pre> 

    (c) y = 3 sin(5x –0.5t) + 4 cos(5x – 0.5t) 

    (d) y = cos x sin t + cos2x sin2t.

    Solution

    (a) The equation represents a stationary wave. 
    (b) The equation does not represent any wave. 
    (c) We have,
     y = 3sin(5x–0.5t) + 4cos(5x – 0.5t) = 5 sin [5x – 0.5t + tan-1 (4/3)].
    This is an equation for a travelling wave. 
    (d) y = cos x sin t + cos 2xsin 2t, is the equation for superposition of two stationary waves. 

    Question 637
    CBSEENPH11018909

    What will be the fundamental frequency of a note emitted by a closed pipe of length 34.8cm at 27°C. The velocity of sound at 0°C is 332 m/s.

    Solution
    The velocity of sound in air at 27°C is, 
    straight v equals straight v subscript straight o square root of fraction numerator 273 plus 27 over denominator 273 end fraction end root 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#6 {main}</pre> 
    The length of pipe L = 34.8cm = 0.348m
    The fundamental frequency of note emitted by one end closed pipe is, 
    v equals fraction numerator straight v over denominator 4 straight L end fraction equals fraction numerator 348 over denominator 4 cross times 0.348 end fraction equals 250 Hz  
    Question 638
    CBSEENPH11018910

    Two closed organ pipes when sounded together produced 2.5 beats per second at temperature 300K. Find the number of beats produced in one second at temperature 312.12K.

    Solution
    Let l1 and l2 be the lengths of two pipes.
    The number of beats produced by two pipes when sounded together is, 
    straight b equals straight v subscript 1 minus straight v subscript 2 equals straight v open parentheses fraction numerator 1 over denominator 4 straight l subscript 1 end fraction minus fraction numerator 1 over denominator 4 straight l subscript 2 end fraction close parentheses 
    where,
    v be the velocity of sound.
    Let v1 and v2 be the velocity of sound at 300K and 312.12K and band b2 be the number of beats produced per second respectively.
    Thus, 
                 space space straight b subscript 2 over straight b subscript 1 equals straight v subscript 2 over straight v subscript 1 equals fraction numerator square root of straight T subscript 2 end root over denominator square root of straight T subscript 1 end root end fraction 
    rightwards double arrow         fraction numerator straight b subscript 2 over denominator 2.5 end fraction equals square root of fraction numerator 312.12 over denominator 300 end fraction end root equals space 1.02  
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#6 {main}</pre>, is the number of beats produced in one second at temperature 312.12 K. 
    Question 639
    CBSEENPH11018911

    An open pipe is emitting third overtone. One end of the pipe is suddenly closed and it is observed that now it starts emitting fourth overtone at frequency 360 higher than earlier. Find the fundamental frequency of the open pipe.

    Solution
    Let l be the length of pipe and v the velocity of sound in air.
    The frequency of the third overtone of open pipe is, 
    space straight v subscript 1 equals 4 fraction numerator straight v over denominator 2 straight l end fraction                   ...(1) 
    When one end is closed, it becomes one end closed pipe.
    The frequency of the fourth overtone of one end closed pipe is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#6 {main}</pre>                    ...(2) 
    It is given that, 
             space straight v subscript 2 minus straight v subscript 1 equals 360
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre> 
    Now the fundamental frequency of open pipe is, 
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#6 {main}</pre> 
                
    Question 640
    CBSEENPH11018912

    Capillaries of length I, 16l and 81l are connected in series. The radii of the capillaries are r, 2r and 3r respectively. If the pressure difference across smallest pipe is P, then find the total pressure difference across the series combination.

    Solution
    Let P1, P2 and P3 be the pressure difference across first, second and third capillary tube respectively.
    Since all the three capillaries are in series.
    Therefore the rate of flow of fluid through all the three capillaries will be same.
    That is,
    dV over dt equals straight pi over 8 fraction numerator straight r to the power of 4 straight P subscript straight I over denominator straight n straight ell end fraction equals straight pi over 8 fraction numerator left parenthesis 2 straight r right parenthesis to the power of 4 straight P subscript 2 over denominator 8 space straight eta cross times 16 straight ell end fraction equals straight pi over 8 fraction numerator left parenthesis 3 straight r right parenthesis to the power of 4 straight P subscript 3 over denominator 8 space straight eta space left parenthesis 8 space straight I straight ell right parenthesis end fraction
     P= P= P3
     
    Since,
    P= P
    Therefore the total pressure across serial combination is, 
    straight P subscript 0 equals straight P subscript 1 plus straight P subscript 2 plus straight P subscript 3 equals 3 straight P
    Question 641
    CBSEENPH11018913

    A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound = 340 m/s)

    Solution

    Length of one end closed pipe is, L =  20 cm = 0.20m 
    Frequency of source is, v = 430 Hz 
    We know that the frequency of nth mode is given by, 
                straight v space equals space left parenthesis 2 straight n minus 1 right parenthesis fraction numerator straight v over denominator 4 straight L end fraction 
    Let nth mode of pipe be in resonance with the source. 
    Therefore, 
                  430 equals left parenthesis 2 straight n minus 1 right parenthesis fraction numerator 340 over denominator 4 cross times 0.2 end fraction 
    rightwards double arrow              straight n equals 1 
    Therefore, the fundamental mode of pipe is in resonance with the source.
    If both the ends of pipe are open then frequency of nth mode is, 
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#6 {main}</pre> 
    Substituting the values, we get 
               430 equals straight n fraction numerator 340 over denominator 2 cross times 0.2 end fraction 
    rightwards double arrow          n = 0.5, which is not possible. Therefore, if both the ends are open, then pipe cannot be in resonance with the source. 

    Question 642
    CBSEENPH11018914

    Liquid is flowing at the rate of V through the capillary tube of radius r and length l under pressure head P. If another tube of radius r/2 and length l/4 is connected in series with first tube, then find the rate of flow of liquid under the same pressure head.

    Solution
    Let η be the coefficient of viscosity of liquid.
    The rate of flow of fluid through capillary tube of radius r and length l under pressure head P is given by, 
    Error converting from MathML to accessible text. 
    Now when another capillary tube is connected, let pressure across first tube decrease to P1 and rate of flow of liquid be V'.
    As the two tubes are in series, therefore
    Error converting from MathML to accessible text. 
    The rate of flow has now become one-fifth of the original rate of flow. 
    Question 643
    CBSEENPH11018915

    Define beats. Give the analytical treatment of beats.

    Solution
    When two waves of nearly equal frequencies, travelling in the same direction superimpose, the intensity of sound at a point changes with time periodically.
    Beats are the periodic variation in the intensity of sound. 

     
     
    The time elapsed between two successive loud sounds or two faint sounds are called beat period.
    Reciprocal of the beat period is called beat frequency. 
    Analytical treatment of beats: 
    Let two waves be represented by, 
    space straight y subscript 1 equals rsinω subscript 1 straight t equals rsin 2 πv subscript 1 straight t                ...(1) 
    straight y subscript 2 equals rsinω subscript 2 straight t equals rsin 2 πv subscript 2 straight t                 ...(2)


    If y is the resultant displacement due to superimposition, then 
    straight y equals straight y subscript 1 plus straight y subscript 2 equals straight r left square bracket sin 2 πv subscript 1 straight t plus sin 2 πv subscript 2 straight t right square bracket 
      equals 2 rsinπ left parenthesis straight v subscript 1 plus straight v subscript 2 right parenthesis straight t space cosπ left parenthesis straight v subscript 1 minus straight v subscript 2 right parenthesis straight t 
      equals left square bracket 2 straight r space cosπ left parenthesis straight v subscript 1 minus straight v subscript 2 right parenthesis straight t right square bracket sin space straight pi left parenthesis straight v subscript 1 plus straight v subscript 2 right parenthesis straight t 
    space space equals straight A space sinπ left parenthesis straight v subscript 1 plus straight v subscript 2 right parenthesis straight t 
    where,
     A = 2r cos straight pi (v1 – v2)t is the amplitude of the resultant wave.
    As intensity is directly proportional to the square of amplitude, therefore, the intensity is maximum when amplitude is maximum. 
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#6 {main}</pre> 
    rightwards double arrow             straight pi left parenthesis straight v subscript 1 minus straight v subscript 2 right parenthesis straight t space equals space nπ 
    rightwards double arrow                  straight t equals fraction numerator straight n over denominator straight v subscript 1 minus straight v subscript 2 end fraction 
    That is at times,
        straight t space equals space 0 comma space space fraction numerator 1 over denominator straight v subscript 1 minus straight v subscript 2 end fraction comma space fraction numerator 2 over denominator straight v subscript 1 minus straight v subscript 2 end fraction space space..... intensity will be maximum. 
    Therefore, the interval between two successive loud sounds is, 
                        straight T equals fraction numerator 1 over denominator straight v subscript 1 minus straight v subscript 2 end fraction 
    Therefore, Beat frequency, b = v1 – v

    i.e. Beat frequency is equal to the difference of frequencies of two interfering harmonic waves.

    Question 644
    CBSEENPH11018916

    What is pressure energy? Derive an expression for it.

    Solution

    Pressure energy is the energy possessed by liquid by virtue of pressure.

    To derive the expression for pressure energy, take a wide tank connected to a narrow pipe fitted with frictionless piston at the base as shown in figure.
                              

    Let A and a be the area of cross-sections of tank and pipe respectively.
    By holding the piston, fill the tank to a height h above the pipe with a liquid of density ρ.
    The pressure of liquid on the piston is,

                            P = ρgh

    Force on the piston in outward direction is,
                           F= ρgha

    Force to be exerted on the piston to keep it in equilibrium is,

                           F = ρgha

    As A >>a, therefore if the piston is pushed inward by a small distance x, the height of liquid in the tank remains constant and hence the force on the piston is constant.
    The work done to push the piston inward by distance x is, W = Fx = ρghax
    This work done against pressure is stored in the form of pressure energy.
    This energy is stored in the volume of liquid pushed inside i.e. ax.
    Therefore, pressure energy stored per unit volume of liquid is,
    Pressure space energy space density space equals space Pax over axρ space equals space straight P over straight rho

    Question 645
    CBSEENPH11018917

    Two notes have wavelength 2.1m and 2.25m in air. Each produce 8 beats per second with a third note of a fixed frequency. Find the frequency of third note. 

    Solution
    Let v be the velocity of the wave in the air and v be the frequency of the third note.
    Let, v1 and vbe the frequencies of waves of wavelength 2.1 m and 2.25m.
    Therefore, 
                  straight v subscript 1 equals fraction numerator straight v over denominator 2.1 end fraction space space space and space space straight v subscript 2 equals fraction numerator straight v over denominator 2.25 end fraction 
    It is clear that space straight v subscript 1 greater than straight v subscript 2. 
    As v produces 8 beats with both space straight v subscript 1 space and space straight v subscript 2 comma thus straight v subscript 1 greater than straight v greater than straight v subscript 2. 

    Therefore,
                       straight v subscript 1 minus straight v equals 8     and    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow                  straight v subscript 1 minus straight v subscript 2 equals 16
    rightwards double arrow               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    rightwards double arrow                   straight v equals 504 straight m divided by straight s
    ∴        straight v subscript 1 equals fraction numerator 504 over denominator 2.1 end fraction equals 240 
    Also,           straight v subscript 1 minus straight v equals 8 
    ∴         straight v equals straight v subscript 1 minus 8 equals 240 minus 8 equals 232 space Hz, is the frequency of the third note. 
    Question 646
    CBSEENPH11018918

    A note produces 7 beats with tuning fork A of frequency 256Hz and 3 beats with tuning fork B of frequency 266hz. Find the frequency of the note.

    Solution

    Let v be the frequency of the note.
    The note produces 7 beats with tuning fork A of frequency 256Hz.
    Therefore, the possible frequencies of note are, v = 256 ± 7
    i.e. 249 or 263 Hz.
    When the note is sounded with tuning fork B of frequency 266 Hz, it produces 3 beats.
    Therefore the possible frequencies of note are, v = 266 ± 3.
    i.e. 263 or 269 Hz.
    The common frequency in both the cases is 263 Hz. 
    Thus, the frequency of the note is 263 Hz.

    Question 647
    CBSEENPH11018919

    Bernoulli derived the relation by assuming the liquid to be non-viscous. How the relation will change if fluid is viscous?

    Solution

    The Bernoulli's equation is given by,
    straight P space plus space ρgh space plus space 1 half ρv squared space equals space constant,
    if liquid is non-viscous.
    But if liquid is viscous, then the quantity on the left hand side will decrease along the direction of the flow of fluid. 

    Question 648
    CBSEENPH11018920

    Two waves of wavelengths 1m and 1.02m produce 13 beats in 2 seconds in air at NTP. Find the velocity of air at NTP.

    Solution
    Let v be the velocity of sound in air at NTP.
    The frequency of wave of wavelength 1 m is, 
                        straight v subscript 1 equals v over 1 
    The frequency of the second wave of wavelength 1.02 m is, 
                       space straight v subscript 2 equals fraction numerator v over denominator 1.02 end fraction 
    The number of beats produced per second is, 
                     straight b equals 13 over 2 equals 6.5 
    ∴       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
                   fraction numerator 0.02 over denominator 1.02 end fraction v equals 6.5 
          v equals 6.5 cross times 51 equals 331.5 space straight m divided by straight s, is the velocity of air at NTP. 

    Question 649
    CBSEENPH11018921

    State and prove Bernoulli's theorem.

    Solution

    Statement: For the streamline flow of non-viscous and incompressible liquid, the sum of potential energy, kinetic energy and pressure energy is constant. 
    straight P over straight rho space plus space gh space plus space 1 half straight v squared space equals space constant space




    Proof:
    Let us consider the ideal liquid of density ρ flowing through the pipe LM of varying cross-section.
    Let P1 and P2 be the pressures at ends L and M and A1 and A2 be the areas of cross-sections at ends L and M respectively.
    Let the liquid enter with velocity V1 and leave with velocity v2.
    Let A1 > A2.
    Now, by equation of continuity,
                         straight A subscript 1 straight v subscript 1 space equals space straight A subscript 2 straight v subscript 2 
    Since, A1 > A2
    Therefore, 
    v2 > v1 and P> P2 
    Let, m be the mass of liquid enetring at end L in time t. 
    The liquid will cover a distance = v1
    Therefore, the work done by pressure on the liquid at end L in time t is, 
    W1 = Force cross times space displacement 
          = P1A1v1t           ... (1) 
    Since same mass m leaves the pipe at end M in same time t, in which liquid will cover the distance given by v2t.
    Therefore, work done by liquid against the force due to pressure P1 is given by, 
     
            W2 = P2A2v2t        ... (2) 
    Net ork done by pressure on the liquid in time t is, 
    W = W1 - W2 
        =P1A1v1t  - P2A2v2t        ... (3) 
    This work done on liquid by pressure increases it's kinetic energy and potential energy. 
    Increase in K.E of liquid is, 
    increment straight K space equals space 1 half straight m space left parenthesis straight v subscript 2 squared space minus space straight v subscript 1 squared right parenthesis space space space space space space space... space left parenthesis 4 right parenthesis thin space

Increase space in space potential space energy space of space liquid space is comma space

increment straight P space equals space mg thin space left parenthesis straight h subscript 2 space minus straight h subscript 1 right parenthesis space space space space space space space space space space space space space space space... space left parenthesis 5 right parenthesis thin space

According space to space work minus energy space relation comma space

straight P subscript 1 straight A subscript 1 straight v subscript 1 straight t space minus space straight P subscript 2 straight A subscript 2 straight v subscript 2 space straight t space

equals space 1 half straight m space left parenthesis straight v subscript 2 squared space minus space straight v subscript 1 squared right parenthesis space space plus space mg thin space left parenthesis straight h subscript 2 space minus straight h subscript 1 right parenthesis space space space... space left parenthesis 6 right parenthesis

If space there space is space no space source space and space sink space of space liquid comma space
then space mass space of space liquid space entering space at space end space straight L
is space equal space to space the space mass space of space liquid space leaving space the space pipe
at space end space straight M space and space is space given space by comma space

straight A subscript 1 straight v subscript 1 ρt space equals space straight A subscript 2 straight v subscript 2 ρt space equals space straight m

rightwards double arrow space straight A subscript 1 straight v subscript 1 straight t space equals space straight A subscript 2 straight v subscript 2 straight t space equals space straight m over straight rho space space space space space space space space space space space space space... space left parenthesis 7 right parenthesis space

From space equation space left parenthesis 6 right parenthesis space and space left parenthesis 7 right parenthesis comma space we space have

straight P subscript 1 space straight m over straight rho minus straight P subscript 2 straight m over straight rho space

equals space 1 half straight m space left parenthesis straight v subscript 2 squared space minus space straight v subscript 1 squared right parenthesis space plus space mg thin space left parenthesis straight h subscript 2 space minus straight h subscript 1 right parenthesis space space
    That is, 
    straight P over straight rho space plus space gh space plus space 1 half straight v squared space equals space constant space 
    Question 650
    CBSEENPH11018922

    A tuning fork gives 3 beats per second with sonometer wire when its length is either 49.8cm or 50.2cm. Find the frequency of the tuning fork.

    Solution
    Let v  be the frequency of tuning fork. 
    Let, v1 and v2 be the frequencies of sonometer wire when its length is 49.8cm and 50.2 cm respectively.
    The frequency of sonometer wire is inversely proportional to the length of the sonometer wire.
    Therefore, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>                    ...(1) 
    From the above equation, it is clear that v1 > v2.
    Since tuning fork produces 3 beats per second with sonometer wire for both lengths, therefore 
                   straight v subscript 1 minus straight v equals 3       
     rightwards double arrow         straight v subscript 1 equals straight v plus 3 
    and         straight v minus straight v subscript 2 equals 3       
    rightwards double arrow           straight v subscript 2 equals straight v minus 3 
    Thus,   straight v subscript 1 over straight v subscript 2 equals fraction numerator straight v plus 3 over denominator straight v minus 3 end fraction                   ...(2) 
    From (1) and (2), 
                     fraction numerator straight v plus 3 over denominator straight v minus 3 end fraction equals 502 over 498 
    On solving, we get 
    space space space straight v equals 750 space Hz, is the frequency of the tuning fork. 
    Question 651
    CBSEENPH11018923

    Why two ships sailing in the same direction tend to come closer?

    Solution
    When two ships are moving in the same direction, the velocity of the water between the two ships increases, because area at A is greater than area at B. 



    Therefore, pressure in the region between the two ships decreases, and water on the outer side exerts more pressure than between two ships. Therefore, the ships tend to come closer towards each other.

    Question 652
    CBSEENPH11018924

    There are three sources of sound of frequency 248 Hz, 250Hz and 252 Hz. Find the number of beats observed per second when all are sounded together.

    Solution
    Let the amplitude of waves from three sources be a.
    Let v = 250 Hz.
    Therefore 248 = v–2 and 252 = v+2.
    When the three waves superimpose, the resultant wave is,
                       straight y equals asin 2 straight pi left parenthesis straight v minus 2 right parenthesis straight t plus asin 2 πvt plus asin 2 straight pi left parenthesis straight v plus 2 right parenthesis straight t 
      equals left square bracket asin 2 straight pi left parenthesis straight v plus 2 right parenthesis straight t plus asin 2 straight pi left parenthesis straight v minus 2 right parenthesis straight t right square bracket plus asin 2 πvt 
    space space equals 2 asin 2 πvt space cos 4 πt plus asin 2 πvt 
                     
      equals straight a left parenthesis 2 cos 4 πt plus 1 right parenthesis sin space 2 πvt 
                         
    The amplitude of resultant wave is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> which is maximum when cos 4 πt equals plus-or-minus 1. 
    rightwards double arrow   4 πt equals nπ       where,   n = 0, 1, 2, 3...
    That is,
                           straight t equals 1 fourth comma space 2 over 4 comma space 3 over 4... 
    Time interval between two maximum is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Thus, the frequency of maximum is 4.
    Hence, the number of beats heard per second when all the three sources are sounded together is 4. 
    Question 653
    CBSEENPH11018925
    Question 656
    CBSEENPH11018928
    Question 657
    CBSEENPH11018929

    What is Magnus effect? Explain.

    Solution
    When a spinning ball is thrown in the blowing air the path of the ball becomes curved. This is known as Magnus effect. 



    This effect arises because layer above the ball moves in the direction of spinning ball. So, the velocity of layer above the ball increases and similarly the velocity of the layer below the ball decreases.
    Hence, due to Bernoulli's theorem the pressure above the ball will be less than below and hence due to this pressure difference the ball follows a curved path. 

    Question 658
    CBSEENPH11018930

    What is the difference between interference and beats?

    Solution

    Interference: The pattern is produced when two waves of same frequency travelling in same direction superimpose.
    Beats are formed when frequencies are slightly different.
    In the interference, the position of maximum and minimum are permanent i.e. intensity at a given point does not change with time.
    In beats the intensity of wave changes at given point in time.

    Question 659
    CBSEENPH11018931

    What is constructive interference?

    Solution
    If two waves reach a point in phase, so that crest of one wave falls on the crest of second and trough falls on the trough. Also, the amplitude of the resultant wave is scalar sum of the amplitude of two waves, then intensity of resultant wave at that point is maximum and constructive interference is said to take place.
    Question 660
    CBSEENPH11018932

    What is destructive interference?

    Solution
    If two waves reach a point in opposite phase, the crest of one wave falls on the trough of second and vice versa. then, the amplitude of the resultant wave is magnitude of scalar difference of the amplitude of two waves.
    Thus, the intensity of resultant wave at that point is minimum and destructive interference is said to take place. 
    Question 661
    CBSEENPH11018933

    State the conditions to obtain interference of sound waves.

    Solution

    The conditions to obtain interference of sound waves is: 
    (i) The two sources of sound must emit the waves of same frequency and the same amplitude.
    (ii) The initial phase difference between the waves emitting from two sources should be constant. 
    (iii) The two wave trains must travel in the same direction. 

    Question 662
    CBSEENPH11018934

    During storms, roofs of huts are sometimes blown up. Why?

    Solution
    During storms, the strong wind blowing over the roof decreases the pressure pto much lower value than the pressure inside the hut pi where there is no blow. Now, according to Bernoulli's principle, this difference of pressure exerts the net force on the roof in upward direction and hence it blows off. 
    Question 663
    CBSEENPH11018935

    What is the principle of atomiser or sprayer?

    Solution
    When the rubber balloon is pressed, the air passes with large velocity at point A. Due to this, the pressure at A decreases. As a result of decrease in pressure, the liquid will rise up in the tube and flow away in the form of fine spray. 
    Question 664
    CBSEENPH11018936

    What is the principle of venturimeter? Derive an expression for the rate of flow of liquid through it.

    Solution

    Venturimeter is based on Bernoulli's theorem. It consists of a two truncated tubes connected by a pipe at narrow ends. The pipe connecting the two tubes is called throat. 

    i) Venturimeter tube is positioned horizontally and the liquid is made to enter in it from end A and after passing through throat BC, it leaves tube at end D.

    ii) Let at A,
    Area of cross-section of tube = a1,
    Pressure of liquid = P1 
    Velocity = v1 
    At the throat,
    Area of cross-section of tube = a2 
    Pressure of liquid = P2 
    Velocity be v2

    According to the equation of continuity, 
    straight a subscript 1 straight v subscript 1 space equals space straight a subscript 2 straight v subscript 2 space space space space space space space space space space... space left parenthesis 1 right parenthesis space

According space to space Bernoulli apostrophe straight s space Theorem comma space

straight P subscript 1 over straight rho plus 1 half straight v subscript 1 squared space plus space gh subscript 1 space equals space straight P subscript 2 over straight rho plus 1 half straight v subscript 2 squared space plus space gh subscript 2 space space space... space left parenthesis 2 right parenthesis

As space the space tube space is space horizontal comma space therefore

straight h subscript 1 space equals space straight h subscript 2 space

Substituting space in space equation space left parenthesis 2 right parenthesis comma space we space get space

straight P subscript 1 over straight rho plus 1 half straight v subscript 1 squared space equals space straight P subscript 2 over straight rho plus 1 half straight v subscript 2 squared space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis

Solving space left parenthesis 1 right parenthesis space and space left parenthesis 3 right parenthesis space for space straight v subscript 1 comma space we space get

straight v subscript 1 space equals space straight a subscript 2 space square root of fraction numerator 2 left parenthesis straight P subscript 1 minus straight P subscript 2 right parenthesis over denominator straight rho space left parenthesis straight a subscript 1 squared space minus space straight a subscript 2 squared right parenthesis thin space end fraction end root space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis space

If space the space difference space in space the space height space of space the space liquid space in space
the space manometer space is space straight h comma space then space straight P subscript 1 space minus space straight P subscript 2 space equals space ρgh

Thus comma space

straight v subscript 1 space equals space straight a subscript 2 square root of fraction numerator 2 gh over denominator left parenthesis straight a subscript 1 squared space minus space straight a subscript 2 squared right parenthesis end fraction end root space space space space space space space space space space... space left parenthesis 5 right parenthesis thin space

Now comma space the space rate space of space flow space of space liquid space through space the space
venturimeter space is comma space

straight V space equals space straight a subscript 1 straight v subscript 1 space equals space straight a subscript 1 straight a subscript 2 space square root of fraction numerator 2 gh over denominator left parenthesis straight a subscript 1 squared space minus space straight a subscript 2 squared right parenthesis end fraction end root

    Question 665
    CBSEENPH11018937

    State and prove Torricelli’s theorem.

    Solution
    Torricelli's Theorem states that the velocity of efflux of the liquid through an orifice is equal to the velocity which a body would attain in free fall from the surface of liquid to orifice.
                     
    Proof:
    Consider a tank containing an ideal liquid of density ρ and having a narrow orifice at L.
    Let the tank be very wide as compared to orifice so that velocity of its free surface can be taken zero.
    Let v be velocity of efflux.
    The pressure at free surface M of liquid and at orifice L is atmospheric.
    i.e., P and let h be the height of liquid above orifice.
    Applying Bernoulli's theorem at L and M, 
    space space space space space space straight P subscript straight L over straight rho space plus space gh subscript straight L space plus space 1 half straight v subscript straight L squared space equals space straight P subscript straight M over straight rho space plus space gh subscript straight M space plus space 1 half straight v subscript straight M to the power of 2 space end exponent

rightwards double arrow fraction numerator straight P subscript straight L space minus space straight P subscript straight M over denominator straight rho end fraction space plus space space 1 half open parentheses straight v subscript straight L squared space minus straight v subscript straight M to the power of 2 space end exponent space close parentheses space equals space straight g open parentheses straight h subscript straight M space minus straight h subscript straight L space close parentheses space

rightwards double arrow space 0 space plus space space 1 half straight v squared space equals space gh space

rightwards double arrow space straight v space equals space square root of 2 gh end root space

When space straight a space body space falls space freely space by space height space straight h
it apostrophe straight s space velocity space straight v apostrophe space is comma space

straight v space equals space square root of 2 gh end root space
    Hence velocity of efflux of liquid is equal to velocity of free falling body. 
    Question 666
    CBSEENPH11018938

    Water flows in a pipe of varying cross section kept horizontally. The mercury stands 12 cm at a point where the velocity of flow is 25 cm/s. Find the pressure at another point where the velocity is 100cm/s.

    Solution
    Bernoulli's theorem for the horizontal flow of liquid can be written as, 
    straight P subscript 1 over straight rho space plus space 1 half v subscript 1 squared space equals space P subscript 2 over rho space plus space 1 half v subscript 2 squared space
    Here,  
    P= 12 cm of Hg 
        = 12 x 13.6 x 980 dyne/cm
    P= ?
    v= 25 cm/s,    
               
    v= 40cm/s
    ρ = 1 gm/cc
    Putting the values in equation (1), we have
    fraction numerator 12 space cross times 13.6 cross times 980 over denominator 1 end fraction space plus space 1 half left parenthesis 24 right parenthesis squared space equals space P subscript 2 over 1 space plus space 1 half left parenthesis 100 right parenthesis squared space

P subscript 2 space equals space 155224 space d y n e divided by c m squared space

space space space space space equals 11.6464 space c m space o f space m e r c u r y space H g


    Question 667
    CBSEENPH11018939

    Two coherent waves of amplitude a and b of intensities I1 and I2 interfere with phase difference ϕ. Find the amplitude and intensity of resultant wave. Deduce the result if both the waves have same amplitude and intensity. Also plot the graph showing the variation of intensity of resultant wave with phase difference.

    Solution

    Let the two coherent waves be, 
                 space straight y subscript 1 space equals space asin left parenthesis ωt right parenthesis space space space space space space space space space space space space.... left parenthesis 1 right parenthesis space

straight y subscript 2 space equals space bsin left parenthesis ωt plus straight ϕ right parenthesis space space space space space space space.... left parenthesis 2 right parenthesis   
    Using principle of superposition, the resultant wave is, 
           straight y equals straight y subscript 1 plus straight y subscript 2 
             equals asin left parenthesis ωt right parenthesis plus bsin left parenthesis ωt plus straight ϕ right parenthesis 
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#6 {main}</pre> 
             equals left parenthesis straight a plus straight b space cosϕ right parenthesis space sinωt plus left parenthesis bsinϕ right parenthesis space cosωt 
             equals rsin left parenthesis ωt plus straight alpha right parenthesis                ...(3) 
    where,
    r is the amplitude of resultant wave. 

    Amplitude of the resultant wave is given by,  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
     equals square root of straight a squared plus straight b squared plus 2 abcosϕ end root               ...(4) 
    If both the waves have same amplitude, then 
         straight r equals square root of straight a squared plus straight a squared plus 2 straight a squared cosϕ end root space equals space 2 acos open parentheses straight ϕ over 2 close parentheses 
    The intensity of  a wave is directly proportional to the square of the amplitude.
    Therefore intensities of waves <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> are
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#6 {main}</pre> 
    where,
     p is constant of proportionality.
    Now from equation (4), 
                   straight r squared equals straight a squared plus straight b squared plus 2 abcosϕ 
               space space pr squared equals pa squared plus pb squared plus 2 pabcosϕ 
                 space straight I equals straight I subscript 1 plus straight I subscript 2 plus 2 square root of straight I subscript 1 straight I subscript 2 end root space cosϕ 
    If straight I subscript 1 equals straight I subscript 2 equals straight I subscript straight o comma space then 
                    straight I equals straight I subscript straight o plus straight I subscript straight o plus 2 square root of straight I subscript straight o straight I subscript straight o end root space cosϕ 
                   space space equals 4 straight I subscript straight o space cos squared open parentheses straight ϕ over 2 close parentheses 
    The variation of intensity of the resultant wave with phase difference is as shown in the figure below. 
               

    Question 668
    CBSEENPH11018940

    A garden sprinkler has 150 small holes, each of 2mm2 in area. If water is supplied at the rate of 0.3 liter/s, then find the average velocity of spray.

    Solution
    Number of holes in the sprinkler, n=150
    Area of each hole, a = 2mm= 0.02 cm
    Total area of 150 holes,
    A=150 X 0.02 = 3 cm
    Rate of flow of water,
    V=0.3  liter/s = 300 cc/s
    By equation of continuity,
                      V = av
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#6 {main}</pre>
    Question 669
    CBSEENPH11018941

    What is the condition for the two waves to interference constructively and destructively:

    (a) in terms of path difference.

    (b) in terms of the phase difference.

    Solution

    The condition for the two waves for constructive interference and destructive interference are:
    (a)
     straight x equals nλ space space left square bracket constructive space interference right square bracket

space space equals fraction numerator 2 straight n minus 1 over denominator 2 end fraction straight lambda space space left square bracket destructive space interference right square bracket 
    (b) straight ϕ equals 2 nπ          (constructive interference) 
           space equals left parenthesis 2 straight n minus 1 right parenthesis straight pi   (destructive interference)

    Question 670
    CBSEENPH11018942

    The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m/min, what is the speed of ejection of the liquid through the holes?

    Solution

    Area of cross-section of cylindrical tube, A = 8 cm

    Velocity of flow in the tube, v = 1.5 m/min = 2.5 cm/sec 
    Diameter of the hole, d = 1 mm = 0.1 cm 
    Area of one hole, aoπd squared over 4 space equals space 2.5 space pi cross times 10 to the power of negative 3 end exponent space c m squared space
    Total area of 40 holes, a = 40 a40 cross times 2.5 straight pi space cross times 10 to the power of negative 3 end exponent space cm squared space
    = 0.1 straight pi space space cm squared
    Let v' be the velocity of the ejection of the liquid through the holes. 
    Using equation of continuity, 
    AV = av
    Putting the values, we get
    straight v space equals space AV over straight a space equals space fraction numerator 8 cross times 2.5 over denominator 0.1 space straight pi end fraction space

space space space space space space space space space space space space space space space space space equals space 200 over straight pi
space space space space space space space space space space space space space space space space space equals space 64 space cm divided by sec 

    Question 671
    CBSEENPH11018943

    Derive an expression for the rate of flow of non-viscous liquid of density ρ through a conical pipe or radii r\ and r1 (r1 > r2) when pressure difference of P is maintained across its ends.

    Solution

    Let P1, a1 and v1 be the pressure, cross-sectional area and velocity at end A respectively.
    P2, a2 and v2 be the same quantities at end B.
     
    From space equation space of space continutiy comma space we space get

straight a subscript 1 straight v subscript 1 space equals space straight a subscript 2 straight v subscript 2 space

straight v subscript 1 space equals space straight v subscript 2 space straight a subscript 2 over straight a subscript 1 space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

Bernoulli apostrophe straight s space theorem space gives comma space

space space space space space space space straight P subscript 1 over straight rho space plus space 1 half straight v subscript 1 squared space equals space straight P subscript 2 over straight rho space plus space 1 half straight v subscript 2 squared space

rightwards double arrow space fraction numerator straight P subscript 1 space minus space straight P subscript 2 over denominator straight rho end fraction space equals space 1 half straight v subscript 2 squared space open parentheses fraction numerator straight a subscript 1 squared space minus space straight a subscript 2 squared over denominator straight a subscript 1 squared end fraction close parentheses space

space space space space space space space space straight v subscript 2 space equals square root of space fraction numerator 2 straight a subscript 1 squared left parenthesis straight P subscript 1 minus straight P subscript 2 right parenthesis over denominator straight rho space left parenthesis straight a subscript 1 squared minus straight a subscript 2 squared right parenthesis end fraction end root space equals space square root of fraction numerator 2 straight r subscript 1 squared straight P over denominator straight rho space left parenthesis straight r subscript 1 to the power of 4 minus straight r subscript 2 to the power of 4 right parenthesis end fraction end root space

Now comma space the space rate space of space flow space of space fluid space is comma space

straight V space equals space straight a subscript 2 straight v subscript 2 space equals space πr subscript 2 squared space square root of fraction numerator 2 straight r subscript 1 to the power of 4 straight P over denominator straight rho space left parenthesis space left parenthesis straight r subscript 1 to the power of 4 minus straight r subscript 2 to the power of 4 right parenthesis end fraction end root

    Question 672
    CBSEENPH11018944

    Two coherent waves of intensities I1 and I2 interfere. Derive the conditions of constructive and destructive interference for the two waves. Also find the intensity of resultant wave at position of constructive and destructive interference.

    Solution
    The intensity of resultant wave at any point, where the two coherent waves of intensities I1 and I2 interfere is given by, 
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#6 {main}</pre>                 ...(1) 
    At the point of constructive interference, the intensity of the resultant wave is maximum.
    From equation (1), the value of I is maximum if 
                    cosϕ equals 1 
    rightwards double arrow                 straight ϕ equals 2 nπ comma    where n = 0, 1, 2, 3...
    The intensity at this point is, 
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#6 {main}</pre> 
                             equals open parentheses square root of straight I subscript 1 end root plus square root of straight I subscript 2 end root close parentheses squared          ...(2) 
    At the point of destructive interference, the intensity of the resultant wave is minimum.
    From equation (1), the value of I is minimum if 
                       space space cosϕ equals negative 1 
    rightwards double arrow                    straight ϕ equals left parenthesis 2 straight n minus 1 right parenthesis straight pi comma where n = 0, 1, 2, 3...
    The intensity at this point is, 
                     straight I subscript min equals straight I subscript 1 plus straight I subscript 2 minus 2 square root of straight I subscript 1 straight I subscript 2 end root 
                          equals open parentheses square root of straight I subscript 1 end root minus square root of straight I subscript 2 end root close parentheses squared                ...(3)
                 
    Question 673
    CBSEENPH11018945

    Is the law of conservation of energy violated in interference?

    Solution
    No, the law of conservation of energy is not violated in interference.
    The energy that disappears at minima appears at maxima.
    In interference, it is just the redistribution of energy which takes place.
    Question 674
    CBSEENPH11018946

    Discuss the theory of Quincke's tube to demonstrate the interference of sound wave.

    Solution
    i) The Quincke's tube consists of a U-tube D1 that can slide into another U-tube D2 as shown in the figure.
    ii) The tube D2 has two openings.
    iii) At one opening a source of sound is placed and at second opening the interference is observed.
                  
    iv) Place a vibrating tuning fork near end A and hear the sound at end C.
    v) The sound waves travel through tube along path ABC and ADC.
    vi) The wave reaching the point C is resultant of two waves.
    vii) The intensity of wave at C depends on the path difference x = ADC – ABC. 
    i.e.           straight x equals nλ;   for  maximum sound 
    and            equals left parenthesis 2 straight n minus 1 right parenthesis straight lambda over 2 ;    for minimum sound 
    viii) Also when the tube D1 is gradually slid inwards or outwards, the change of intensity can be observed and positions of maximum and minimum can be noticed at regular interval of time.
    ix) Let for the position of D1, the intensity of sound is maximum.
    x) On moving the tube by a distance left parenthesis 2 straight n minus 1 right parenthesis straight lambda over 2, the intensity of sound changes to a minimum and moving the tube by space space space straight n straight lambda over 2 the intensity remains maximum.
    Question 675
    CBSEENPH11018947

    A source emitting sound of wavelength 2m is placed in front of a wall 2m from it. A detector is also placed in front of the wall at a distance 2m on the same side. Find the minimum distance between source and detector to hear the maximum sound.

    Solution
    The question is illustrated in the figure below.
    Let the minimum distance between source and detector to hear the minimum sound be 2x. 
     
    The path travelled by reflected wave is AO + OB.
    The path travelled by direct wave is AB.
    After reflection, the wave suffers a phase change of π radian which is equivalent to an extra path of λ/2.
    The net path difference between two waves reaching the detector is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre> 
                       equals 2 AO minus 2 straight x plus straight lambda over 2 equals 2 left parenthesis AO minus straight x right parenthesis plus straight lambda over 2 
                       equals 2 left parenthesis square root of straight x squared plus 4 end root minus straight x right parenthesis plus straight lambda over 2 
    For maximum intensity at B, the value of incrementshould be nλ. 
    For AB to be minimum increment should be <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    i.e.         increment equals 2 left parenthesis square root of straight x squared plus 4 end root minus straight x right parenthesis plus straight lambda over 2 equals straight lambda 
     
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow       space space space 2 left parenthesis square root of straight x squared plus 4 end root minus straight x right parenthesis space equals space 1
    rightwards double arrow                          2 straight x space equals space 7.5 straight m 
    Question 676
    CBSEENPH11018948

    In a large room, Devyendu receives direct sound wave from a source 120m away from her. She also receives waves from same source which reach her after being reflected from the 25m high ceiling. For which wavelengths will she observe constructive interference of sound?

    Solution
    The situation is as shown in the figure.
     
    The path travelled by direct wave is, 
         
                      SD = 120m 
    The path reflected by reflected wave is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                  equals 2 square root of left parenthesis 60 right parenthesis squared plus left parenthesis 25 right parenthesis squared end root 
                  equals 130 straight m 
    After reflection, the wave suffers a phase change of π radian which is equivalent to an extra path of 7/2.
    The net path difference, between two waves reaching Devyendu is,
      space space space increment equals 130 plus straight lambda over 2 minus 120 equals 10 plus straight lambda over 2 
    For constructive interference,
                         increment equals nλ. 
    Thus,
                  10 plus straight lambda over 2 equals nλ 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    ∴            space space straight lambda equals 20 comma space 20 over 3 comma space 20 over 5........
    Question 677
    CBSEENPH11018949

    A tank is filled with liquid upto height H as shown. There is an orifice at a depth h below the surface of water. Find the horizontal range of water on the ground.


    Solution
    We have, the escaping liquid analogous to horizontal projectile.
    The relation between vertical distance y by which projectile falls and horizontal distance x where it falls is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    where,  u is horizontal velocity of projection.
    Here,  
    Vertical distance, y = (H-h)
    Initial velocity,space straight u space equals space square root of 2 gh end root
    Horizontal range, x = R = ?
    therefore space space space space space space space space left parenthesis straight H minus straight h right parenthesis space equals space 1 half fraction numerator g R squared over denominator 2 g h end fraction

rightwards double arrow space space space space space space space space space straight R squared equals left parenthesis straight H minus straight h right parenthesis left parenthesis 4 straight h right parenthesis

rightwards double arrow space space space space space space space space straight R space equals space 2 square root of left parenthesis straight H minus straight h right parenthesis left parenthesis straight h right parenthesis end root
    Question 680
    CBSEENPH11018952

    What is an echo?

    Solution
    Echo is the phenomenon of repetition of sound produced due to reflection by a distant extended obstacle like cliff, hill, wall etc.
    Question 681
    CBSEENPH11018953

    What is the time of persistence of sound on our ear?

    Solution
    It is 0.1 second.
    Question 682
    CBSEENPH11018954

    What is the minimum time after which the reflected sound should be back to hear the echo?

    Solution
    The minimum time after which the reflected  sound should be back to hear the echo is 0.1s.
    Question 683
    CBSEENPH11018955

    What is the minimum distance of reflector from observer to hear an echo of articulate sound?

    Solution
    The minimum distance of reflector from observer to hear an echo of articulated sound is 17m.
    Question 684
    CBSEENPH11018956

    What is the minimum distance of reflector from observer to hear an echo of monosyllabic, disyllabic, trisyllabic sound?

    Solution
    The minimum distance of reflector from observer to hear an echo of monosyllabic sound = 34m
    The minimum distance of reflector from observer to hear an echo of disyllabic sound = 68m
    The minimum distance of reflector from observer to hear an echo of trisyllabic sound = 102m
    Question 686
    CBSEENPH11018958

    A tank is filled with liquid upto height H. There is an orifice at a depth h below the surface of water as shown. Find the ratio H/h so that horizontal range is maximum.


    Solution
    Horizontal distance at which liquid strikes the ground is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    If R (horiozontal range) is maximum, then R2 is  also maximum.
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
         rightwards double arrow space space space space straight d over dh open square brackets 4 straight h left parenthesis straight H minus straight h right parenthesis close square brackets equals 0

rightwards double arrow space space space space 4 straight d over dh left parenthesis hH minus straight h squared right parenthesis equals 0

rightwards double arrow space space space space space space space straight H minus 2 straight h equals 0

rightwards double arrow space space space space space space space straight H over straight h equals 2 
    Question 687
    CBSEENPH11018959

    What is the principle of measuring the depth of sea at a place using SONAR method?

    Solution
    The principle of measuring the depth is based on echo.
    Question 688
    CBSEENPH11018960

    What is Doppler effect?

    Solution
    Doppler effect is the change in pitch of sound when there is relative motion between the source of the sound, the observer, and the media. 
    Question 689
    CBSEENPH11018961

    What is the cause of Doppler effect when the source of sound is in motion?

    Solution
    When the source of the sound is in motion then Doppler effect is due to the change in the apparent wavelength as received by observer.
    Question 691
    CBSEENPH11018963

    What is the cause of Doppler effect when observer is in motion?

    Solution
    Doppler effect is due to the change in the number of waves received by the observer.
    Question 693
    CBSEENPH11018965

    What is the apparent frequency of wave received by stationary observer when source is moving away from observer?

    Solution

    Apparent frequency of wave is given by, 
                         straight v subscript app equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v 

    Question 696
    CBSEENPH11018968

    What is the change in frequency of wave received when source is moving away from stationary observer?

    Solution

    Change in frequency of wave received when source is moving away from stationary observer is given by, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

    Question 697
    CBSEENPH11018969

    What is the apparent frequency of wave received by stationary observer when source is moving towards observer?

    Solution

    The apparent frequency of wave received by stationary observer when source is moving towards observer is given by, 
                     straight v subscript app equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v

    Question 698
    CBSEENPH11018970

    What is the change in frequency of wave received when source is moving towards stationary observer?

    Solution

    Change in frequency of wave received when source is moving towards stationary observer is given by, 
                            increment straight v equals fraction numerator straight u over denominator straight V plus straight u end fraction straight v

    Question 699
    CBSEENPH11018971

    What is the apparent frequency of wave received by an observer when it approaches towards stationary source?

    Solution

    The apparent frequency of wave received by an observer when it approaches towards stationary source is given by, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Question 700
    CBSEENPH11018972

    What is the change in frequency of wave received when observer is moving:
    (i) towards stationary source;
    (ii) away from stationary source?

    Solution

    The change in frequency of wave received when observer is moving given by, 
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#6 {main}</pre> ; when observer is moving towards stationary source. 
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#6 {main}</pre>; when observer is moving away from the stationary source. 

    Question 701
    CBSEENPH11018973

    What is the apparent frequency of wave received by an observer when it moves away from stationary source?

    Solution

    Apparent frequency of wave received by an observer when it moves away from stationary source is given by, 
                          straight v subscript app equals fraction numerator straight V minus straight u over denominator straight V end fraction straight v

    Question 702
    CBSEENPH11018974

    Suppose a blind person is standing on a platform and train at rest is whistling. Suddenly the blind person observes the high pitch of whistle from train. What will he conclude about the motion of train?

    Solution
    Because of the high pitch of whistle from the train, the blind person will conclude that the train is moving towards him. 
    Question 703
    CBSEENPH11018975

    What do you mean by statement “Doppler effect in sound is asymmetric'?

    Solution
    When a source moves with certain velocity towards a stationary observer, the apparent frequency of wave received is different from that of when the observer moves with the same velocity towards the stationary source. This is what we mean by Doppler effect in sound is asymmetric. 
    Question 704
    CBSEENPH11018976

    Why Doppler effect in sound is asymmetric?

    Solution
    The velocity of observer or source is comparable to the velocity of sound. Hence, the doppler effect in sound is asymmetric. 
    Question 705
    CBSEENPH11018977

    Can Doppler effect in sound be symmetric?

    Solution
    Doppler effect in sound can be symmetric if the velocity of the observer or source is very small as compared to the velocity of sound. 
    Question 706
    CBSEENPH11018978

    What is the change in the frequency of wave received by observer, when an observer moves towards a stationary source of light with velocity u or the source of light moves towards stationary observer with same speed?

    Solution

    When an observer moves with velocity u or the source of light moves towards stationary observer with same speed is, 
    Change in frequency, nu space equals spaceplus straight u over straight c straight v 

    Question 707
    CBSEENPH11018979

    What is the change in the frequency of wave received by observer, when an observer moves away from a stationary source of light with velocity u or the source of light moves away from stationary observer with same speed?

    Solution

    Change in the frequency, when an observer moves away with velocity 'u' or the source of light moves away from stationary observer is given by, 
                            increment nu space equals negative space straight u over straight c straight v

    Question 708
    CBSEENPH11018980

    What do you mean by statement “Doppler effect in light is symmetric'?

    Solution
    Doppler effect in light is symmetric - When a source of light moves with certain velocity towards a stationary observer, the apparent frequency of wave received is same as that, when the observer moves with the same velocity towards the stationary source of light. 
    Question 709
    CBSEENPH11018981

    Why Doppler effect in light is symmetric?

    Solution
    Doppler effect in sound is symmetric because the velocity of light is very high. 
    Question 711
    CBSEENPH11018983

    A source emitting waves at frequency v is approaching a high wall with speed u. Find the number of beats heard by an observer moving along with source of wave.

    Solution

    Frequency of the wave = nu
    Speed of the wave with which the wave is approaching = u
    Number of beats observed along with with the
    source of the wave is, space fraction numerator 2 straight u over denominator straight V minus straight u end fraction straight v 

    Question 712
    CBSEENPH11018984

    One train is approaching and other is leaving the platform with same speed u. Both the trains whistle and emit the waves of same frequency v. What is the number of beats heard by stationary observer on the platform?

    Solution

    Speed with which the train is approaching and leaving the platform = u
    Frequency of waves = straight nu 
    The number of beats heard by stationary observer on the platform = fraction numerator 2 av over denominator straight v squared minus straight u squared end fraction straight v 

    Question 713
    CBSEENPH11018985

    Write an expression for Doppler effect when both source and observer are in motion.

    Solution

    The formula of doppler effect when both source and observer are in motion is given by,
                                straight v apostrophe equals fraction numerator straight V plus straight u subscript straight o over denominator straight V minus straight u subscript straight s end fraction straight v 
    where,
     uo and us are positive if they move towards each other.
    The values are taken as negative if they move away from each other.

    Question 714
    CBSEENPH11018986

    A sound source and listener both are at rest on the earth, but a strong wind is blowing. Is there any Doppler effect?

    Solution
    No, because there is no relative velocity between the source and observer. 
    Question 715
    CBSEENPH11018987

    A whistle is being revolved in a circle. What changes in the frequency of the whistle appear to a person standing at the center of circle?

    Solution
    Person will not observe any change in frequency, if a person is standing at the centre of circle. 
    Question 716
    CBSEENPH11018988

    What conclusion can we draw from the red shift from stars?

    Solution
    Redshift signifies that our universe is expanding.
    Question 717
    CBSEENPH11018989

    Blue shift is observed from a distant star on the earth. What is the interpretation of this observation?

    Solution
    The blue shift means that the star is moving towards the earth. 
    Question 718
    CBSEENPH11018990

    What are the characteristics of musical sound?

    Solution

    The characteristics of musical sound are: 
    (i) Loudness 
    (ii) Pitch, and
    (iii) Quality.

    Question 719
    CBSEENPH11018991

    Why the roaring of lion is different from sound of mosquito?

    Solution
    The roaring of the lion is louder and has low pitch than the sound produced by a mosquito. Hence, they sound so different. 
    Question 720
    CBSEENPH11018992

    How the notes emitted by two wires of same length and diameter but of different material can be distinguished?

    Solution
    The two wires produce notes of different pitch. Hence, the material can be distinguished. 
    Question 721
    CBSEENPH11018993

    What are the musical intervals:

    (i) major tone?

    (ii) minor tone?

    Solution

    The music intervals are:
    (i) The major tone has interval 9/8.
    (ii) The minor tone has interval 10/9. 

    Question 722
    CBSEENPH11018994

    What is the value of interval of semi tone?

    Solution
    The semitone has an interval which is equal to 16/15.
    Question 723
    CBSEENPH11018995

    What is meant by unison?

    Solution
    Two notes are said to be in unison if both of them have the same frequency. 
    Question 724
    CBSEENPH11018996

    How many keys are there in diatonic scale?

    Solution
    In a diatonic scale there are eight keys and 7 intervals. 
    Question 725
    CBSEENPH11018997

    Oil of density 0.8gm/cc is flowing through a pipe having one end A at a height 2m from ground and other end B at a height 6m from ground. The radii of ends A and B are 1.0 cm and 0.8 cm respectively. The velocity and pressure of oil at end A is 80 cm/s and 2x106dyne/cm2. Find the velocity and pressure at end B.

    Solution

    Given,
    Height of pipe from ground, h= 2m = 200 cm
    Height of the pipe, h= 6m = 600 cm
    Velocity, v= 80 cm/s
    P= 2 x 10dyne/cm
     
     straight a subscript 1 equals πr subscript 1 squared equals straight pi left parenthesis 1 right parenthesis squared equals straight pi space cm squared

straight a subscript 2 equals πr subscript 2 squared equals straight pi left parenthesis 0.8 right parenthesis squared equals 0.64 straight pi space cm squared

straight rho equals 0.8 space gm divided by cc
    
    From equation of continuity,

                a1v1=a2v
    rightwards double arrow space space space space space space space space straight pi cross times 80 equals 0.64 straight pi cross times straight v subscript 2 space

rightwards double arrow space space space space space space space space space space straight v subscript 2 equals 125 space space c m divided by straight s
    v2 is the velocity at end B.
    The bernoulli's theorem  is,
    space space space space space space space straight P subscript 1 over straight rho plus 1 half straight v subscript 1 squared plus gh subscript 1 equals straight P subscript 1 over straight rho plus 1 half straight v subscript 2 squared plus gh subscript 2
    Substituting the respective values, we have
     fraction numerator space 2 cross times 10 to the power of 6 over denominator 0.8 end fraction plus 1 half left parenthesis 80 right parenthesis squared plus 980 cross times 200

equals fraction numerator straight P subscript 2 over denominator 0.8 end fraction plus 1 half left parenthesis 125 right parenthesis squared plus 980 cross times 600

space straight P subscript 2 equals 1.683 cross times 10 to the power of 6 space dyne divided by cm squared 
    P2 is pressure at end B. 

    Question 726
    CBSEENPH11018998

    What are different intervals in diatonic scale?

    Solution
    The different intervals in diatonic scale are major tone, minor tone and semi tone.
    Question 727
    CBSEENPH11018999

    What are the intervals of equally tempered scale?

    Solution
    The intervals of equally tempered scale is equal to (2)1 /12.
    Question 728
    CBSEENPH11019000

    How many keys are there in equally tempered scale?

    Solution
    There are thirteen keys and twelve intervals in the equally tempered scale.
    Question 729
    CBSEENPH11019001

    What is the reverberation time of dead room?

    Solution
    Zero. 
    Question 730
    CBSEENPH11019002

    What is the intensity level of the painful sound?

    Solution
    The intensity level of the painful sound is 120 dB. 
    Question 731
    CBSEENPH11019003

    What is the intensity of sound having intensity level 0 bel?

    Solution
    The intensity of sound having intensity level 0 bel is 10–12 W/m2
    Question 732
    CBSEENPH11019004

    What is the intensity level of threshold of hearing?

    Solution
    The intensity level of threshold of hearing is zero bel.
    Question 733
    CBSEENPH11019005
    Question 734
    CBSEENPH11019006

    What is articulate sound? What is the minimum distance of reflector from the source to hear the echo of articulate sound?

    Solution
    The sound which requires negligible sound to produce is called as articulate sound. 
     
    Let the minimum reflector be at a distance d from the source and observer.
    The time after which the reflected sound is received back is, 
                           straight t equals fraction numerator 2 straight d over denominator straight v end fraction 
    where,
    v is the velocity of sound.
    To hear an echo, t ≥ 0.1s.
    That is, 
                      fraction numerator 2 straight d over denominator straight v end fraction greater or equal than 0.1 
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#6 {main}</pre>
    Thus, the minimum distance of reflector to produce echo should be 17m. 
    Question 735
    CBSEENPH11019007

    What is periodic motion?

    Solution
    The motion that repeats after regular interval of time is called periodic motion.
    Question 736
    CBSEENPH11019008

    What is an oscillatory motion?

    Solution
    To and fro motion of an object is called an oscillatory motion.
    Question 737
    CBSEENPH11019009

    What is time period?

    Solution
    Time period is the time taken by a particle to complete one oscillation.
    Question 738
    CBSEENPH11019010

    Define frequency.

    Solution
    Frequency is defined as the number of periodic motions executed by a body per second. 
    Question 739
    CBSEENPH11019011

    What is the condition for motion to be simple harmonic motion?

    Solution
    When the acceleration of particle is directly proportional to the displacement and directed towards the mean position, the motion is a simple harmonic motion. 
    Question 740
    CBSEENPH11019012

    In the displacement equation y = r sin(ωt/ + ϕ) what does ϕ called and what information does it give?

    Solution
    In equation y = r sin(ωt + ϕ),
    ϕ is the initial phase of particle and is known as epoch.
    It tells us about the position of particle at time, t = 0.
    Question 741
    CBSEENPH11019013

    What determines the natural frequency of an oscillator?

    Solution
    The frequency of an oscillator is determined by the inertial and elastic properties of system (oscillator). 
    Question 742
    CBSEENPH11019014

    Is there any relation between uniform circular motion and simple harmonic motion?

    Solution

    Simple harmonic motion is produced by the projection of uniform circular motion onto one of the axes in the x-y plane. 

    Question 743
    CBSEENPH11019015

    What is syllable sound? What is the minimum distance of reflector from the source to hear the echo of monosyllabic sound?

    Solution
    A sound  which requires some time to produce is known as syllable sound.
    Monosyllabic sound takes 1/5 second to produce.
    Thus the echo of monosyllabic sound will be heard only if it is received back after 1/5 second.
     
    Let the reflector be at a minimum distance of d from the source and observer.
    The time after which the reflected sound is received back is, 
                              straight t equals fraction numerator 2 straight d over denominator straight v end fraction 
    where,
    v is the velocity of sound.
    To hear echo of monosyllabic sound, t ≥ 0.2s.
    i.e.,
                fraction numerator 2 straight d over denominator straight v end fraction greater or equal than 0.2 
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#6 {main}</pre> 
    Thus the minimum distance of reflector should be 17m. 

    Question 744
    CBSEENPH11019016

    We cannot hear echo in small rooms? Explain why?

    Solution
    The minimum distance between speaker and wall should be 17m, to hear the echo of simple sound t.
    The distance between two walls of a small room is less than 17m. Hence, no echo can be heard. 
    Question 745
    CBSEENPH11019017

    A boy stands between two vertical walls, near to one wall when he claps his hands together, the echo of handclap continues for a few seconds with decreasing loudness. Why?

    Solution
    Echo continues for some time due to multiple reflections of sound from two walls. The decreasing loudness is due to the absorption of sound by walls.
    Question 746
    CBSEENPH11019018

    What is periodic motion?

    Solution
    The motion that repeats after a fixed interval of time is called periodic motion.
    Question 747
    CBSEENPH11019019

    Write down the expression for the instantaneous position of particle vibrating in simple harmonic motion.

    Solution
    Instantaneous position of the particle vibrating in S.H.M is given by,
                           y = A sin ωt
    where, 
    y = displacement of the body from the mean position, 
    a = maximum displacement or amplitude of the body, 
    omega is the angular frequency of the wave, and 
    t  is the time of oscillation. 
    Question 748
    CBSEENPH11019020

    Write down the expression for the instantaneous velocity and acceleration of particle vibrating in simple harmonic motion.

    Solution
    Instantaneous velocity, v = A ω cos ωt, and
    Acceleration of particle, a = -Aω2 sin at. 
    Question 749
    CBSEENPH11019021

    How do you define the mean position of particle executing simple harmonic motion?

    Solution
    The mean position of the particle is the position where kinetic energy and speed is maximum and acceleration of the particle is zero. 
    Question 750
    CBSEENPH11019022

    Define amplitude.

    Solution
    The maximum displacement on either side of mean position is called amplitude of motion. 
    Question 751
    CBSEENPH11019023
    Question 752
    CBSEENPH11019024

    Are all the periodic motions simple harmonic?

    Solution
     No, all the periodic motions is a simple harminic motions. In periodic motion, the particle necessarily does not go to and fro about it's mean position.  
    Question 753
    CBSEENPH11019025

    Does the time period of particle vibrating in simple harmonic motion depend upon its displacement?

    Solution
    No, time period of a particle exhibiting S.H.M does not depend upon displacement. 
    Question 754
    CBSEENPH11019026

    Where will the particle executing simple harmonic motion have acceleration maximum and speed zero?

    Solution

    For a particle exhibiting Simple Harmonic motion, acceleration is maximum and speed is zero at extreme position.

    Question 755
    CBSEENPH11019027

    Are echoes and resonant vibrations same phenomena?

    Solution
    No, both echoes and resonant vibrations are different phenomena.
    Echo is due to the reflection of sound from a distant obstacle, while resonant vibrations take place when the frequency of the two vibrating bodies becomes equal. 
    Question 756
    CBSEENPH11019028

    The bats do not have eyes, even then they turn back without striking physically against the wall. Explain how?

    Solution
    The phenomenon of echo is used by bats to turn back without striking physically against the wall.
    Bats emit ultrasonic waves, which after reflection from an obstacle is received back by them. 
    Question 757
    CBSEENPH11019029

    What is the average velocity of the particle executing simple harmonic motion over one complete oscillation?

    Solution
    Over one complete oscillation, the average velocity of the particle is zero.
    Question 758
    CBSEENPH11019030

     If the heart beats for 75 times in a minute, then what is the beat frequency of heart?

    Solution

    Number of heart beat = 75 
    Time, t = 1 minute = 60 seconds
    The beat frequency of heart =  75 over 60 equals 1.25 space Hz

    Question 759
    CBSEENPH11019031

    What is differential form of simple harmonic motion?

    Solution
    The differential form of simple harmonic motion is,
                           fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight omega squared straight y equals 0
    Question 760
    CBSEENPH11019032

    What is the net force acting on a particle when it passes through the mean position?

    Solution
    The net force acting on a particle when it passes through the mean position is zero.
    Question 761
    CBSEENPH11019033

    How long the particle takes to move from mean position to 0.707r, where r is the amplitude of oscillation?

    Solution
    The particle takes one eighth of the total time taken by particle to make one complete oscillation.
    Question 762
    CBSEENPH11019034

    An engine approaches a high vertical wall with constant speed of 90 m/sec. When it is at a distance of 1050 m from the wall, it blows a whistle, whose echo is heard after 5 second. Calculate the speed of sound in air.

    Solution
    Velocity with which the engine is approaching the wall = 90 m/s
    Let V be the velocity of sound and u be the velocity of the source.
    Echo is heard after time t.
    Let the engine whistle when it is at P and hear the echo at Q. 
     
    ∴   Distance travelled by engine in time t. 
                     PQ = ut                      ...(1) 
    Distance travelled by sound in time t. 
                     PR + RQ = Vt             ...(2) 
    On adding (1) and (2), we get
                PR + RQ + PQ = 2PR
                                      = (V+u)t 
    Here, we have
    Initial velocity, u = 90 m/s
    Distance travelled, PR = 1050 m
    Time taken to hear the echo, t = 5s 
    Therefore, 
    Speed of sound in air is given by, 
               2 x 1050 = (V + 90)x 5
    rightwards double arrow                 5V = 1650
    rightwards double arrow                   V = 330 m/s 

    Question 764
    CBSEENPH11019036
    Question 765
    CBSEENPH11019037

    A ball is dropped on a rigid floor and bounces back elastically. Will the ball execute periodic motion?

    Solution
    Yes, since the ball bounces back, the ball will execute periodic motion.
    Question 766
    CBSEENPH11019038

    A ball is dropped on a rigid floor and bounces back elastically. Will the ball execute simple harmonic motion?

    Solution
    No, the ball will not execute simple harmonic motion. 
    It is not necessary that the ball should move to and fro around it's mean position. So, it will not undergo simple harmonic motion. 
    Question 767
    CBSEENPH11019039

    Time period of oscillation of spring pendulum is 2s on the surface of the earth. What will be its time period on the surface of the moon?

    Solution
    Time period of oscillation of spring pendulum is independent of g.
    Therefore, time period remains the same i.e. 2s on the surface of the moon.
    Question 768
    CBSEENPH11019040

    What is the frequency of oscillation of the particle, if it completes one oscillation in 0 025s?

    Solution
    Frequency of oscillation of the particle = 1 over straight t equals40 Hz
    Question 769
    CBSEENPH11019041

    Can we use pendulum watch on the satellite moon?

    Solution
    Yes, we can use pendulum watch on the moon, because it is independent of the value of g. Hence, it will show correct time. 
    Question 770
    CBSEENPH11019042

    Compressional wave impulses are sent to the bottom of sea from the ship and the echo is heard after 1.8s. Find the depth of sea. Assume that the density of sea water is 1000kg/m3 and bulk modulus of water is 2.25 x 109 N/m2.

    Solution

    Given here,
    Density space of space sea space water comma space straight rho equals 1000 space kg divided by straight m cubed space

Bulk space modulus space of space water comma space straight E space equals space 2.25 space cross times space 10 to the power of 9 straight N divided by straight m squared 
    Therefore velocity of sound in sea water is, 
       straight v equals square root of straight E over straight rho end root space equals space square root of fraction numerator 2.25 cross times 10 to the power of 9 over denominator 1000 end fraction end root space equals space 1500 space straight m divided by straight s 
    Let d be the depth of the sea.
    As the echo is heard after 1.8s, therefore the wave travels 2d distance in water in time 1.8s,
    i.e.       space space space space space space space space space space space 2 straight d space equals space vt 
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#6 {main}</pre>

    Question 771
    CBSEENPH11019043

    If we cut the spring into two parts, how does the spring constant of each part change?

    Solution
    The spring constant of each part is greater than the spring constant of the original spring.
    Question 772
    CBSEENPH11019044

    The spring when loaded by 1 kg extends by 1 cm. What is the spring constant of the spring?

    Solution

    Mass of the load = 1 kg 
    Distance by which spring extends, x = 1 cm
    We have, 
                       kx = mg
    Therefore, 
    The spring constant of the spring is,
    straight k equals mg over straight x equals fraction numerator 1 cross times 9.8 over denominator 0.01 end fraction equals 980 straight N divided by straight m 

    Question 773
    CBSEENPH11019045

    What is the frequency of oscillation of a simple pendulum mounted in a cabin falling freely?

    Solution
    The frequency of oscillation of a simple pendulum mounted in a cabin falling freely is zero.
    Question 774
    CBSEENPH11019046

    Does the time period of oscillation of simple pendulum depend on the mass of bob? 

    Solution
    No, the time period of oscillation of simple pendulum does not depend on the mass of the bob. time period depends on acceleration and displacement. 
    Question 775
    CBSEENPH11019047

    A wrist watch falls freely from the top of a tower. Does the watch give correct time during free fall?

    Solution
    Time period of wrist watch is independent of g.
    Therefore, watch gives correct time during free fall.
    Question 776
    CBSEENPH11019048

    A pendulum clock is in a lift that descends at constant velocity. Does it keep correct time?

    Solution
    Acceleration of the lift is zero, therefore the pendulum clock will show correct time. 
    Question 777
    CBSEENPH11019049

    By what factor the time period of simple pendulum changes when the length of pendulum is decreased to one fourth?

    Solution
    When length of the pendulum is decreased to one-fourth, the time period of simple pendulum will decrease by factor two. 
    Question 778
    CBSEENPH11019050

    What is an ideal simple pendulum?

    Solution
    An ideal simple pendulum is a point mass attached with massless, flexible and inextensible string suspended from a rigid support.
    Question 779
    CBSEENPH11019051

    When the oscillations of simple pendulum are simple harmonic oscillations?

    Solution
    The oscillations of the simple pendulum are simple harmonic oscillations if the amplitude of oscillations is very small. 
    Question 780
    CBSEENPH11019052
    Question 781
    CBSEENPH11019053

    What should be the minimum distance between source and reflector to hear the distinct echo of a simple sound?

    Solution

    The time period of persistence of ear for simple sound =  0.1s.
    Therefore, to hear the distinct echo of simple sound, the minimum time taken by the sound to reach the listener after reflection should be 0.1s.
    Let d be the distance between source and reflector.
    The time taken by sound to reach the listener after reflection = t
    Sound will travel the total distance of 2d. 
    ∴           2d = vt 
    rightwards double arrow          d = vt/2 
    Now, putting the values,
               v = 340 m/s,    t = 0.1s 
    We get,
                   d = 17 m 
    Therefore minimum distance of reflector from the source should be 17m. 

    Question 782
    CBSEENPH11019054

    N springs each of spring constant K are arranged in parallel. What is the spring constant of the parallel combination?

    Solution
    The spring constant of parallel combination, keq= k1 + k2.
    Therefore, spring constant of parallel combination = NK. 
    Question 783
    CBSEENPH11019055

    What is the effect on the time period of oscillation of a simple pendulum, when the bob is immersed in a liquid?

    Solution
    When the bob is immersed in a liquid, time period of oscillation of a simple pendulum increases. 
    Question 784
    CBSEENPH11019056

    What is the effect on the time period of oscillation of a spring pendulum, when the mass of the pendulum is immersed in a liquid?

    Solution
    Immersing the mass of the pendulum in a liquid does not affect the period of oscillation of a spring pendulum. Therefore, the time period of the pendulum remains the same. 
    Question 785
    CBSEENPH11019057

    What is the effect on the time period of oscillation of a simple pendulum, when the temperature of the wire from which the bob is suspended is increased?

    Solution
    When temperature of the wire from which the bob is suspended is increased, time period of simple pendulum will increase.
    Question 786
    CBSEENPH11019058

    What is the change in the time period of oscillation of spring pendulum when taken on the Jupiter?

    Solution

    When pendulum is taken on jupiter, there will be no change in the time period of oscillation of spring pendulum. 
    The value of 'g' is negligible on jupiter. 

     
    Question 787
    CBSEENPH11019059

    What provides the restoring force for simple harmonic oscillations in case of cylinder floating in the liquid?

    Solution
    In case of a cylinder floating in the liquid,  upward thrust provides the necessary restoring force for oscillations. 
    Question 788
    CBSEENPH11019060

    A man standing between two parallel cliffs clap and hears two echoes, one 2-4s and second at 3.6s after clapping. If the velocity of sound is 340m/s, find the distance between the two cliffs.

    Solution
    Let the man be at a distance of d1 from one cliff and d2 from the other cliff.
    The first echo is heard after reflection from the first cliff and the second reflection is heard after reflection from the second cliff. 

    As the first echo is heard at time t1 = 2.4s,
    ∴            2 straight d subscript 1 equals vt subscript 1 
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#6 {main}</pre> 
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#6 {main}</pre> 
    ∴             2 straight d subscript 2 equals vt subscript 2 
                   space space space straight d subscript 2 equals vt subscript 2 over 2 equals fraction numerator 340 cross times 3.6 over denominator 2 end fraction equals 612 straight m 
     Thus, the distance between two cliffs is given by, 
         straight d equals straight d subscript 1 plus straight d subscript 2 equals 408 plus 612 equals 1020 straight m 
    Question 789
    CBSEENPH11019061

    Is it true that potential energy at mean position is zero?

    Solution
    The potential energy at mean position may or may not be zero. But P.E has minimum value at mean position. 
    Question 790
    CBSEENPH11019062

    What is the kinetic energy of particle at extreme position?

    Solution
    At the extreme position, kinetic energy of the particle is zero. 
    Question 791
    CBSEENPH11019063

    How does the total energy of a particle executing simple harmonic motion depend on the instantaneous position of the particle?

    Solution
    The total energy of a particle executing simple harmonic motion does not depend on the instantaneous position of the particle. 
    Question 792
    CBSEENPH11019064

    How does the potential energy of a particle executing simple harmonic motion depend on the instantaneous position of the particle?

    Solution
    The potential energy of a particle executing simple harmonic motion at any instant is directly proportional to the square of the instantaneous position of the particle. 
    Question 793
    CBSEENPH11019065

    A particle is vibrating with frequency v. What is the frequency of variation of kinetic energy?

    Solution
    Vibrating frequency of the particle = v
    Frequency of variation of kinetic energy = 2v 
    Question 794
    CBSEENPH11019066

    A particle is vibrating with frequency v. What is the frequency of variation of total energy?

    Solution
    The frequency of vibration of total energy is equal to zero. 
    Question 795
    CBSEENPH11019067
    Question 797
    CBSEENPH11019069

    A particle of mass 2kg is executing simple harmonic motion. The total energy of particle is 16J. What is potential energy of particle when the velocity of particle is 4 m/s?

    Solution
    For a Simple Harmonic Motion, the potential energy of the particle when the velocity is 4 m/s is zero. 
    Question 798
    CBSEENPH11019070

    What are free vibrations? 

    Solution
    The body is said to execute free vibrations, if a body after being disturbed from its mean position of equilibrium is allowed to vibrate freely without interference of any other vibrating body. 
    Question 799
    CBSEENPH11019071

    How is the amplitude of free oscillations change with time?

    Solution
    The amplitude of free oscillations remains constant, with time. 
    Question 800
    CBSEENPH11019072

    What are forced vibrations?

    Solution
    When a body is continuously driven by an external force, and if it is maintained in a state of vibration, it is known as forced vibrations. 
    The body is maintained in a state of vibration by periodic force of frequency other than natural frequency of vibration of body. 

    Question 801
    CBSEENPH11019073

    What is the nature of motion of forced oscillations?

    Solution
    Forced oscillations are simple harmonic in nature. 
    Question 803
    CBSEENPH11019075

    What are resonant vibrations?

    Solution
    Resonant vibrations is the phenomenon of producing vibratory motion in a body by external period force.
    The frequency of the vibrating body is same as that of the natural frequency of the body and this condition is called as resonance. 
    The vibrations are called as resonant vibrations. 
    Question 804
    CBSEENPH11019076

    What is Doppler’s effect? Derive an expression for the apparent frequency received by observer when alone source is in motion. (Assume media to be at rest).

    Solution

    Doppler's effect is the change in the pitch when there is a relative motion between source and observer.
    Let a source which is at rest be emitting waves of frequency f and wavelength λ.
    Let V be the velocity of the wave emitted by source,
                                V=fλ. 
            
    Let the distance between source and observer be V.
    Therefore, the wave will take one second to reach the observer.
    Now let the source start moving with velocity u towards the observer.
    The wave, which was just emitted by the source, starts moving will take one second to reach the observer and in a second source, will travel 'u' distance. 
    The source will be at a distance V–u just after 1 second and there will be f waves in V–u distance.
    Therefore wavelength of wave when source starts moving is, 
                        straight lambda apostrophe equals fraction numerator straight V minus straight u over denominator straight f end fraction 
    The apparent frequency space straight f apostrophe is, 
                        straight f apostrophe equals fraction numerator straight V over denominator straight lambda apostrophe end fraction equals fraction numerator straight V over denominator straight V minus straight u end fraction straight f 
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#6 {main}</pre>               ...(1) 
    When source starts moving away from observer then replace u by –u.
    Therefore, the apparent frequency is, 
    straight f apostrophe equals fraction numerator straight V over denominator straight V plus straight u end fraction straight f                                    ...(2) 
    From equation (1), the apparent frequency increases when the source moves towards the observer.
    From equation (2) the apparent frequency decreases when the source moves away from the observer. 

    Question 805
    CBSEENPH11019077

    Show that y(t) = A sin ωt is periodic function of period 2 ϕ / ω.

    Solution
    Given,
    y(t) = A sin ωt
    We have to prove that y(t) is a  periodic function of period 2 ϕ / ω.
    Therefore, 
    straight y left parenthesis straight t right parenthesis space equals space straight y left parenthesis straight t plus fraction numerator 2 straight pi over denominator straight omega end fraction right parenthesis
Now comma

space space straight y left parenthesis straight t plus fraction numerator 2 straight pi over denominator straight omega end fraction right parenthesis equals straight A space sinω left parenthesis straight t plus fraction numerator 2 straight pi over denominator straight omega end fraction right parenthesis
space space space space space space space space space space space space space space space space space space space space equals space straight A space sin space left parenthesis ωt plus 2 straight pi right parenthesis

space space space space space space space space space space space space space space space space space space space space equals space straight A space sin space ωt

space space space space space space space space space space space space space space space space space space space space equals space straight y left parenthesis straight t right parenthesis
    Therefore, y(t) is periodic function of period 2π /ω.
    Question 806
    CBSEENPH11019078

    What is a periodic motion? Give examples of periodic motion.

    Solution

    Periodic motion is a motion which repeats itself after a regular interval of time.
    Examples of periodic motion are:

    (i) Revolution of planets around the sun.

    (ii) Motion of the simple pendulum.

    (iii) Motion of the moon around the earth.

    (iv) Motion of the wings of a bee.

    Question 807
    CBSEENPH11019079

    Derive an expression for the apparent frequency received by observer when alone observer is in motion. (Assume media to be at rest). 

    Solution
    Let a source at rest be emitting the waves of frequency f and wavelength X.
     
    When the observer is also at rest, he is receiving f waves in one second.
    Now let the observer start moving with velocity u towards a stationary source.
    When the observer starts moving with velocity u, he will receive wavesspace straight u over straight lambda more than f. 
    Total number of waves received by observer in 1 second is, 
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#6 {main}</pre>  
    Frequency is the number of waves received by the observer in 1 second.
    Therefore, apparent frequency of wave received by observer, 
                           straight f apostrophe equals fraction numerator straight V plus straight u over denominator straight V end fraction straight f 
    If the observer moves away from the source, the observer will receive wavesstraight u over straight lambda equals straight u over straight V straight f less than f. 
    Therefore apparent frequency, when observer moves away is, 
                            straight f apostrophe equals fraction numerator straight V minus straight u over denominator straight V end fraction straight f
    Question 808
    CBSEENPH11019080

    How does the apparent frequency change when there is relative motion between the source of sound and observer?

    Solution
    When the source of sound approaches towards stationary observer or observer approaches towards the source, the apparent frequency increases.
    When the observer and the source of the sound, recedes away from each other, the apparent frequency decreases. 
    Question 809
    CBSEENPH11019081

    What is oscillatory motion? Give examples.

    Solution

    To and fro motion about its mean position is called oscillatory motion.
    Examples of oscillatory motion are:

    (i) Motion of a simple pendulum.

    (ii) Motion of a body floating in liquid.

    (iii) Motion of liquid in a U-tube.

    Question 810
    CBSEENPH11019082

    Are all the periodic motions oscillatory? Is the vice versa true?

    Solution
    All the periodic motions are not necessarily oscillatory.
    For example, motion of planet around the sun is periodic. But the motion is not to and fro about a mean position. Hence, the motion of planet around the sun is not oscillatory.
    All the oscillatory motions are periodic because oscillatory motion repeats after regular interval of time. Therefore vice versa is true.
    Question 811
    CBSEENPH11019083

    State some application of Doppler’s effect.

    Solution

    Some application of Doppler's effect are: 
    (i) It is used to estimate the speed of distant stars, the speed of aeroplane or submarines.
    (ii) Observing red shift or blue shift, we can find the direction of motion of the object.
    (iii) Using Doppler effect, we can find the magnitude and direction of revolution of the sun.

    Question 812
    CBSEENPH11019084

    Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

    Solution

    (a) It is not periodic motion because motion does not repeat itself after a regular interval of time. 

    (b) the given graph illustrates a periodic motion, which is repeating itself after every 4 seconds. 

    (c) The given graph does not exhibit a periodic motion because the motion is repeated in unequal intervals of time.

    (d) It is periodic motion which is repeating itself in 5 seconds.

    Question 813
    CBSEENPH11019085

    The whistle of an approaching engine appears to be shriller than that of a preceding engine. Why?

    Solution
    When the source moves towards the observer the apparent frequency is, 
                         space space space space straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u subscript straight s end fraction straight v 
    When source moves away from the observer, then apparent frequency is, 
                             straight v subscript 2 equals fraction numerator straight V over denominator straight V plus straight u subscript straight s end fraction straight v 
    The ratio of this frequency is given by, 
        space space straight v subscript 1 over straight v subscript 2 equals fraction numerator straight V plus straight u subscript straight s over denominator straight V minus straight u subscript straight s end fraction greater than 1 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Question 814
    CBSEENPH11019086

    If we move away from the source of sound with speed of sound, we hear nothing. Why?

    Solution
    Apparent frequency when the observer moves away from the source is, 
                        straight v apostrophe equals fraction numerator straight V minus straight u subscript straight o over denominator straight V end fraction straight v 
    when,
         straight V equals straight u subscript straight o comma space then 
         straight v apostrophe equals 0 
    A note of zero frequency does not produce any sound.
    Therefore, we will hear nothing. 
    Question 815
    CBSEENPH11019087

    Show that when an observer moves towards or away from the source with same speed, the apparent change in frequency is same.

    Solution
    Let a source of sound be emitting the waves of frequency v.
    Let V be the velocity of sound.
    The apparent frequency of wave received by observer when observer moves with velocity u is, 
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>         [moves towards source] 
            straight v subscript 2 equals fraction numerator straight V minus straight u over denominator straight V end fraction straight v          [moves away from source] 
    ∴               increment straight v subscript 1 equals straight v subscript 1 minus straight v equals straight u over straight V straight v           ...(1) 
    and    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>        ...(2) 
    Hence the result. 
    Question 816
    CBSEENPH11019088

    Show that when source moves towards the stationary observer, the apparent increase in frequency is more than the apparent decrease in frequency if the source moves away from the source with same speed.

    Solution
    Let a source of sound be emitting the waves of frequency v.
    Let V be the velocity of sound.
    The apparent frequency of wave received by observer when source moves with velocity u is, 
                straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v        [moves towarads source] 
                straight v subscript 2 equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v      [moves away from source] 
    ∴         space space space increment straight v subscript 1 equals straight v subscript 1 minus straight v equals fraction numerator straight u over denominator straight V minus straight u end fraction straight v                ...(1) 
    and        increment straight v subscript 2 equals straight v subscript 2 minus straight v equals negative fraction numerator straight u over denominator straight V plus straight u end fraction straight v        ...(2) 
                  fraction numerator open vertical bar increment straight v subscript 1 close vertical bar over denominator open vertical bar increment straight v subscript 2 close vertical bar end fraction equals fraction numerator straight V plus straight u over denominator straight V minus straight u end fraction 
    rightwards double arrow            open vertical bar increment straight v subscript 1 close vertical bar space greater than space open vertical bar increment straight v subscript 2 close vertical bar 
    Hence the result.
    Question 817
    CBSEENPH11019089

    A whistle is being revolved in a circle. What changes in the frequency of the whistle appear to a person standing outside the circle?

    Solution
    The person will observe the frequency to be changing between maximum and minimum values.
    During one complete rotation of whistle, the person will observe maximum frequency higher than actual frequency once.
    The minimum frequency lower than the actual frequency will be observed once 
    Once minimum frequency lower than actual frequency and two times actual frequency. 
    Question 818
    CBSEENPH11019090

    The motion of particle moving in a circle with constant speed is not simple harmonic. Show that the motion of its shadow on axis parallel to diameter is simple harmonic.

    Solution
    Let a particle P be revolving in a circle (known as reference circle) of radius ‘r’ with constant angular velocity straight omega.
                  
    If T is the time taken by the particle to complete one revolution, then
    Angular velocity is given by, 
                          space space space space space straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction

    Let the particle P start from S.
    The angular position of the particle at any instant t is given by,

                            θ = ωt + ϕ
    Since the motion of the particle in the circle is not to and fro, it is not a simple harmonic motion. 
    Let a parallel beam of light be incident on the particle P and take the shadow on Y1Y2 axis.
    As the particle revolves, its shadow vibrates between K and L.
    If another particle Q is allowed to move along with the shadow, it will move to and fro along the shadow.
    So, position of the particle Q along Y1Y2 axis w.r.t. O is given by,


    space space space space space space space O Q equals straight O apostrophe straight N equals r s i n left parenthesis omega t plus straight ϕ right parenthesis

rightwards double arrow space space space space space space straight y equals r s i n left parenthesis omega t plus straight ϕ right parenthesis
    Since the motion of the particle Q along the shadow is represented by simple harmonic functions, hence the motion of the particle Q along the shadow is simple harmonic motion. 
    Question 819
    CBSEENPH11019091

    Two trains A and B on platform are at a certain distance apart and sounding horn at same frequency. When a person standing between two trains starts running from train A towards B, he hears beats. Explain why.

    Solution
    When the person starts running from A to B, the frequency of the sound of the horn from train A received by the person will be more, and that of from train B will be less. The person hears the beats, due to the difference in the frequency of sound received.
    Question 820
    CBSEENPH11019092

    A particle is constrained to move along the shadow of particle revolving in a circle with constant velocity on its diameter. How will the amplitude, frequency and phase of vibrating particle be related with the parameters of particle revolving in the circle?

    Solution
    Given, a particle is revolving in a circle with constant angular frequency.
    Then the amplitude of vibrating particle is equal to the radius of circle.
    And, the frequency of oscillation is equal to the frequency of revolution.
    Phase of vibrating particle is equal to the angular displacement of revolving particle.
    Question 821
    CBSEENPH11019093

    Two sirens A and B are at a certain distance apart and sounding at same frequency. When a person standing between two trains starts running along the right bisector of AB, he does not hear any beat. Explain why.

    Solution
    The apparent frequency of sound observed by him from both the sources will be same, when the person starts running along the right bisector of AB.
    That is why the person does not hear any echo.
    Question 822
    CBSEENPH11019094

    When the source of sound and observer do not move relatively, but the media moves along the line in which the sound travels, the frequency of sound heard by the observer is different from actual. Explain why. 

    Solution
    When the media moves along the line in which the sound travels, then the velocity of sound is, 
                           v = V±u

    where,
    v is the velocity of sound in still media,
    uis the velocity of media.
    Due to change of velocity of sound, the apparent frequency of sound is different from actual frequency. Hence the frequency of the sound observed is different from the actual frequency. 
    Question 823
    CBSEENPH11019095

    What do you mean by simple harmonic motion? Write down the differential equation for simple harmonic motion.

    Solution
    To and fro motion of an object in which the acceleration of particle is always directly proportional to displacement, and directed towards mean position is a simple harmonic motion.
    The differential equation of simple harmonic motion is given by, 
                            fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight omega squared straight y equals 0
    Question 824
    CBSEENPH11019096

    What is the difference between epoch and phase?

    Solution
    Epoch is the initial phase of a particle and it remains constant all the time for a given motion.
    Phase is the angular position of the particle on the circle which represents the given Simple Harmonic Motion.
    Phase would increase continuously with time going up by 2π radians for every oscillation.
    Question 825
    CBSEENPH11019097

    Car A is following the car B. Both are moving with same velocity. The driver of car A claps at regular interval T. What will be the time interval between two successive claps as heard by the driver of car B?

    Solution
    The relative velocity of car B w.r.t. car A is zero.
    Therefore, the apparent frequency of clapping is same as that of actual.
    Hence, the time interval between two successive claps heard by the driver of car B is also T.
    Question 826
    CBSEENPH11019098

    What type of force is necessary for the particle to execute simple harmonic motion?

    Solution
    A particle executing simple harmonic motion has acceleration directly proportional to displacement and is directed towards the mean position.
    Mathematically, we can write, 
     space space space space space space space space space space space space space space space straight alpha proportional to negative straight y

U s i n g space N e w t o n apostrophe s space f i r s t space L a w space o f space m o t i o n comma space

space space space space space space space space space space space space space space space space space space straight F equals m a

therefore space space space space space space space space space space space space straight F space proportional to negative m y space

rightwards double arrow space space space space space space space space space space space space space straight F space proportional to negative straight y 

    Therefore, the force required to execute simple harmonic motion is directly proportional to displacement and directed towards the mean position.
    The force directed towards mean position is called restoring force.
    Thus a restoring force is directly proportional to displacement and is necessary for the particle to execute simple harmonic motion.
    Question 827
    CBSEENPH11019099

    What is the change in frequency when both source and observer move with the same velocity in the same direction?

    Solution
    The apparent frequency when both source and observer move in the same direction is, 
                        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Here,
    Velocity of the observer and source are equal.
    That is,
                             space straight u subscript straight o equals straight u subscript straight s 
    ∴                         straight v apostrophe equals straight v 
    i.e. there is no change in frequency. 
    Question 828
    CBSEENPH11019100

    What do you mean by red shift?

    Solution
    The spectral lines are observed to shift towards the higher wavelength when the stars are moving away from the earth. That is, the stars are moving towards the red part of the spectrum due to doppler effect. 
    This is called as red shift. 
    Question 829
    CBSEENPH11019101

    What do you mean by blue shift?

    Solution
    When the source moves towards the observer, the spectral lines received by an observer from the source are observed to shift towards the shorter wavelength. That is, towards the blue part of the spectrum due to Doppler effect.
    This process is called blue shift. 
    Question 830
    CBSEENPH11019102

    A car approaching with velocity u towards the stationary observer crosses past him. What is the apparent change in frequency of the note when car is crossing the observer?

    Solution
    Consider,
    v = frequency of note, and
    V =  velocity of note in air.
    The apparent frequency of note while car is approaching before crossing, 
                    straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v 
    The apparent frequency of note after crossing, 
                     straight v subscript 2 equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v 
    The change in the apparent frequency of note on crossing, 
                       increment straight v equals straight v subscript 2 minus straight v subscript 1
                           equals open parentheses fraction numerator straight V over denominator straight V plus straight u end fraction straight v close parentheses minus open parentheses fraction numerator straight V over denominator straight V minus straight u end fraction straight v close parentheses
                          space space equals negative fraction numerator 2 uV over denominator straight V squared minus straight u squared end fraction straight v 
    Question 831
    CBSEENPH11019103

    Why are soldiers ordered to move out of step while crossing a bridge?

    Solution
    Every object in the universe has a natural frequency of it's own. When frequency of external force becomes equal to natural frequency of oscillation, resonance takes place. During the condition of resonance, amplitude of vibration becomes maximum.
    Bridge can be considered as stretched string. If soldiers march in steps and frequency of marching steps is the same as that of frequency of bridge, then bridge will be set into resonant vibrations and will collapse. That is why soldiers are asked to break their steps while moving on the bridge. 
    Question 832
    CBSEENPH11019104

    An engine blowing a whistle of frequency 288Hz moves with speed 15m/s. Find the apparent frequency of whistle heard by an observer at rest to which the engine is approaching. The velocity of sound in air is 330 m/s.

    Solution
    The apparent frequency of wave received by stationary observer when source moves towards the observer is given by, 
                              straight v apostrophe equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v 
    We have,
    Velocity space of space sound space in space air comma space space straight V equals 330 straight m divided by straight s

Speed space of space engine comma space straight u equals 15 straight m divided by straight s 
    Frequency of the engine, straight v equals 288 HZ 
    Apparent frequency of whistle heard by an observer is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Question 833
    CBSEENPH11019105

    At a certain speed of a bus, the body of the bus starts vibrating strongly. Why?

    Solution
    The frequency of the bus engine becomes equal to the natural frequency of it's frame at a certain speed. This is the condition of resonance and resonant vibrations of the frame takes place. Therefore, we feel strong vibrations of the bus. 
    Question 834
    CBSEENPH11019106

    The pitch of sound of a source at rest appears to drop by 6% to a moving person. Find the velocity of person. The velocity of sound is 330m/s

    Solution
    The apparent frequency is less than actual frequency.
    Therefore, the person is moving away from the source.
    We know, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre>
    Here,   fraction numerator straight v apostrophe over denominator straight v end fraction equals 94 over 100 space space space and space space space space straight V space equals space 330 straight m divided by straight s
    ∴                space space space 94 over 100 equals fraction numerator 330 minus straight u over denominator 330 end fraction 
    i.e.,  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>, is the velocity of the person. 
    Question 835
    CBSEENPH11019107

    Why is loud sound heard at resonance?

    Solution
    During the condition of resonance, compression falls on compression and rarefaction falls over rarefaction, due to which amplitude of vibration of particle increases.
    Since I proportional toA2, therefore maximum sound is heard at resonance.
    Question 836
    CBSEENPH11019108

    Glass window panes are some time broken by an explosion even taking place far away. Why?

    Solution
    When explosion takes place at a far away place, the disturbance reaching the glass panes sets the glass panes into vibration.
    If the frequency of vibration is equal to the natural frequency of window pane, then resonant vibration takes place and hence window panes break.
    Question 837
    CBSEENPH11019109

    An observer at rest is standing near the train which blows the signal of frequency 400 Hz. Another train is approaching the observer with velocity 5 m/sec which also blows the signal with frequency 400 Hz. If the observer hears 6 beats in one second then what is the velocity of sound?

    Solution
    The apparent frequency of signal from train approaching towards observer is given by, 
    straight v apostrophe equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v                         ...(1) 
    where,
    v is the velocity of sound and u is the velocity of the source.
    The actual frequencies of the signal from both the sources are same.
    Therefore, number of beats heard by the observer is, 
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow         straight b equals fraction numerator uv over denominator straight V minus straight u end fraction 
    rightwards double arrow         bV minus bu equals uv  
    rightwards double arrow          straight V equals fraction numerator straight b plus straight v over denominator straight b end fraction straight u 
    Here, we have
         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> ,         straight u equals 5 straight m divided by straight s 
    Therefore,
    Velocity of the sound is, 
    straight V equals fraction numerator 6 plus 400 over denominator 6 end fraction cross times 5 equals 338.33 space straight m divided by straight s 
    Question 838
    CBSEENPH11019110

    An aeroplane passing over the building some time causes rattling of windows of the building. Explain why.

    Solution
    When the frequency of the sound produced by the engine of aeroplane passing over building is equal to the natural frequency of windows of the building, then resonance takes place and causes the rattling of window.
    Question 839
    CBSEENPH11019111

    How earthquakes some times cause disaster?

    Solution
    When the waves produced during earthquake have same frequency as that of the natural frequency of buildings, resonance takes place.
    As a result, the buildings start vibrating with large amplitude and the amplitude is so large that the buildings collapse. In this way the earthquakes cause big disaster.
    Question 840
    CBSEENPH11019112

    Why does the motion of simple pendulum in due course stop while oscillating in air?

    Solution
    While the simple pendulum is oscillating in air, as a result of air friction, the energy of oscillator is dissipated into heat energy.
    When whole of the energy of simple pendulum is dissipated, the simple pendulum stops oscillating. 
    Question 841
    CBSEENPH11019113

    Two engines cross each other travelling in opposite directions at 108km/hr. One engine sounds a whistle of frequency 540Hz. What is the frequency of sound heard by the other engine before they cross each other? Does the frequency of sound heard by the other engine remain same after crossing? Take speed of sound 330m/s.

    Solution
    Before crossing, both the engines are approaching each other.
    Thus, the apparent frequency of sound is, 
            straight v subscript 1 equals fraction numerator straight V plus straight u subscript straight o over denominator straight V minus straight u subscript straight s end fraction straight v 
    Here,   straight V equals 330 space straight m divided by straight s 
               straight u subscript straight o equals straight u subscript straight s equals 108 space km divided by hr space equals space 30 space straight m divided by straight s space

straight v space equals space 540 space Hz 
    ∴ Apparent frequency, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    After crossing, both the engines are moving away from each other.
    Thus, the apparent frequency of sound is, 
                   straight v subscript 1 equals fraction numerator straight V minus straight u subscript straight o over denominator straight V plus straight u subscript straight s end fraction straight v 
    ∴          space space space space straight v subscript 1 equals fraction numerator 330 minus 30 over denominator 330 plus 30 end fraction cross times 540 equals 450 Hz 
    The frequency of sound heard by the other engine changes after crossing. 

    Question 844
    CBSEENPH11019116

    Can we use pendulum watch on the satellite moon? Explain.

    Solution
    The swing of the pendulum watch acts downwards because of the force of gravity. Hence, we cannot use the pendulum watch on the satellite moon as 'g' is absent on the moon. 
    Question 845
    CBSEENPH11019117

    At what position the tension in the string of simple pendulum is (i) maximum (ii) minimum when oscillating?

    Solution
    Tension in the string of a simple pendulum is, 
    (i) maximum at the mean position
    (ii) minimum at the extreme position.
    Question 846
    CBSEENPH11019118

    The wavelength of yellow sodium line (5896A) emitted by a star is red shifted to 6010 Å. What is the component of the star's recessional velocity along the line of sight? (For small recessional speeds, you may use a formula for Doppler effect analogous to that for sound). Speed of light is 30x108m/s.

    Solution
    Change in wavelength, because of red-shift, measured by observer (stationary) when the source moves away is given by,
                           <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    where,
     u is the velocity of the source.
    Therefore,
    Component of star's recessional velocity along the line of sight is,
     straight u equals straight c fraction numerator increment straight lambda over denominator straight lambda end fraction equals fraction numerator 6010 minus 5896 over denominator 5896 end fraction cross times 3 cross times 10 to the power of 8
                               almost equal to space 5.8 space cross times space 10 to the power of 6 space straight m divided by straight s
    Question 847
    CBSEENPH11019119

    What is the phase difference between displacement and acceleration of particle executing simple harmonic motion?

    Solution
    Phase difference between displacement and acceleration of particle executing S.H.M is straight pi
    Question 848
    CBSEENPH11019120

    A particle is executing simple harmonic motion with amplitude A. At what displacement from the mean position, the energy of particle is half kinetic and half potential?

    Solution

    Amplitude = A
    Displacement from the mean position, where the energy is half kinetic energy and half potential =space plus-or-minus fraction numerator A over denominator square root of 2 end fraction

    Question 849
    CBSEENPH11019121

    A source of sound produces waves of wavelength 28cm. The source is moving with velocity 12 m/s due east. Find the apparent wavelength of wave received by observer on: (i) east (ii) west of source. Take velocity of sound in air as 336 m/s.

    Solution

    Here, we have
    Wavelength,  straight lambda equals 28 cm space equals space 0.28 straight m 
    Velocity of the sound in air, straight V equals 336 straight m divided by straight s comma space space space space space straight u space equals space 12 space straight m divided by straight s 
    (i) When the observer is on the east of train, the train moving due east will approach the observer.
    The apparent wavelength of wave received by observer is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    (ii) When the observer is on the west of train, the train moving due east will move away from the observer.
    The apparent wavelength of wave received by observer is, 
              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
                 equals 0.29 space straight m space equals space 29 space cm 
                   

    Question 850
    CBSEENPH11019122

    Where will the particle executing simple harmonic motion have acceleration maximum and speed zero?

    Solution
    At the extreme position, acceleration is maximum and speed is zero. 
    Question 851
    CBSEENPH11019123

    What is the kinetic energy of particle at extreme position?

    Solution
    The kinetic energy of particle at extreme position is zero.
    Question 852
    CBSEENPH11019124

    The position of the particle oscillating along X-axis is given by x = 4(sin100 πt) + 2. Find the extreme positions in which the particle is oscillating.

    Solution
    The extreme positions of the oscillating particle are the maximum and minimum value of x.
    Maximum value of sin100 πt is +1 and minimum value is – 1.
    Therefore,
    straight X subscript min equals 4 left parenthesis negative 1 right parenthesis plus 2 equals negative 2
straight X subscript max equals 4 left parenthesis plus 1 right parenthesis plus 2 equals 6
    Thus the particle oscillates between the position -2 and 6.
    Question 853
    CBSEENPH11019125

    A policeman on duty detects a decrease of 12% in the pitch of horn from a car crossing him. Find the velocity of car. The velocity of sound in air is 340m/s.

    Solution
    The apparent pitch of horn while car is approaching before crossing is, 
                        straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v 
    The apparent pitch of horn after crossing is, 
     
                        straight v subscript 2 equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v 
    The percentage drop in pitch is given by,
           fraction numerator straight v subscript 1 minus straight v subscript 2 over denominator straight v subscript 1 end fraction cross times 100 equals open parentheses 1 minus straight v subscript 2 over straight v subscript 1 close parentheses cross times 100 
                               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                               equals fraction numerator 2 straight u over denominator straight V plus straight u end fraction 100 
    We have, the percentage drop in pitch is 12%.
    So,
                       
                      12 equals fraction numerator 2 straight u over denominator 340 plus straight u end fraction 100 

    rightwards double arrow           <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow                      188 straight u equals 12 cross times 340 
    rightwards double arrow                         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>, is the velocity of the car. 
    Question 854
    CBSEENPH11019126

    The sirens of two fire engines have a frequency of 650 Hz each. A man hears the sirens from two engines, one moving with speed 10 m/sec and second with speed 54 km/h. What is the difference in the frequency of two sirens heard by the man, when:
    (i) both the engines are approaching him,
    (ii) both the engines are receding from him?
    Take the speed of sound to be 330 m/s.

    Solution

    The apparent frequency of siren from approaching engine is, 
                     straight v apostrophe equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v 
    and that from receding engine is, 
                      straight v apostrophe equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v 
    (i) The apparent frequency of sirens from approaching engines are, 
                  straight v subscript 1 apostrophe equals fraction numerator straight V over denominator straight V minus straight u subscript 1 end fraction straight v equals fraction numerator 330 over denominator 330 minus 10 end fraction 650 
                                     equals 330 over 320 650 equals 670.31 space Hz 

    and         straight v subscript 2 apostrophe equals fraction numerator straight V over denominator straight V minus straight u subscript 2 end fraction straight v equals fraction numerator 330 over denominator 330 minus 15 end fraction 650 
                                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    The difference in the frequencies of wave received from two engines is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    (ii) The apparent frequency of sirens from receding engine are given by, 
          straight v subscript 1 apostrophe apostrophe space equals space fraction numerator straight V over denominator straight V plus straight u subscript 1 end fraction straight v space equals space fraction numerator 330 over denominator 330 plus 10 end fraction 650 
                                  equals 330 over 340 650 equals 630.88 Hz
    and  straight v subscript 2 apostrophe apostrophe space equals space fraction numerator straight V over denominator straight V plus straight u subscript 2 end fraction straight v equals fraction numerator 330 over denominator 330 plus 15 end fraction 650 
                                 equals 330 over 345 650 equals 621.74 Hz
    The difference in the frequency of two sirens as heard by the man is given by, 
                    space space space straight v subscript 1 apostrophe apostrophe minus straight v subscript 2 apostrophe apostrophe space equals space 630.88 minus 621.74 space equals space 9.14 Hz

    Question 855
    CBSEENPH11019127

    The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of source. Find the velocity of source. Take the velocity of sound as 350m/s.

    Solution
    The apparent frequency of wave perceived by observer when source of sound approaches the observer is, 
    straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v                     ...(1) 
    The apparent frequency of wave perceived by observer when source of sound moves away from the observer is, 
    straight v subscript 2 equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v                      ...(2) 
    Change in the frequency of wave received is, 

              straight v subscript 1 minus straight v subscript 2 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v minus fraction numerator straight V over denominator straight V plus straight u end fraction straight v
       space space space space space space space space space space space space space equals fraction numerator 2 Vu over denominator straight V squared minus straight u squared end fraction straight v 
                        almost equal to fraction numerator 2 straight u over denominator straight V end fraction straight v                [∵ u<<V] 
    Percentage change in frequency is, 
              fraction numerator straight v subscript 1 minus straight v subscript 2 over denominator straight v end fraction 100 equals fraction numerator 2 straight u over denominator straight V end fraction cross times 100 
    It is given that the percentage change in frequency is 2%.
    Therefore,
                     fraction numerator 2 straight u over denominator straight V end fraction cross times 100 equals 2 
    rightwards double arrow             straight u equals straight V over 100 equals 3.5 straight m divided by straight s, is the velocity of the source. 
    Question 857
    CBSEENPH11019129

    The displacement of a particle executing simple harmonic motion is represented by = r sin ωt. Find the velocity and acceleration of the particle in terms of displacement.

    Solution

    Displacement of particle executing S.H.M is given by,
     y=  space space r space sinωt
    Velocity of the particle is, 

    straight v equals dy over dt equals straight d over dt left parenthesis straight r space sin space ωt right parenthesis space equals space rω space cos space ωt
    This implies, 
    straight v space equals space straight omega space square root of straight r squared cos squared ωt end root space

space space equals space straight omega space square root of straight r squared left parenthesis 1 minus sin squared ωt right parenthesis end root

space space equals space straight omega space square root of straight r squared minus straight r squared sin squared ωt end root space

straight v space equals space straight omega square root of straight r squared minus straight y squared end root space

And comma space

Acceleration comma space straight a space space equals space dv over dt space equals space straight d over dt open parentheses rω space cos space ωt close parentheses space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space minus space rω squared space sin space ωt space equals space minus straight omega squared straight y space

That space is comma space

straight a space equals space minus straight omega squared straight y

    Question 858
    CBSEENPH11019130

    A source of sound of frequency v is moving towards a wall with speed u. Find the number of beats heard by a stationary observer on the ground when:
    (i) observer is between source and wall.
    (ii) source is between observer and wall.

    Solution
    (i) When observer is between source and wall, then source is approaching both observer and wall. 

     
    The apparent frequency of wave received by observer is, 
                          straight v subscript 1 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v 
    The apparent frequency of wave incident on wall is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    As the wall is stationary, therefore the frequency of wave reflected is same as that of wave incident in it.
    Thus, the frequency of wave reflected by the wall is v2.
    Now the observer receives a wave of frequency v1 from source and a wave of v2 after reflection from the wall.
    Therefore the number of beats heard per second is, 
                            straight b equals straight v subscript 1 minus straight v subscript 2 equals 0
    (ii) When the source is between observer and wall, then source is moving away from the observer and approaching towards the wall. 
     
    The apparent frequency of wave received by observer is,
                          straight v subscript 1 equals fraction numerator straight V over denominator straight V plus straight u end fraction straight v 
    The apparent frequency of wave incident on wall is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Therefore number of beats heard per second is, 
                       straight b equals straight v subscript 2 minus straight v subscript 1 equals fraction numerator 2 uV over denominator straight V squared minus straight u squared end fraction straight v 
    Question 859
    CBSEENPH11019131

    A particle is vibrating in a straight line with amplitude A. What is the total distance travelled and displacement of particle in one complete oscillation?

    Solution

    Let the particle be vibrating between points L and M.
    In one complete vibration the particle moves from O to L, then from L to O and then from O to M and finally from M to O as shown in figure.

    Therefore the total distance travelled by particle in one complete oscillation is, 
    S = SOL + SLO + SOM + SMO 
       = A + A + A + A = 4A
    As the particle comes back to same position after one complete oscillation, therefore the net displacement of particle is zero.

    Question 860
    CBSEENPH11019132

    Derive the differential equation of simple harmonic motion.

    Solution

    The acceleration of particle executing simple harmonic motion, a = – ω2y

    We know that the acceleration is second derivative of displacement.
    Therefore, 
     space space space space space space space space space space space space space space straight a equals fraction numerator straight d squared straight y over denominator dt squared end fraction
therefore space space space fraction numerator straight d squared straight y over denominator dt squared end fraction equals negative straight omega squared straight y
space space space space space space fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight omega squared straight y equals 0 
    The above equation represent the differential equation of simple harmonic motion.

    Question 861
    CBSEENPH11019133

    A sound from a stationary siren is reflected by a high wall approaching with velocity u towards a stationary person. The siren emits the wave at frequency v travelling in air with speed V. Find the apparent frequency of wave reflected by wall and number of beats heard per second. Siren and person both are on the same side of wall.

    Solution
    The apparent frequency of wave received by wall approaching the siren is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Wall reflects back the received wave.
    Therefore the wall behaves like a source of wave emitting wave at a frequency v1.
    The wall is approaching the person.
    Therefore, apparent frequency of reflected wave received by observer is, 
                   straight v subscript 2 equals fraction numerator straight V over denominator straight V minus straight u end fraction straight v subscript 1 space equals space fraction numerator straight V plus straight u over denominator straight V minus straight u end fraction straight v 
    The number of beats heard per second is given by, 
            space space space space space space space space straight b equals straight v subscript 2 minus straight v equals fraction numerator 2 straight u over denominator straight V end fraction straight v 
    Question 862
    CBSEENPH11019134

    What is the difference between musical sound and noise?

    Solution

    Musical sound is pleasant to the ears, and it is produced by a series of similar pulses following each other regularly and rapidly at equal intervals without any sudden change in amplitude. 

    Noise is the sound which produces a displeasing effect to our ears, and it is produced by a series of different pulses following each other after unequal intervals of time.

    Question 863
    CBSEENPH11019135

    Is there any difference between note and tone?

    Solution

    A tone is a sound of a single frequency.
    For example, tuning fork, when struck gently, produces a pure tone. 
    Note is formed by combination of many tones. That is note has a combination of different frequencies. 

    Question 864
    CBSEENPH11019136

    Does the pitch only depend upon frequency?

    Solution
    No. The pitch also depends upon relative motion between source, observer and media through which sound is propagated. 
    Question 865
    CBSEENPH11019137

    A large auditorium has a curved back. Why?

    Solution
    A large auditorium has a curved back to render the sound waves parallel so that these can be heard over a large distance.
    The speaker is placed at the focus of curved surface and the sound after reflection from curved surface becomes parallel. 
    Question 866
    CBSEENPH11019138

    What is the difference between echo and a reverberation?

    Solution
    Echo: An echo is formed when sound wave reflects and comes back after an interval ≥ 0.1s. Therefore in an echo, the original and reflected sound is heard separately.
    Reverberation: It consists of successive reflections at intervals very small as compared to 0.1 second so that they cannot produce separate echoes. 
    Question 867
    CBSEENPH11019139

    Why should reverberation period be neither too high nor too low for good hearing?

    Solution
    If reverberation period is too high, it results in the mixing of speech and will reduce the clarity of speech.
    If reverberation period is too small, the sound would suddenly end up and therefore reduce the audibility.
    Question 868
    CBSEENPH11019140

    Two sound notes can be distinguished by their quality, pitch and loudness. Explain which one is the distinguishing factor from the notes emitted by two different tuning forks.

    Solution
    Pitch is the distinguishing factor because the notes emitted by two different tuning forks have different frequencies.
    Pure note is produced by a tuning fork without overtones, therefore notes cannot be distinguished from quality.
    When tuning forks are struck with same force, the forks produce the sound of same loudness. Hence, notes from two tuning forks also cannot be distinguished from their loudness. 
    Question 869
    CBSEENPH11019141

    What factors distinguish note and its echo?

    Solution
    Note is louder than its echo because on reflection, the intensity of sound decreases without changing its pitch and quality.
    Thus, loudness is the distinguishing factor for note and its echo.
    Question 870
    CBSEENPH11019142

    How many keys are there in diatonic scale and what are the intervals of diatonic scale?

    Solution
    There are eight keys and seven intervals in the diatonic scale.
    The intervals are 
    9 over 8 comma space 10 over 9 comma space 16 over 15 comma space 9 over 8 comma space 10 over 9 comma space 9 over 8 comma space 16 over 15
    Question 871
    CBSEENPH11019143

    What are the factors on which the intensity of sound depends?

    Solution

    Intensity of sound depends on the following factors:
    (i) Amplitude of vibration (I ∝ A2)  
    (ii) The frequency of vibration (I ∝ v2). 
    (iii) The density of media (I ∝ ρ).
    (iv) Surface area of a vibrating body. 
    (v) The distance of listener from the source. 

    Question 872
    CBSEENPH11019144

    What is the change in intensity level when the intensity of sound increases by 10times the original intensity?

    Solution
    Increase in intensity level is given by, 
               space space space increment straight L space equals space 10 Log straight I subscript 2 over straight I subscript 1 db 
    where,
    I1 and I2 are the original and final intensities. 
    We have,   
                         straight I subscript 2 over straight I subscript 1 equals 10 to the power of 7 
    Therefore,
               increment straight L space equals space 10 Log 10 to the power of 7 space equals space 70 db
    Question 873
    CBSEENPH11019145

    If displacement of a particle at any instant is represented by y = A cos ωt + B sin ωt, where A and B are constants, prove that the motion of particle is simple harmonic motion. What is the amplitude and epoch of the motion?

    Solution

    Position of particle is given by, 
    y = A = cosomega t + B sinomega t               ...(1)
    The velocity of particle is given by, 
     straight v space equals fraction numerator d y over denominator d t end fraction equals fraction numerator straight d over denominator d t end fraction left parenthesis straight A space c o s space omega t space plus space straight B space s i n space w t right parenthesis
space space space equals space minus space A omega space s i n space omega t plus B omega space c o s space omega t right parenthesis 
    Acceleration of particle is given by,
    straight a equals dv over dt equals straight d over dt left parenthesis negative Aω space sin space ωt plus space Bω squared space sin space ωt right parenthesis

space space space equals space minus space Aω squared cosωt minus Bω squared sin space ωt

space space space equals negative straight omega squared left parenthesis Acos space ωt plus straight B space sin space ωt right parenthesis

space space space equals negative straight omega squared straight y
    Since the acceleration of particle is directly proportional to displacement and directed towards mean position, therefore the motion is simple harmonic motion.
    Now, 
    Let space amplitude comma space straight A equals straight r space sinϕ space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

and space space space space space space space space space space space straight B space equals space straight r space cosϕ space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis space

Substituting space space straight A space and space straight B space in space left parenthesis 1 right parenthesis comma space we space get comma

straight y space equals space straight r space sin space straight ϕ space cos space ωt space plus space straight r space cos space straight ϕ space space sin space ωt

space space space equals straight r left parenthesis cos space ωt space sin space straight ϕ plus sin space ωt space cosϕ right parenthesis

space space space equals space rsin left parenthesis ωt plus straight ϕ right parenthesis
    Squaring (2) and (3) and adding, we have
    space space space space space space straight A squared plus straight B squared equals straight r squared

rightwards double arrow straight r square root of straight A squared plus end root straight B squared  
    Dividing (2) (3), we have
    space space space space space space fraction numerator space straight A over denominator straight B end fraction equals tan space straight ϕ space space
rightwards double arrow space space space space space straight r equals square root of straight A squared plus straight B squared end root 
    Therefore comma space
space A m p l i t u d e space equals space straight r equals space square root of straight A squared plus straight B squared end root
    Epoch = straight ϕ space equals space tan to the power of negative 1 end exponent left parenthesis straight A over straight B right parenthesis.

    Question 874
    CBSEENPH11019146

    A particle is executing simple harmonic motion. What is phase relationship between:

    (i) displacement and velocity

    (ii) velocity and acceleration?

    Solution
    Let the position of particle be given by,
    
straight y space equals space r s i n space omega t space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

therefore space space space straight v equals r omega space c o s space omega t
space space space space space space space space space equals r omega space s i n space open parentheses omega t straight pi over 2 close parentheses space space space space space space space space space space space... left parenthesis 2 right parenthesis space
a n d space

A c c e l e r a t i o n comma space straight a equals space minus r omega squared s i n omega t

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals r omega squared s i n left parenthesis omega t plus straight pi right parenthesis space space space space... left parenthesis 3 right parenthesis
    From equation (1) and (2),
    Velocity leads the displacement by phase <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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    From equation (2) and (3), we have
    The acceleration leads the velocity by phase pi divided by 2.
    Question 875
    CBSEENPH11019147

    A particle is executing simple harmonic motion with amplitude r and frequency ω. Find the velocity and acceleration of particle at mean and extreme positions.

    Solution
    Given a particle is executing Simple Harmonic Motion. 
    The position, velocity and acceleration of particle executing simple harmonic motion is given by,
    Position comma space straight y space equals space straight r space s i n space omega t

V e l o c i t y comma space straight v space equals space straight omega square root of straight r squared minus straight y squared end root

a n d space space space
A c c e l e r a t i o n comma space straight a equals negative straight omega squared straight y
    Question 876
    CBSEENPH11019148

    Show that total energy of particle executing simple harmonic motion is constant.

    Solution
    A particle executing simple harmonic motion possesses both kinetic energy and potential energy.

    Total energy of particle executing simple harmonic motion at any point is equal to the sum of kinetic energy and potential energy.
    That is, 
    Total energy, E = K.E + P.E
    Let at any instant, the particle be at P at a distance y from mean position.
    Let, v be the velocity of particle at P.

    Kinetic energy:
    Kinetic energy of particle executing simple harmonic oscillation at any instant is given by,
                            
space space straight K equals 1 half mv squared
    The velocity of particle at a distance y from the mean position is,
    space space space space space space straight v equals straight omega square root of straight A squared minus straight y squared end root

therefore space straight K equals 1 half mv squared equals 1 half mω squared left parenthesis straight A squared straight Y squared right parenthesis
    Potential energy:
    Potential energy stored in the particle is equal to the work done in displacing the particle from mean position to y.
    Let the particle be displaced through a distance x from mean position.
    The restoring force F acting on particle is,
    F with rightwards harpoon with barb upwards on top equals negative mω squared x with rightwards harpoon with barb upwards on top
    Therefore, when the particle is moved through a distance x, work done against the restoring force is given by, 
    dW equals F with rightwards harpoon with barb upwards on top. stack d x with rightwards harpoon with barb upwards on top

space space space space space space equals straight F space dxcos space 180 degree

space space space space space space equals negative straight F space dx space

dW equals negative mω squared straight x space dx
    Therefore, total work done by restoring force in displacing the particle from mean position to P is,
    straight W equals integral dW space

space space space equals integral presubscript 0 presuperscript straight y minus mω squared straight x space dx space

space space space equals negative 1 half mω squared straight y squared
    The potential energy stored in the body is equal in magnitude and opposite in sign of the work done by restoring force. Thus
    space space space space space straight U minus negative straight W equals 1 half mω squared straight y squared
    Total energy is,
    E = K + U
    equals 1 half mω squared left parenthesis straight A squared minus straight y squared right parenthesis plus 1 half mω squared straight y squared
equals space 1 half mω squared straight A squared
    From the above equation, total energy is independent of the position of particle during its motion.
    Thus, total energy is constant. 
    Hence proved. 

    Question 877
    CBSEENPH11019149

    What do you mean by postulates of kinetic theory of gases?

    Solution
    The postulates of Kinetic Theory was developed in the nineteenth century explaining the behaviour of gases. Kinetic Theory gives:

    i) molecular interpretation of pressure and temperature of a gas, and is consistent with gas laws and Avogadro's hypothesis.

    ii) explains specific heat capacities of many gases.

    iii) relates measurable properties of gases such as viscosity, conduction and disffusion with molecular parameters. 

     
    Question 878
    CBSEENPH11019150

    According to kinetic theory of gases what type of collision the molecules of gases undergo?

    Solution
    Collision between gas molecules are perfectly elastic and instantaneous, according to the Kinetic Theory of gases. 
    Question 879
    CBSEENPH11019151

    According to kinetic theory of gases, does the number density of gas molecules change with time?

    Solution
    No. The number of gas molecules per unit volume does not change with time, according to the Kinetic Theory of gases.
    Question 880
    CBSEENPH11019152

    According to kinetic theory of gases, do the gas molecules interact due to intermolecular forces?

    Solution
    Intermolecular force between gas molecules is zero, according to the Kinetic Theory of gases. 
    Question 881
    CBSEENPH11019153

    What type of energy is taken into account while finding the pressure exerted by a gas?

    Solution
    Kinetic energy of molecules is taken into account while finding the pressure exerted by gas. 
    Question 882
    CBSEENPH11019154

    Define r.m.s. speed of gas.

    Solution
    Root Mean Square speed of gas is the square root of the mean of the square of the speed of individual molecules. 
    Question 883
    CBSEENPH11019155

    Define mean free path.

    Solution
    Molecular motion is random motion and molecules travel in a straight path between two successive collisions. 

    Mean free path is defined as the average distance travelled by molecules between two successive collisions.
    Question 884
    CBSEENPH11019156

    Write the expression for mean free path of a gas molecule.

    Solution
    The mean free path of a gas molecule is given by, 

                            λ=12πd2n, 

    where 'd' is the diameter of gas molecule and 'n' is the number of gas molecule per unit volume.
    Question 885
    CBSEENPH11019157

    How does r.m.s. speed depend on molecular mass?

    Solution
    The r.m.s. speed of gas is inversely proportional to the square root of molecular mass.

    Relation between r.m.s and molecular mass is given by, 
                              v2¯ = 3 kBTm
    where, m is the molecular mass.
    Question 886
    CBSEENPH11019158

    How does the number density of gas molecules change with temperature at constant pressure?

    Solution
    The number density of gas molecules decreases with the increase of temperature at constant pressure. 
    Question 887
    CBSEENPH11019159

    What is SI unit of gas constant R?

    Solution
    J mol-1 K-1 or Joule per mole per Kelvin. 

    Question 888
    CBSEENPH11019160

    What is Boltzman’s constant?

    Solution
    Boltzman's constant is defined as the ratio of universal gas constant and Avogadro’s number and is equal to 

    k = RN = 8.316.023 ×1023               = 1.38 × 1023 J/mol/K

    Question 889
    CBSEENPH11019161

    What is absolute zero of temperature?

    Solution
    Absolute zero of temperature is the temperature at which all the molecular motion stops. The atoms are completely at rest and transmits no thermal energy. 
    Question 890
    CBSEENPH11019162

    Under what conditions the real gas obeys the ideal gas equation PV = nRT most closely?

    Solution
    Ideal gas equation is given by, 

                          PV = nRT

    The real gas obeys the ideal gas equation at low pressure and high temperature. 
    Question 891
    CBSEENPH11019163

    Plot the graph of the Maxwell’s distribution of speed of gas molecules in equilibrium.

    Solution
    The graph below shows the Maxwell's distribution of speed of gas molecules in equilibrium. 


    Maxwellian distribution gives the number of molecules between the speeds v and v + dv.
    Question 892
    CBSEENPH11019164

    If kinetic energy per unit volume of gas is E, then what is the pressure of the gas?

    Solution
    The pressure of the gas is two-third of the kinetic energy per unit volume of gas. 
    Question 893
    CBSEENPH11019165

    Show that for a particle in linear simple harmonic motion, the average kinetic energy over a period of oscillation is half the total energy.

    Solution
    Let a particle of mass m execute simple harmonic motion with frequency ω. Let a be the amplitude of vibration.
    space therefore space space space Total space energy space of space particle comma space straight E equals 1 half mω squared straight A squared
    The position of particle at any instant during simple harmonic motion is given by,
    space straight x space equals space straight A space s i n space omega t
therefore space thin space K i n e t i c space e n e r g y space straight K space equals 1 half m v squared

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half m omega squared straight A squared c o s squared omega t

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight E space c o s squared omega t 

    Now the average kinetic energy of particle over complete cycle is,
    top enclose straight K space equals 1 over straight T integral presubscript 0 presuperscript straight T straight E space cos squared space ωt space dt space

space space space space equals straight E over straight T integral presubscript 0 presuperscript straight T fraction numerator straight I plus cos 2 ωt over denominator 2 end fraction dt space

space space space space equals fraction numerator straight E over denominator 2 straight T end fraction open square brackets straight t plus fraction numerator sin space 2 space ωt over denominator 2 straight omega end fraction close square brackets subscript 0 superscript straight T space

space space space space equals fraction numerator straight E over denominator 2 straight t end fraction left square bracket straight T right square bracket equals straight E over 2
    Hence proved. 
    Question 894
    CBSEENPH11019166

    What is an ideal gas equation?

    Solution
    The ideal gas equation is,

                             PV = nRT 
    where, 
    n is the number of moles and R is the universal gas constant. 

    Question 895
    CBSEENPH11019167

    What is kinetic energy per unit volume of gas at pressure P?

    Solution
    The kinetic energy of gas per unit volume is equal to 3/2 P where, P is the pressure of the gas.

                              K.E = 32P
    Question 896
    CBSEENPH11019168

    Name the gas which possesses the maximum r.m.s. velocity at a given temperature.

    Solution
    The r.m.s velocity of hydrogen gas is maximum at a given temperature.
    Question 897
    CBSEENPH11019169

    The average kinetic energy of translation of H2 gas molecule at temperature T is 0.01 eV. What is the average kinetic energy of translation of N2 gas molecule at same temperature T?

    Solution
    Average Kinetic Energy = 3 over 2 K T
    At the same temperature, the average translational K.e is not dependent on the mass of the matter. That is, it remains the same for the same temperature. 
    Therefore, Average translational K.E of N2 gas = 0.01 eV
    Question 898
    CBSEENPH11019170

    What is Boyle’s law?

    Solution
    Boyle's law states that at constant temperature, pressure of a given mass of gas varies inversely with volume. 

    The law holds precisely true at high temperature and low pressure.
    Question 899
    CBSEENPH11019171

    What is Charle’s law?

    Solution
    Charles Law states that for a fixed pressure, the volume of the gas is directly proportional to the absolute temperature of gas.
    Question 900
    CBSEENPH11019172

    What is Regnault’s law or Law of pressure?

    Solution
    Law of pressure states that for a fixed mass of gas, at constant volume, the pressure (P) of the gas varies directly with absolute temperature (T) of gas.
    Question 901
    CBSEENPH11019173

    State Avogadro’s hypothesis.

    Solution
    Avogadro’s hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure have the same number of molecules.
    Question 902
    CBSEENPH11019174

    What is Graham’s law of diffusion?

    Solution
    Graham’s law of diffusion states that the rate of diffusion of gas molecules is inversely proportional to the square root of it's molecular weight.
    Question 903
    CBSEENPH11019175

    Which of the two occupies more volume– one mole of H2 gas at STP or one mole of N2 at STP?

    Solution
    One mole of H2 and one mole of N2 at STP, both gases occupy same volume, that is 22.4 litres.
    Question 904
    CBSEENPH11019176

    What is standard temperature and pressure?

    Solution
    Standard temperature is 273.15 K and a pressure of 100000 Pa (1 bar). 
    Question 905
    CBSEENPH11019177

    What is difference between STP and NTP?

    Solution
    STP is standard temperature and pressure and NTP is normal temperature and pressure.

    In STP condition, temperature is 0o C and pressure at 1 bar (105 Pa).
    In NTP conditions, the temperature is 20o C and pressure is 1.01325 × 105 Pa.
    Question 906
    CBSEENPH11019178

    Plot the graph for the variation of kinetic energy, potential energy and total energy with displacement of particle executing simple harmonic motion.

    Solution

    Kinetic energy of a particle executing S.H.M is given by, 
    straight K equals 1 half mω squared left parenthesis straight A squared minus straight y squared right parenthesis
space space equals 1 half mω squared straight A squared minus 1 half mω squared straight y squared 
    Potential energy is given by,
    
straight U equals 1 half mω squared straight y squared
    Therefore the graph for kinetic energy versus displacement is a parabola, which is shown below.

    The graph for potential energy versus displacement is also parabola.
    Total energy is given by, 
    straight E equals straight K plus straight U equals 1 half mω squared straight A squared 
    Graph is straight line parallel to displacement axis.

    Question 907
    CBSEENPH11019179

    Plot the graph for the variation of kinetic energy, potential energy and total energy with time for a particle executing simple harmonic motion.

    Solution

    The kinetic energy of the particle executing simple harmonic motion is given by, 

     
    K.E = 1 half m v squared space equals space 1 half m omega squared A squared space cos squared omega t
    Potential energy is given by, 
    P.E, U = 1 half m omega squared y squared space equals space 1 half m omega squared A squared sin squared omega t space 
    Total energy of the particle is given by, 
    T.E = K.E + P.E 
         = 1 half m omega squared A squared space left parenthesis cos squared omega t space plus space sin squared omega t right parenthesis thin space 
         =1 half m omega squared A squared

    Question 908
    CBSEENPH11019180

    A particle is executing simple harmonic motion with frequency f. Find the frequency of oscillation of kinetic energy.

    Solution
    Kinetic energy of particle executing simple harmonic motion is given by,
    straight K equals 1 half mv squared

space space equals 1 half mω squared straight A squared cos squared 2 πft

space space equals straight E space cos squared ωt equals straight E open parentheses fraction numerator 1 plus cos 4 πft over denominator 2 end fraction close parentheses

space space equals straight E over 2 plus fraction numerator Ecos space 2 straight pi left parenthesis 2 straight f right parenthesis straight t over denominator 2 end fraction 

    The first term is constant and second term is oscillatory having frequency 2f.
    Thus the frequency of oscillation of kinetic energy is 2f. 
    Question 909
    CBSEENPH11019181

    Define Avogadro number.

    Solution
    Avogadro number is the number of atoms in 12 gm of 6C12 isotope of carbon.

    Numerical value of Avogadro number is 6.023 x 10
    23.
    Question 910
    CBSEENPH11019182

    How much volume does Avogadro number of molecules of a gas at STP occupy?

    Solution
    One Avogadro number of molecules of a gas at STP occupies 22.4 litre of volume.
    Question 912
    CBSEENPH11019184

    What is the relation between gas constant R, Avogadro’s number N and Boltzmann’s constant k?

    Solution
    The relation between gas constant R, Avogadro's number NA and Boltzman's constant KB is given by,

                                R = kBNA
    Question 913
    CBSEENPH11019185

    A particle executes simple harmonic motion with amplitude 0.6m. Its velocity at mean position is 12m/s. Find the frequency of oscillation and acceleration at extreme position.

    Solution

    Given that,
    Amplitude of the motion, r= 0.6cm
    Velocity at mean position, v= 12m/s
    We space know comma space straight nu subscript straight omicron equals rω
    Therefore, '

    space space space space space space space space space space space space space straight omega equals straight v subscript straight omicron over straight r equals fraction numerator 12 over denominator 0.6 end fraction equals 20 rad divided by straight s
and
space space space space space space space space space space space space space straight f equals fraction numerator straight omega over denominator 2 straight omega end fraction equals fraction numerator 20 over denominator 2 straight pi end fraction equals 3.183 Hz
    The acceleration of particle at extreme position is, 
    straight alpha subscript straight omicron equals r omega squared
space space space space equals 0.6 left parenthesis 20 right parenthesis squared
space space space space equals 240 c m divided by straight s squared 

    Question 914
    CBSEENPH11019186

    An air bubble is released from the bottom of tank. What happens to the size of air bubble as it rises upward in the tank?

    Solution
    As the bubbles rises up, the pressure inside the bubble decreases and hence the size of bubble increases.

    The size of the bubble increases because as the bubble rises up, pressure inside the bubble decreases.
    Question 915
    CBSEENPH11019187

    A vessel contains a mixture of one mole of O2 and two moles of H2 at 310K. What is the ratio of the average kinetic energy per oxygen molecule to that of hydrogen?

    Solution
    Average kinetic energy per gas molecule is independent of the nature of gas but depends only on its temperature. Thus kinetic energy per molecule of both the gases is same. Hence, their ratio has value one.
    Question 916
    CBSEENPH11019188

    What are the basic postulates of kinetic theory of gases?

    Solution

    The basic postulates of kinetic theory of gases are:

    (i) All gases consist of atoms or molecules. The atoms or molecules of one gas are all similar to one another and different from the molecules of the other gas.

    (ii) Molecules of a gas are in random motion.

    (iii)The volume occupied by gas molecules is negligibly small as compared to volume of the container.

    (iv) Molecules collide with each other. The collisions are elastic and instantaneous.

    (v) Between the two consecutive collisions, the path followed by molecules is straight line.

    (vi) Gas molecules do not exert any force of attraction or repulsion upon each other. 

    Question 917
    CBSEENPH11019189

    Which of the basic postulates of kinetic theory of gases do not fit exactly for ideal gas?

    Solution

    The postulates of Kinetic Theory which do not fit exactly for ideal gas are:

     (i) Gas molecules do not exert any force of attraction or repulsion upon each other.

    (ii) Volume occupied by gas molecules is negligible.

    Question 918
    CBSEENPH11019190

    State two postulates of kinetic theory of gases which are obeyed by all the gases.

    Solution

    The two postulates of kinetic theory of gases which are obeyed by all the gases are:

    (i) All the gases consist of molecules that are similar to one another for one gas and different from the molecules of other gas.

    (ii) Molecules of all the gases are in random motion with speed ranging from zero to infinity. 

    Question 919
    CBSEENPH11019191

    A gas when heated at constant volume is found to exert greater pressure. Explain why?

    Solution
    When a gas is heated, the r.m.s velocity of the molecules increases which resulting in increase of the change in momentum of molecules on collision. Hence gas molecules exert more force and hence more pressure. 
    Question 920
    CBSEENPH11019192

    A body oscillates simple harmonically according to equation y=5 cos(3 pi t plus pi over 3 right parenthesis c m What is displacement, velocity and acceleration at t= 3s?


    Solution

    Simple Harmonic Equation is given by,
                 y = 5cos(3pi t plus pi over 3 right parenthesis c m
    Therefore, 
    Velocity, v equals space dy over dt space equals space minus 15 space straight pi space left parenthesis 3 πt space plus space straight pi over 3 right parenthesis space cm space divided by straight s
    Acceleration comma space straight a space equals space dv over dt space equals space minus 45 space straight pi squared space cos space open parentheses 3 πt space plus space begin inline style straight pi over 3 end style close parentheses space cm divided by straight s squared
    Now, y (t=3) = 5 cos (9straight pi space plus space straight pi over 3 right parenthesis thin space cm
                         = negative 5 space cos space open parentheses straight pi over 3 close parentheses space cm space equals space minus 2.5 space space cm space
    Therefore, 
    v|t=3= -15straight pi space sin space open parentheses begin inline style straight pi over 3 end style close parentheses space cm divided by sec space equals space fraction numerator 15 straight pi square root of 3 over denominator 2 end fraction space cm divided by sec 
    Acceleration, a|t=3negative 45 space straight pi squared space cos space left parenthesis 9 straight pi space plus space straight pi over 3 right parenthesis space cm divided by straight s squared 
                                 equals space 45 space straight pi squared space cos space open parentheses straight pi over 3 close parentheses space cm divided by straight s squared 
                                  = fraction numerator 45 space straight pi squared over denominator 2 end fraction space c m divided by s squared 

    Question 921
    CBSEENPH11019193

    What is kinetic interpretation of temperature?

    Solution
    According to the kinetic theory of gases pressure of an ideal gas is given by, 

                    P=13ρv2 = 13MVv2 
    where, v is the mean of the square speed.

              PV=13Mv2 

    Now, average translational kinetic energy of the molecule is given by, 

          12Mv2=32PV=32nRT 

          12Mv2=32nNkT 

        12MnNv2=32kT

    or         12mv2=32kT 

    T: absolute temperature.

    This is the relation between average kinetic energy of a molecule of gas and absolute temperature of the gas.

    According to this relation average kinetic energy of a molecule of an ideal gas is proportional to its absolute temperature; is indeppendent of the pressure, volume or the nature of the ideal gas. This is called kinetic interpretation of temperature.

    Question 922
    CBSEENPH11019194

    Show that pressure of gas is equal to two-third of kinetic energy per unit volume of gas.

    Solution
    Pressure of a gas is given by P = 13ρv2       ... (1)
    Density is equal to mass per unit volume. 
    Density, ρ = MV

    Now, multiplying and dividing equation (1) by 2, we get

    P = 231V12Mv2    = 23V(Kinetic energy of gas)   = 23Kinetic energy of gasVolumeP= 23( K.E per unit volume of gas)  

    Hence proved. 



    Question 923
    CBSEENPH11019195

    Derive an expression for average kinetic energy of a molecule of an ideal gas.

    Solution
    Consider one mole of an ideal gas of molecular weight Mo at absolute temperature T. 
    Let, 
    m : mass of one molecule of gas
    v : r.m.s speed of the gas molecule
    P : Pressure exerted by an ideal gas

    Therefore, 

    P = 13ρv2 = 13MVv2

    That is,

    PV = 13 Mv2   ... (1)

    Also, PV = nRT
    ∴  13Mv2 = nRT

     12Mnv2 = 32RT12Mov2 = 32RT 12mNv2 = 32RT 12mv2 = 32RNT = 32kT     

    Thus, the average kinetic energy of a molecule of an ideal gas is 32kT



    Question 924
    CBSEENPH11019196

    Deduce Boyle’s law on the basis of kinetic theory of gases.

    Solution
    Boyle's law states that, keeping temperature constant, pressure of a given mass of gas varies inversely with volume. 

    According to kinetic theory of gases, 
    pressure of an ideal gas is given by,

                        P=13ρv2 = 13MVv2      [ ρ = MV]

    So,             PV=13Mv2 

    According to kinetic interpretation of temperature,  v2T

    If temperature is constant then,

                         PV = constant 

    This proves the Boyle's law. 
    Question 925
    CBSEENPH11019197

    A particle executing simple harmonic motion has acceleration 5.12 cm/s2 when at 8 cm from the mean position. Find the time period of oscillation.

    Solution

    Using the formula of acceleration,
    We have
                     a = straight omega squared straight y
    So, time period of the oscillation is given by, 
    therefore space space space space space space straight T space equals space 2 straight pi square root of 1 over straight omega squared end root equals 2 straight pi square root of straight y over straight a end root
    Now, putting the values of acceleration and displacement give, we get
    straight T equals 2 straight pi square root of fraction numerator 8 over denominator 5.12 end fraction end root equals 7.854 space straight s, is the time period of the oscillation. 

    Question 926
    CBSEENPH11019198
    Question 927
    CBSEENPH11019199

    Deduce Charle’s law on the basis of kinetic theory of gases.

    Solution
    Charles law states that, for a fixed pressure, the volume of gas is proportional to it's absolute temperature. 

    Now, according to kinetic theory of gases, pressure of an ideal gas is given by, 

                           P=13ρv2 = 13MVv2 

    i.e.,                               V=13MPv2 
    where, 
    V : is the volume of the gas
    v2 : mean of the squared speed.

    At constant pressure, for a given mass of gas
                             Vv2 

    And,                  v2T 

    Therefore, at constant pressure VT

     
    Hence, Charles' law is proved. 
    Question 928
    CBSEENPH11019200

    A particle executes simple harmonic motion of amplitude 16cm and time period 2.4s. What is the minimum time taken by the particle to move between two points 8cm on either side of mean position?

    Solution

    The position of the particle executing simple harmonic motion is given by
    space space space space space space space space space space space space y equals A space sin fraction numerator 2 pi over denominator T end fraction t
G i v e n space h e r e comma space

A m p l i t u d e space o f space t h e space p a r t i c l e comma space A equals 16

T i m e space p e r i o d comma space T equals 2.4 space s space

therefore space y equals 16 space sin fraction numerator 2 pi over denominator 2.4 end fraction t

    Let the position of particle be +8 cm at tand -8cm at t2. 

    therefore space space space 8 equals 16 space sin space fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 1 space

rightwards double arrow space space space space 1 half equals space sin fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 1 space

rightwards double arrow space space space space fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 1 equals straight pi over 6 or fraction numerator 5 straight pi over denominator 6 end fraction

rightwards double arrow space space space space space space straight t subscript 1 equals 0.2 straight s space or space 1.0 straight s

and space minus 8 equals 16 space sin fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 2
rightwards double arrow space space minus 1 half equals sin fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 2 space end subscript

rightwards double arrow space space fraction numerator 2 straight pi over denominator 2.4 end fraction straight t subscript 2 equals fraction numerator 7 straight pi over denominator 6 end fraction or fraction numerator 11 straight pi over denominator 6 end fraction

    Therefore the time taken by particle to move between two points 8cm on either side of mean position is,

    straight t subscript 2 minus straight t subscript 1 equals 1.2 straight s space or space 0.4 straight s space or space 2 straight s

    The minimum time is 0.4s.

    Question 929
    CBSEENPH11019201

    Two particles A and B execute simple harmonic motions given by the equations yA = 10 sin12 πt, yB = 12 sin10 πt Which has greater velocity at mean position?

    Solution
    The velocity at the mean position is given by,
    space space space space space space space space straight nu subscript ring operator equals straight omega

therefore space space space space space space straight v subscript 0 straight A end subscript equals 10 cross times 12 straight pi equals 120 straight pi

a n d space space space straight v subscript omicron B end subscript equals 12 cross times 10 straight pi equals 120 straight pi
    Therefore, both the particles have same velocity at the mean position.
    Question 930
    CBSEENPH11019202

    Deduce Avogadro’s law from the kinetic theory of gases.

    Solution

    Avogadro law states that under similar physical conditions of temperature and pressure, equal volume of all the gases contains equal number of molecules. 

    Let P be the pressure, V be the volume and T be the absolute temperature. 

    Let us consider two gases having same P, V and T. Let one gas contains n1 molecules, each of mass m1 and second gas contains n2 molecules each of mass m2. Let v1, and v2 be the r.m.s. velocities of two gases.

    Now, According to kinetic theory of gases, pressure of an ideal gas is given by 

    P = 13ρv2 = 13MVv2 = 13 nmVv2For gas having mass m1,P= 13 n1m1V v12         ... (1)For gas having mass m2,P= 13 n2m2V v22         ... (2) 

    Now, from equations (1) and (2), we have

    m1n1v12 = m2n2 v22     ... (3)

    Now, since temperature of both the gases is the same, the average K.E of both the gases will be same.

    That is, 12m1v12 = 12m2v22       ... (4) 

    So, from equation (3) and (4), we have

                          n1 = n2

    That, is the number of molecules for both gases is the same. 

    Hence, Avogadro's law is proved. 




                                               

    Question 931
    CBSEENPH11019203

    Drive the relation between r.m.s. velocity of air molecules and the speed of sound in air.

    Solution
    Speed of sound in air is given by, 
                       straight v subscript straight s equals square root of γP over straight rho end root 
    Root mean square speed of air molecules is, 
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#6 {main}</pre> 
    Relation between r.m.s. velocity of air molecules and the speed of sound in air is given by, 
    ∴          straight v subscript straight s over straight v subscript straight r equals square root of straight gamma over 3 end root 
    Question 932
    CBSEENPH11019204

    Is the temperature less than absolute zero possible? Justify your answer.

    Solution
    No, temperature less than absolute zero is not possible. Temperature is due to molecular motion and at zero Kelvin the molecular motion stops. The molecular motion cannot be decreased further. This is the reason why temperature cannot be less than the absolute temperature. 
    Question 933
    CBSEENPH11019205

    Explain why cooling is caused after evaporation on the basis of kinetic theory?

    Solution
    According to the kinetic theory, the molecules of a liquid are moving with different velocities. The r.m.s velocity of the molecules is determined by the temperature of the liquid. Further, due to the intermolecular force of attraction acting downwards in the molecules, they do not leave the surface film of the liquid. But molecules having high velocity are able to overcome the force of attraction and hence escape. Therefore, the r.m.s velocity of the molecules which did not escape decreases and hence the temperature. So, cooling is followed by evaporation.

    Question 934
    CBSEENPH11019206

    Though the velocity of air molecules is of the order of 500 m/sec, even then the fragrance or smell spreads at a much slower rate. Why?

    Solution

    The velocity of air molecules is of the order of 500 m/s. But, the fragrance spreads at a much slower rate because the vapour molecules do not travel straight with this velocity. Due to collisions the molecules follow zig-zag path and the effective displacement of molecules per unit time i.e drift motion is very less. 

    Question 935
    CBSEENPH11019207

    Does evaporation take place only at boiling temperature?

    Solution
    All the molecules having sufficient escape velocity can undergo evaporation. According to the Maxwellian distribution of velocities, some of the molecules have very high velocities at all the temperatures. Thus, evaporation can take palce at all temperatures.
    Question 936
    CBSEENPH11019208

    How do the molecules exert pressure on the walls of the container?

    Solution
    The molecules which are moving possess momentum. When the molecules strike with the walls of container, they undergo change in momentum and hence exert force and pressure. 

    Using the principle of conservation of momentum, the momentum imparted to the walls during the collision = 2mvx
    Question 937
    CBSEENPH11019209

    Absolute zero temperature is not the zero energy temperature. Explain.

    Solution
     The absolute zero is the temperature at which molecular motion ceases i.e translational kinetic of molecules become zero, but the molecules possess the potential energy, according to the Kinetic theory of gases. Thus, the absolute zero is not the zero energy temperature.
    Question 938
    CBSEENPH11019210

    The maximum acceleration of particle executing simple harmonic oscillations is a and the maximum velocity is v. What is the frequency and amplitude of oscillation?

    Solution
    Let straight omega be the frequency and A the amplitude of oscillation of a particle executing simple harmonic oscillation.

    Maximum acceleration is,

    straight a subscript straight omicron equals straight omega squared straight A space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    Maximum velocity is given by,
     straight v subscript ring operator equals omega A space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    Squaring (2) and dividing by(1), we get
    straight v subscript 0 squared over straight a subscript 0 equals fraction numerator straight omega squared straight A squared over denominator straight omega squared straight A end fraction equals straight A

therefore space space space straight A equals straight v subscript 0 squared over straight a subscript 0 comma space i straight s space the space amplitude space of space the space oscillation.          
    Also dividing (1) by (2), we have,
    straight a subscript 0 over straight v subscript 0 equals fraction numerator straight omega squared straight A over denominator ωA end fraction equals straight omega

Therefore comma space
space space space space straight omega equals straight a subscript 0 over straight v subscript 0 comma space is space the space frequency space of space the space oscillation. space

     
    Question 939
    CBSEENPH11019211

    Why do the molecules of an ideal gas possess only kinetic energy?

    Solution
    The molecular interaction between two gas molecules of an ideal gas is zero. Therefore, the molecules of an ideal gas do not posess the potential energy. So, the molecules only posess kinetic energy due to the intra-molecular motion.

    Question 940
    CBSEENPH11019212

    The vertical motion of a huge piston having a small block of mass m placed over it in simple harmonic motion with a frequency is ω . What is the maximum amplitude of piston’s simple harmonic motion for block and piston to remain together?

     

    Solution

    If the acceleration of piston is less or equal to acceleration due to gravity at that point, the block will not leave the piston at any point. 
    The value equal to 'g' for the block not to leave the piston is the maximum acceleration at the extreme point. 
    Therefore, 


    space space space space space straight omega squared straight A equals straight g
space space space space space space space space space straight A equals straight g over straight omega squared
    where A is amplitude of vibration.  

    Question 941
    CBSEENPH11019213

    Explain on the basis of kinetic theory that pressure of gas increases with the increase of temperature.

    Solution
    Pressure of gas is equal to the rate of change of momentum of molecules striking per unit area. According to kinetic theory of gases, r.m.s. speed of molecules is directly proportional to the square root of temperature. Therefore, with the increase in temperature, the r.m.s speed increases, hence increase in the change in momentum. With increase in temperature, collision frequency also increases.

    Therefore, both the factors increase the rate of change of momentum and hence pressure increases with the increase in temperature.
    Question 942
    CBSEENPH11019214

    Time period of simple harmonic oscillator is T. In what time will it traverse a distance equal to half of amplitude from mean position?

    Solution

    The displacement of particle executing simple harmonic oscillation is given by, 
                          y = A sin ωt

    Let the particle take to time to go from mean position to half of extreme position. 
    Therefore, 
    therefore space space space space space space space space space straight A over 2 equals straight A space s i n space omega t subscript 0
rightwards double arrow space space space space space space space space space space s i n space omega t subscript 0 equals 1 half
rightwards double arrow space space space space space space space space space space space omega t subscript straight omicron equals straight pi over 6
rightwards double arrow space space space space space space space space space straight t subscript 0 equals fraction numerator straight pi over denominator 6 straight omega end fraction equals fraction numerator straight pi cross times straight T over denominator 6 cross times 2 straight pi end fraction equals straight T over 12
    So, at a time of straight T over 12 the simple harmonic oscillator is equal to half of amplitude from the mean position. 

    Question 943
    CBSEENPH11019215

    The velocity of a particle executing simple harmonic motion is v when displacement is x1 and v2 when displacement is x2. What is the amplitude of vibration of the particle?

    Solution

    The velocity of the particle executing simple harmonic motion when at a distance x from mean position is given by, 
    straight v space equals space straight omega space square root of straight r squared minus straight y squared end root

Here comma space

straight v space equals space straight v subscript 1 space at space space straight x space equals space straight x subscript 1 space comma space and space
straight v space equals space straight v subscript 2 space end subscript at space straight x space equals space straight x subscript 2 space

Therefore comma space

straight v subscript 1 space end subscript equals space straight omega space square root of straight r squared minus straight x subscript 1 squared end root space space space space space space space... space left parenthesis 1 right parenthesis space and

straight v subscript 2 space equals space straight omega space square root of straight r squared minus straight x subscript 2 squared end root space

Squaring space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space and space dividing comma space

straight v subscript 1 squared over straight v subscript 2 squared space equals space fraction numerator straight omega squared left parenthesis straight r squared minus straight x subscript 1 squared right parenthesis over denominator straight omega squared left parenthesis straight r squared minus straight x subscript 2 squared right parenthesis end fraction space

space space space space space space space space space space space equals space fraction numerator left parenthesis straight r squared minus straight x subscript 1 squared right parenthesis over denominator left parenthesis straight r squared minus straight x subscript 2 squared right parenthesis end fraction space

space space space space space space space space straight v subscript 1 squared straight r squared space minus space straight v subscript 1 squared straight x subscript 2 squared space equals space straight v subscript 2 squared straight r squared space minus space straight v subscript 2 squared straight x subscript 1 squared space space

rightwards double arrow space space straight r squared space left parenthesis straight v subscript 2 squared space minus space straight v subscript 1 squared right parenthesis thin space equals space straight v subscript 2 squared straight x subscript 1 squared space minus space straight v subscript 1 squared straight x subscript 2 squared space

straight i. straight e. comma space space space space space space space space space space space space space space space space space space space space space straight r space equals space square root of fraction numerator straight v subscript 2 squared straight x subscript 1 squared space minus space straight v subscript 1 squared straight x subscript 2 squared over denominator straight v subscript 2 squared minus straight v subscript 1 squared end fraction end root

 

    Question 944
    CBSEENPH11019216

    If the intermolecular forces vanish then what will be the volume occupied by 1 kg of H2O at N.T.P?

    Solution
    Mass of H2O = 1 kg
    Volume occupied = ?

    If the intermolecular force vanishes, water in liquid form will change it's state to gas. 

    Number of moles in 1kg of water is given by,

    n = 100018 = 5009 molesUsing the ideal gas equation, we have      PV = nRT   V = nRTP = 5009×8.31×2731.01 × 105= 1.2479 m3

    1.2479 m3 is the required volume occupied by 1 kg of H2O at NTP.
    Question 945
    CBSEENPH11019217

    If r.m.s speed of molecule of hydrogen at N.T.P is 2 km/sec., then what is r.m.s speed of oxygen molecules at N.T.P?

    Solution
    Given, 
    r.m.s speed of molecule of hydrogen at NTP = 2 km/sec
    Find - rms speed of oxygen molecule at NTP?

    RMS speed of molecules is given by,

                          vrms=3RTM 

    ∴              (vr)H2(vr)o2=Mo2MH2 = 322 = 4

    ∴                (vr)o2 = 14(vr)H2 = 0.5 km/sec
    Question 946
    CBSEENPH11019218

    The velocity of a particle executing simple harmonic motion is V when displacement is x1 and v2 when displacement is x2. What is the frequency of oscillation?

    Solution
    The velocity of the particle executing simple harmonic motion when at a distance x from mean position is given by, 
    space space space space space space space space space straight v equals straight omega square root of straight r squared minus straight y squared end root space

Here space we space have comma space

space space space straight v equals space straight v subscript 1 space space at space straight x equals straight x subscript 1 space comma space and space

space space straight v equals straight v subscript 2 space space at space straight x equals straight x subscript 2 space

therefore space space space space space space space space space space space space straight v subscript 1 equals straight omega square root of straight r squared minus straight x subscript 1 squared end root space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space

and space space space space space space space space space space straight v subscript 2 equals straight omega square root of straight r squared minus straight x subscript 2 squared end root space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    Squaring (1) and (2) and subtracting,
    straight v subscript 1 squared minus straight v subscript 2 squared equals straight omega squared left parenthesis straight x subscript 2 squared minus straight x subscript 1 squared right parenthesis

rightwards double arrow space space space space space straight omega squared equals fraction numerator straight v subscript 1 squared minus straight v subscript 2 squared over denominator straight x subscript 2 squared minus straight x subscript 1 squared end fraction

rightwards double arrow space space space space straight omega equals square root of fraction numerator straight v subscript 1 squared minus straight v subscript 2 squared over denominator straight x subscript 2 squared minus straight x subscript 1 squared end fraction end root
    straight omega is the frequency of the oscillation. 
    Question 947
    CBSEENPH11019219

    Molar volume is the volume occupied by molecules of any (ideal) gas at N.T.P. Show that it is 22.4 liters.

    Solution
    Ideal gas equation is given by, 
                      PV = nRT

    For 1 mole of gas, n=1

    So,             PV = RT
     V = RTP

    At N.T.P, T = 273 K and P = 1.013 × 105 Pa
    Therefore, 
    Molar volume is given by,

    V = RTP =8.3×2731.01×105 = 22.4 ×10-3 m3 = 22.4 l


    Question 948
    CBSEENPH11019220

    Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27°C and 1 atmospheric pressure. ( k = 1.38 x 10 2–3 J/K).

    Solution
    The ideal gas equation is given by, 
                      PV = nRT
    Therefore, 
                       n = PVRT = PVnKT

    So, total number of air molecules in volume V is given by

                      nN = PVKT 

    Here,
    Pressure, P = 1 atm = 1.01 × 105 N m-2
    Volume, V = 25 m3
    Temperature, T = 27o C = 300 K
    Boltzman constant, k = 1.38 × 10-23 J/K

    So, total number of air molecules is given by, 

    nN = PVkT = 1.01 × 105× 251.38×10-23×300 = 6.10 × 1026 molecules
    Question 949
    CBSEENPH11019221

    At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the r.m.s. speed of helium gas atom at –20°C? [Atomic mass of Ar = 39.9, He = 4.0]

    Solution
    R.M.S speed of atom is given by,

               v=3kTm = 3RTM                  ...(1)
    where m is mass of atom and M atomic mass. 

    Here we have, 

    Atomic mass of Argon, MAr=39.9,       

    Atomic mass of Helium, 
    MHe = 4.0  

    Temperature, THe =-20°C = 253K,

    Temperature of Argon gas, TAr=? 

    Now usign equation (1), we have 

                      3RTHeMHe = 3RTArMAr  

                                TAr = THeMHeMAr      = 253×39.94      =2.5×103K

    is the required temperature at which r.m.s speed of an argon gas cylinder is equal to the r.m.s speed of helium gas atom.
    Question 950
    CBSEENPH11019222

    Estimate the average thermal energy of a helium atom at:
    (i) room temperature (27°C),
    (ii) the temperature on the surface of the sun (6000°K),
    (iii) the temperature of 10 million degree Kelvin

    (the typical core temperature in the case of a star) k = 1.38x102–3J/K.

    Solution

    The average thermal energy of atom/molecule is given by ,
                       U=32kT
    (i) Given, 

    Room temperature = 27°C = 300 K.

    Therefore the average thermal energy of He atom at room temperature is, 

                 u=32×1.38×10-23×300   = 6.2×10-21J   

    (ii) Given, 

    Temperature at surface of the sun=6000K

    Therefore the average thermal energy of He atom at temperature of the sun is, 

             u=32×1.38×10-23×6000  = 1.24 × 10-19J 

    (iii)  Thermal energy at temperature "
    So, the average thermal energy of the He atom is, 

           u = 32kT = 32×1.38×10-23×107   =2.1 × 10-16 J

    Question 951
    CBSEENPH11019223

    n1 moles of gas A at temperature T1 and n2 moles of gas B at temperature T2 are mixed together without loss of energy. Find the temperature of mixture.

    Solution
    Gas A:

    Number of moles of gas = n1 
    Temperature = T1
    K.E of 1 molecule of gas = 32KT1
    Total K.E = n1N32KT1 

    Gas B

    Number of moles of gas = n2
    Temperature = T2 
    K.E of 1molecue = 32KT2 
    Total K.E = n2 N32KT2

    Now, total kinetic energy of the mixture is,

                   K.E = 32kNn1T1 + n2T2   ... (1)
    Let, the temperature of the mixture at equilibrium be T.  The total K.E of the mixture at equilibrium is given by, 

                    K.E = (n1 + n2) N32kT        ...(2) 

    Now, using equations (1) and (2), we have 

                           T = n1T1 + n2T2n1 + n2
    Question 952
    CBSEENPH11019224

    Two particles vibrating simple harmonically with amplitude A and frequency co along the same straight line. They pass one another when going in opposite direction. Each time when they cross are at a distance 0.707A. What is the phase difference between them?

    Solution

    Let the equation of motion of two particles be, 

    straight y equals straight A space space sin space ωt space and space space space straight y space equals space straight A space sin open parentheses ωt plus straight ϕ close parentheses
    where,
    ϕ is the phase difference between two particles.
    Each time the particles cross at a distance 0.707A.
    Therefore, 
              0.707A=A sinomega t 

    and    0.707A=A sin(wt+straight ϕ


    rightwards double arrow       0.707=sinomega t and, 
              0.707= snspace space space open parentheses ωt plus straight ϕ close parentheses 
    Now,

         sin(left parenthesis ωt plus straight ϕ right parenthesis=sin(left parenthesis wt plus straight ϕ right parenthesis=sin 
          left parenthesis wt right parenthesis cos left parenthesis straight ϕ right parenthesis plus cos left parenthesis wt right parenthesis sin left parenthesis straight ϕ right parenthesis
    Since,  
    s i n space omega t equals 0.707

space rightwards double arrow space c o s space omega t space equals 0.707 space

therefore space 0.707 equals 0.707 space c o s left parenthesis straight ϕ right parenthesis plus 0.707 space space s i n left parenthesis straight ϕ right parenthesis space

rightwards double arrow space space space space c o s left parenthesis straight ϕ right parenthesis plus s i n left parenthesis straight ϕ right parenthesis equals 1 space

rightwards double arrow space space space space c o s open parentheses straight ϕ close parentheses equals 1 minus s i n left parenthesis straight ϕ right parenthesis space

rightwards double arrow space space space space c o s squared left parenthesis straight ϕ right parenthesis equals 1 plus s i n squared left parenthesis straight ϕ right parenthesis minus 2 s i n left parenthesis straight ϕ right parenthesis space

rightwards double arrow space space space 1 minus s i n squared left parenthesis straight ϕ right parenthesis equals 1 plus s i n squared left parenthesis straight ϕ right parenthesis minus 2 s i n left parenthesis straight ϕ right parenthesis space

rightwards double arrow space space space space space space 2 space s i n squared left parenthesis straight ϕ right parenthesis equals 2 space s i n left parenthesis straight ϕ right parenthesis space

rightwards double arrow space space space space s i n left parenthesis straight ϕ right parenthesis space equals space 1 space o r space 0 space

rightwards double arrow space space space space space space straight ϕ space equals space 0 to the power of degree space space o r space 90 degree 
    We can see that ϕ = 0 is not possible.
    Therefore the phase difference between two particles is 90°. 

    Question 953
    CBSEENPH11019225

    Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?

    Solution
    Using the ideal gas equation, we can say the number of moles of gas is given by

                                 n=PVRT

    i.e.  number of moles in a gas is independent of the nature of gas but depends upon P, V and T. Since P, V and T for all gases are same, therefore all the three vessels contain equal number of moles and hence molecules.

    At a given temperature, the r.m.s speed is inversely proportional to mass of atom or molecules, therefore v
    rms is largest for lightest gas i.e. neon.
    Question 954
    CBSEENPH11019226

    Two containers of same size are at the same temperature. One of them contains 1 kg of H2 gas and other 1 kg of O2 gas.
    (a) Which vessel contains more molecules?
    (b) Which vessel is under greater pressure?

    Solution
    (a) Since molecular weight of H2 is less than that of oxygen, therefore the number of moles and hence molecules in 1 kg of H2 are greater than in O2.

    b)  Pressure of a gas in terms of density is given by,
     
                         P = 13ρ v2 

    Here, density of both gases is the same. 

    So,                 P v2 

    Now, according to kinetic interpretation of temperature, 

    12m v2 = 32 K T

    i.e., P v2 1m

    Mass of hydrogen molecules is less than that of oxygen molecule. Therefore, more pressure is exerted on the H2 gas container than the O2 gas container. 
    Question 955
    CBSEENPH11019227

    A particle is executing simple harmonic motion. What fraction of the total energy is kinetic when the displacement is half of amplitude?

    Solution
    Total energy of particle executing simple harmonic motion is,
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#6 {main}</pre>
    Therefore kinetic energy at position is,
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#6 {main}</pre>
    Therefore kinetic energy at position half the amplitude of oscillation is,
         straight K space equals space 1 half m omega squared open square brackets straight A squared minus open parentheses straight A over 2 close parentheses squared close square brackets

space space space space space equals 1 half m omega squared open parentheses 3 over 4 straight A squared close parentheses

space space space space equals 3 over 4 straight E 
    Now fraction of the total energy, which is kinetic is,
    straight K over straight E equals 3 over 4 
    Question 956
    CBSEENPH11019228

    If the absolute temperature of gas is increased by 300%, then by what percentage the r.m.s. velocity of molecules will increase?

    Solution
    According to the kinetic interpretaion of temperature, r.m.s velocity is directly proportional to square root of absolute temperature. 

    i.e.,                          vr  T    ... (1)

    Let the initial temperature of gas be T.

    If the temperature is increased by 300% then the temperature of gas becomes 4T. 

    So, using equation (1), we have

                                  vr'  4T    ... (2) 

    Therefore, from equations (1) and (2), we have

                                   vr' = 2v

    The r.m.s. velocity at 4T temperature is double the r.m.s. velocity at temperature T. 

    So, percentage increase in velocity is given by, 

              %age = 2v-vv×100 = 100 % 

    Question 957
    CBSEENPH11019229

    What is the ideal gas equation for 20 g of gas whose molecular weight is 50g?

    Solution
    Given, molecular weight of the gas = 50 g
    Mass of gas given = 20 g

    The ideal gas equation is given by, 

               PV = nRT                      ...(1) 

    where n is the number of moles. 

    Here the number of moles of gas is, 

                     n=2050=0.4 

    ∴  The gas equation becomes 

                   PV = 0.4RT
    Question 958
    CBSEENPH11019230

    A particle is executing simple harmonic motion. What is the ratio potential energy to kinetic energy when the displacement is 0.6A, where A is the amplitude of oscillation?

    Solution

    Kinetic energy of particle at a distance y from the meain position is,
    K equals 1 half m omega squared left parenthesis A squared y squared right parenthesis
    Potential energy of particle at a distance y from the mean  position is,
    straight U equals 1 half mω squared straight y squared space
    The ratio of potential energy to kinetic energy at a distance y from the mean position is,
    space space space space space space space straight U over straight K equals fraction numerator left parenthesis 0.6 straight A right parenthesis squared over denominator straight A squared minus left parenthesis 0.6 straight A right parenthesis squared end fraction equals fraction numerator 0.36 over denominator 0.64 end fraction equals 9 over 16

    Question 959
    CBSEENPH11019231

    A particle of mass 2kg executing simple harmonic motion is 9J at a distance 0-8m from mean position. If the amplitude of vibration of particle is 1 m, then find the frequency of oscillation of particle.

    Solution
    We know that the kinetic energy of particle at a distance y from the mean position is,
    space space space space space space space space space
space space space space space space space space straight K equals 1 half mω squared left parenthesis straight A squared minus straight y squared right parenthesis
    Given that,
    Mass, m=2kg
    Energy, K.E =9J 
    Amplitude, A = 1m
    Dispalcement, y = 0.8m
    therefore space space space space space 9 equals 1 half cross times 2 cross times straight omega squared open square brackets open parentheses 1 close parentheses squared minus left parenthesis 0.8 right parenthesis squared close square brackets space
space space space space space space space space space space equals 0.36 space straight omega squared space

rightwards double arrow space space space space space straight omega equals 5 space r a d divided by straight s space

rightwards double arrow space space space space space straight v space equals space left parenthesis 5 divided by 2 straight pi right parenthesis space h z 
    This is the frequency of the oscillation of the particle. 
    Question 960
    CBSEENPH11019232

    The kinetic energy of particle executing simple harmonic motion is 6J at mean position and 4J at a distance 4 cm from the mean position. What is the amplitude of oscillation of particle?

    Solution

    Let the particle of mass m vibrate with frequency straight omega and A be the amplitude of oscillation.
    The kinetic energy of particle at a distance y from mean position is, 
    space space space space space space space space space space space space space space space straight K. straight E space equals space 1 half mω squared open parentheses straight A squared minus straight y squared close parentheses space

Here comma space straight K. straight E space at space mean space position space is space 6 space straight J. space

6 space equals space 1 half mω squared space left parenthesis straight A squared space minus 0 squared right parenthesis space equals space 1 half mω squared straight A squared space space space space space space space... space left parenthesis 1 right parenthesis thin space

straight K. straight E space at space straight a space distance space 4 space cm space from space mean space position space is space 4 straight J

4 space equals space 1 half mω squared space left square bracket straight A squared space minus space left parenthesis 0.04 right parenthesis squared right square bracket space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space

Dividing space left parenthesis 1 right parenthesis space by space left parenthesis 2 right parenthesis comma space we space get

space space space space space space fraction numerator 6 over denominator space 4 end fraction space equals space fraction numerator straight A squared over denominator straight A squared minus left parenthesis 0.04 right parenthesis squared end fraction space

rightwards double arrow space space 6 left square bracket straight A squared space minus space left parenthesis 0.04 right parenthesis squared right square bracket space equals space 4 straight A squared space

rightwards double arrow space space straight A squared space equals space 3 space left parenthesis 0.04 right parenthesis squared space

rightwards double arrow space straight A space equals space 0.04 space square root of 3 space straight m space equals space 4 square root of 3 space cm

    A is the amplitude of the oscillation. 

    Question 961
    CBSEENPH11019233

    In a certain region of space there are 5 molecules per cc on an average. The temperature there is 100K. What is the average pressure of this gas?

    Solution
    Given, number of molecules per cc = 5 = 5× 106molecules /m3

    Pressure of the ideal gas is given by,

                                   P = nKT 

    Temperature, T = 100 K
    Boltzman constant, k = 1.38 × 10-23 J/m/K

    So, average pressure of the gas is P = 5×106× 1.38×10-23×100 = 6.9×10-15 N/m2





    Question 962
    CBSEENPH11019234

    A body of mass 0.2kg executes simple harmonic motion of period 4s. The body acquires the velocity 4m/s after 0.6s it passes through its mean position. Find the potential energy of particle at t = 1s.

    Solution
    The velocity of body at any instant is given by, 
    space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight v equals Aω space cos space ωt
therefore space space straight A equals fraction numerator straight v over denominator straight omega space cos space ωt end fraction
    Time period of oscillation of body = 4s.
    Therefore at t = 1s, the body will be at extreme position.
    Potential energy stored in the body at t = 1s is,
    straight U equals 1 half m omega squared straight A squared equals 1 half m omega squared open parentheses fraction numerator straight v over denominator straight omega space c o s space omega t end fraction close parentheses squared
space space space
space space equals 1 half straight m open parentheses fraction numerator straight v over denominator c o s space omega t end fraction close parentheses squared
    We have,
    Velocity of the body, v=4 m/s at t=0.6s
    Mass of the body, m=0.2kg
    
a n d space space space space space straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction equals fraction numerator 2 straight pi over denominator 4 end fraction equals straight pi over 2
therefore space space space space space space straight U equals 1 half cross times 0.2 open parentheses fraction numerator 4 over denominator c o s space 0.3 straight pi end fraction close parentheses squared

space space space space space space space space space space space space equals 4.631 straight J 
    Question 963
    CBSEENPH11019235

    Which of the following function represents:

    (a) simple harmonic motion,

    (b) periodic but not simple harmonic motion,

    (c) non-periodic motion: (i) a cosωt + b sin ωt (ii) sin ωt + sin 2 ωt

    (iii) sin2 ωt ± cos 2 ωt

    (iv) log |cos (ωt) |

    (V) e–ωt

    (vi) cos2 ωt + cos ωt

    Solution

    (i) Simple harmonic

    (ii) periodic but not simple harmonic motion

    (iii) non-periodic

    (iv) periodic but not simple harmonic motion

    (v) non-periodic

    (vi) periodic but not simple harmonic motion

    Question 965
    CBSEENPH11019237

    Find average increase in kinetic energy of a molecule of gas per unit rise in temperature.

    Solution

    At temperature t, 
    Average kinetic energy of gas molecule is given by E = 3 over 2kT

    Question 966
    CBSEENPH11019238

    A particle is executing simple harmonic motion. What is phase relationship between velocity and acceleration?

    Solution
    Let the position of particle be given by,
    space space space space space space space space space space space space space space space space space space straight y equals straight r space sin space ωt space

therefore space straight v equals straight r space straight omega squared space sin space ωt space

space space space straight a space equals space minus straight r space straight omega squared space sin space ωt space

space space space space space equals space straight r space straight omega squared space cos space left parenthesis ωt space plus space straight pi divided by 2 right parenthesis 
    Therefore, velocity lags behind the acceleration by phase ϕ/2.
    Question 967
    CBSEENPH11019239

    Which of following conditions are not sufficient for simple harmonic motion? Explain:

    (a) Acceleration ∝ displacement.

    (b) Force acting on particle should be directed towards mean position.

    Solution

    (a) Acceleration is directly proportional to displacement. But for a motion to execute S.H.M, the direction of acceleration must be directed towards mean position is not indicated here.
    Therefore thiscondition is not sufficient for the particle to execute simple harmonic motion.

    (b) Force is directed towards mean position and hence the acceleration. But this does not give the variations of acceleration w.r.t. displacement.
     Therefore this condition is also not sufficient.

    But both the conditions jointly are sufficient for simple harmonic motion.

    Question 968
    CBSEENPH11019240

    For oscillating simple pendulum:

    (i) What is the direction of acceleration of the bob at: (a) mean position (b) end points?

    (ii) Is the tension in the string constant throughout the oscillation? If not, when it is, 
    (a) the least, (b) the greatest?

    Solution

    (i)
    (a) At mean position the acceleration of bob is radial towards the point of suspension.

    (b) At end position, the acceleration of bob is tangential towards the mean position.

    (ii) Tension in the string is different at different positions.

    (a) Tension is least at the end points.

    (b) Tension is greatest at the mean position.

    Question 969
    CBSEENPH11019241

    A particle is in equilibrium at O. When it is displaced from O, it experiences a force directly proportional to displacement and directed towards O. Explain the motion of particle.

    Solution
    Consider a particle of mass m placed at O.
    Let it be displaced from O by OB = r.
    When the particle is released, it will experience a force F = – kr and starts accelerating towards O.
              
    Let at any instant, the particle be at x from O.
    Let the acceleration of the particle be a and velocity be v. 
    straight a space equals space minus kx over straight m space

and space straight v space equals space square root of straight k over straight m left parenthesis straight r squared minus straight x squared right parenthesis end root space
When space particle space reaches space straight O comma space

Force comma space straight f space equals space minus kx space equals space minus space straight k space left parenthesis 0 right parenthesis space equals 0 space

Velocity comma space straight v space equals space square root of straight k over straight m left parenthesis straight r squared minus 0 squared right parenthesis end root space equals space square root of straight k over straight m end root space straight r space
    That is, the force acting on the particle is zero but velocity is maximum.
    Because of inertia, it will not stop at O, but continue to move in the same direction, i.e. right towards A. 
    As the particle escapes from O towards right, it experiences the force in the left direction and hence retards.
    Ultimately the particle will stop at A.
    Restoring force acts at point A and the particle starts accelerating towards left. 
    When the particle reaches O, force on it will be 0 and velocity will be maximum. 
    Due to inertia, it will again overshoot O, and due to resorting force it retards and comes to rest at B. Thus, the particle oscillates between B and A. 
    Question 970
    CBSEENPH11019242

    Show that motion of body dropped in a tunnel dug along the diameter of the earth is simple harmonic motion. Find the time period of motion.

    Solution
    Let the earth be a homogenous sphere of radius R and density ρ.
    Let the bore be dug along the diameter of the earth.
                     
    The acceleration due to gravity at point P at a distance y from the centre of the earth is,
    straight a equals 4 over 3 πGpy equals 4 over 3 πGρ straight R over straight R straight y equals straight g over straight R straight y
    w here comma space straight g equals 4 over 3 πGρR, is the acceleration due to gravity at the surface of the earth.
    Since acceleration 'a' at point P is directed towards the centre of the earth, therefore in vector form we can write, 
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#6 {main}</pre>
    Therefore the motion is simple harmonic motion.
    
space space space space space space straight omega squared equals straight g over straight R
therefore space space space straight T equals 2 straight pi square root of 1 over straight omega squared end root
space space space space space space space space space equals 2 straight pi square root of straight R over straight g end root 
    Substituting R and g, we get,
    T equals 2 pi square root of fraction numerator 6.4 cross times 10 to the power of 6 over denominator 9.8 end fraction end root equals 5079 s, is the time period of the motion. 
    Question 971
    CBSEENPH11019243

    Why the simple pendulum does not oscillate in free falling lift?

    Solution
    The restoring force for the simple pendulum to oscillate is provided by the gravity.  In free falling lift, the effect value of gravity is zero. Therefore no restoring force acts on the bob and hence the pendulum does not oscillate. 
    Question 972
    CBSEENPH11019244

    An ideal gas has pressure 2 atmospheric at 300 K. If the number of molecules are decreased to half, then at what temperature the pressure of gas will be again 2 atmospheric?

    Solution
    Given, 
    Ideal gas = 2 atm
    Temperature, T = 300 K 

    Ideal gas equation is given by, 

                                 PV = nRT
    So, we have
                                 2V = n R 300  ... (1)

    Now number of molecules of the gas is decreased to half. So, the pressure of the gas will decrease to half.

    Let T' be the increased temperature to increase the pressure to 2 atm. 

                              2V = n2RT'    ... (2)

    From equations (1) and (2), we have

                                 T' = 600 K 



    Question 973
    CBSEENPH11019245

    A man having a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during free fall?

    Solution
    The watch works on the action of spring. The restoring force of the spring does not depend on the gravity. Therefore the watch gives correct time even during free fall. 
    Question 974
    CBSEENPH11019246

    What is molar volume of gas in ft3 at N.T.P.?

    Solution
    The molar volume is 22.4 l at Normal Temperature and Pressure.  

    So, V = 22.4 l = 22.4 × 103 cm3 

          V = 22.4 × 103 ×130.48 ft3 
             = 6.791 ft3 
    Question 975
    CBSEENPH11019247

    Why does the motion of simple pendulum in due course stop while oscillating in air?

    Solution
    A simple pendulum oscillates in air, due to air friction.
    The energy of the oscillator is dissipated into heat energy.
    When whole of the energy of simple pendulum is dissipated, the simple pendulum stops oscillating. 
    Question 976
    CBSEENPH11019248

    Find the rms velocity of the molecules of gas of density 1.67kg/m3 at 2 atmospheric pressure. [Take 1 atm=105 N/m2.]

    Solution
    Given, 
    Density of gas, ρ = 1.67 kg/m3 
    Pressure, P = 2 atm = 2× 105 N/m2 

    R.M.S velocity of molecules = ?

    vr = 3Pρ = 3×2×1051.67 = 600 m/s 
    Question 977
    CBSEENPH11019249

    An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35° C? 

    Solution

    At the bottom of the lake,

    Volume of air bubble, V =  1.0 cm3 
    Depth of lake = 40 m
    Pressure at the bottom of the lake, PB = 40 m of H2O + 1 atm                                                                    =40×1000×9.8 + 1.01 ×105                                                                     = 4.93×105 Pa
    Temperature, T = 12o C = 12+273 K = 285 K

    At the top of the lake

    Volume of air bubble = ?
    Changed temperature, T' = 35o C
    Pressure at the top of the lake, P= 1 atm = 1.01× 105 Pa

    Volume at the top, Vt = ?

    Now using the gas equation, we have

    PbVbTb = PtVtTt   Vt = PbPtTtTbVb              = 4.93×1051.01 ×105×308285×1              = 5.3 cm3 

    is the required volume of the air bubble at the surface.




    Question 978
    CBSEENPH11019250

    The black oxide of copper is prepared in four different ways and the following observations noted:

    Copper (in g)

    Copper oxide (in g)

    5.015

    6.284

    8.286

    10.357

    18.443

    23.169

    12.011

    15.051

     

    What do you infer from this data?

    Solution
    Fraction of the copper present in the oxide of copper is:

    (a) 5.015/6.284 = 0.798

    (b) 8.286/10.357 = 0.800

    (c) 18.443/23.169 = 0.796 

    (d) 12011/15.051 = 0.798

    Since the fraction of the copper present in the oxide of copper is constant, this proves the law of constant proportions. 
    Question 979
    CBSEENPH11019251

    A spring is fixed at one end and at the free end a mass is attached. The system is placed on a horizontal frictionless table. Show that when the mass is displaced from its equilibrium position, it will execute simple harmonic motion. Also find the time period of oscillation.

    Solution
    Let a spring have its one end fixed to a rigid support and mass m attached at the other end as shown in figure. 
                   
    Let the mass m be pulled towards right by distance y.
    The spring will also extend by distance y.
    The restoring force set up in the spring is,
                           F=-ky
    where,
    k is the spring constant of the spring.
    Negative sign implies restoring force is opposite to the direction of extension.
    When the mass is released, it will move under the action of restoring force.
    The acceleration produced in the mass due to restoring force is, 
                        straight a equals straight f over straight m equals negative straight k over straight m straight y
    Thus, the acceleration is directly proportional to displacement and directed towards the mean position. 
    Hence, the motion of the mass is simple harmonic motion.
    The time period is given by, 
    straight T equals 2 straight pi square root of displacement over acceleration end root space

space space equals 2 straight pi square root of straight y over kyIm end root

space space equals 2 straight pi square root of straight m over straight k end root 
    Question 980
    CBSEENPH11019252

    A vessel of volume V = 5 litre contains 1.4gm of N2 at temperature 1800K. If at this high temperature, 30% of the gas dissociates into atoms, then find the pressure of gas.

    Solution
    Given, 

    Volume of the vessel, V = 5 l 

    Molecular weight of nitrogen = 28 g 

    Temperature, T = 1800 K

    Number of moles in 1.4 gm of N
    2 is, no=1.428=120moles 

    Also, given that at high temperature 30% of the gas dissociates into atoms.

    Number of moles of gas dissociated into atoms is,
    n'=0.3no=0.3×120=3200moles 

    Number of moles of atomic nitrogen is,   n1=2n'=3200×2=0.03 moles 

    Number of moles of molecular nitrogen is,
    n2=0.7×120=0.035 moles

    Thus total number of moles of nitrogen gas is given by n = n+ n2 = 0.065 moles

    Now, using the ideal gas equation, 
                             PV = nRT

    We have, n = 0.065 
    T = 1800 K
    V = 5 litre = 5× 10-3 m3 

    Therefore, 

     Pressure, P = nRTV = 0.065×8.314×18005×10-3                      = 1.945 × 105 N / m2


                  
    Question 981
    CBSEENPH11019253

    Calculate the number of molecules in 1cc of ideal gas at 27°C and 10mm pressure of mercury. Mean kinetic energy of molecule at 27°C is 6.21x10–14 ergs.

    Solution
    Given, 1 cc of ideal gas at 27o C and 10 mm pressure of mercury.

    Mean K.E of molecule = 6.21x10–14 ergs.

    We know that,

    Pressure of a gas is equal to two-third of the average kinetic energy per unit volume of the gas.

    Thats is, 

                 P=23Ed Ed = 32P 

    Here we have, 

       P=10mm of mercury = 1 cm of mercury       = 1×13.6×980=1.33×104 dyne/cm2.

    Therefore, 

       Ed=32P = 32×1.33×104         = 2×104 erg/cc                            ....(1)   
          
    Let n be the number of molecules in 1 cc of gas, therefore the energy per cm3 of gas is,

       Ed=n×6.21×10-14 ergs/cc           ...(2)

    From equations (1) and (2),

    Number of molecules is = n=2×1046.21×10-14=3.22×1017 molecules
    Question 982
    CBSEENPH11019254

    Five molecules of an ideal gas have their speed 2km/s, 2.4km/s, 3km/s, 1.6 km/s and 2 km/s respectively. Find the r.m.s. velocity of gas molecules.

    Solution

    The r.m.s velocity of gas molecule is given by,
    vr = v12+v22+v32+v42+v525    = (2)2 + (2.4)2+ (3)2+(1.6)2+(2)25   = 2.25 km/s 

      
              

    Question 983
    CBSEENPH11019255

    Show that motion of a loaded spring is simple harmonic motion and calculate its time period of oscillation.Show that motion of a loaded spring is simple harmonic motion and calculate its time period of oscillation.

    Solution
    Let a point mass m be suspended from a massless spring suspended from a rigid support O.

                        
    Let due to load m the spring extend by length l to acquire the equilibrium.
    The restoring force set up in the spring is,
                              straight F subscript 1 equals negative straight k straight ell 
    where,
    k is the spring constant of the spring.
    Negative sign is because restoring force is in the upward direction opposite to the direction of extension which is in downward direction.
    As the mass is in equilibrium, therefore
    space space space space mg space equals space straight k straight ell space space space space space space

rightwards double arrow space space space space straight k equals mg divided by straight ell
    Let the mass be now pulled further by a distance y.
    Now the restoring force set up in the spring is,
                       straight F subscript 2 equals negative straight k left parenthesis straight ell plus straight y right parenthesis
    The net force on the mass in this position is,
    straight f equals straight F subscript 2 plus mg

space space equals negative straight k left parenthesis straight ell plus straight y right parenthesis plus mg space

space space equals left parenthesis negative straight k straight ell plus mg right parenthesis minus ky

space space space equals negative ky
    The acceleration produced in the mass is,
    straight a equals straight f over straight m equals negative straight k over straight m straight y
    Thus, the acceleration is directly proportional to displacement and directed towards mean position.
    Hence, the motion is simple harmonic motion.
    The time period is given by, 

    space space space straight T equals 2 straight pi square root of displacement over acceleration end root space

space space space straight T equals 2 straight pi square root of fraction numerator straight y over denominator straight k divided by straight m end fraction end root space

space space space straight T equals 2 straight pi square root of straight m over straight k end root space

space space space straight T equals 2 straight pi square root of straight ell over straight g end root space space space space space space space space space space space space space space open square brackets because space mg space equals straight k straight ell close square brackets 
    Question 984
    CBSEENPH11019256

    An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder. (R= 8.3 J/mol/ K, molecular mass of O2=32].

    Solution
    Ideal gas equation is given by, PV=nRT
    ∴                         n=PVRT     ... (1)

    Volume of oxygen gas cylinder, V = 30 l = 30×10-3 m3 
    Pressure, P = 15 atm + 1 atm = 16 atm = 16×1.01×105 Pa
    Temperature, T = 27o C = 300 K

    Putting this in equation (1), we have
    n = 16×1.01×105×30×10-3300×8.3 = 19. 5 moles

    After some amount of oxygen is withdrawn from the cylinder, 

    New pressure, P' = 11 atm + 1 atm = 12 atm
    Temperature, T' = 17o C = 290 K 

    So, number of moles = 12×1.01×105×30×10-3290×8.3 = 15.1 moles

    So, number of moles withdrawn from the cylinder is given by,
    n = 19.5-15.1 = 4.4 moles 

    So, mass of O2 taken out is given by, 
    m = 4.4 × 32 gm = 140.8 gm 

    Question 985
    CBSEENPH11019257

    Calculate the mean free path of a gas molecule, given that the molecular diameter js 2 x 10–8cm and number of molecules per cc is 3 x 1019.

    Compare the mean free path with mean wavelength of visible light.

    Solution

    Given,
    Molecular diameter, d=2×10-8cm,Number of molecules per cc, n=3×1019 per cc


    Mean free path of the molecule is given by,

                         λ=12πd2n
                           =12×3.14(2×10-8)2×3×1019

                           =1.876×10-5 cm=1.876 Ao 

    The average wavelength of visible light is 5800 Ao.

    Thus the mean free path is less than the average wavelength of visible light. 

    Question 986
    CBSEENPH11019258

    An ideal gas is enclosed in a horizontal cylinder of area of cross-section A fitted with an airtight and frictionless piston of mass M. The piston is in equilibrium with atmospheric pressure P. The volume of gas enclosed in the cylinder is V. The cylinder is slightly displaced from equilibrium position and released. Show that the motion of piston is simple harmonic motion and find the time period of oscillation. Assume that the system is completely isolated from surrounding.

    Solution
    The system is as shown in the figure below.
                       
    Let the piston be displaced by y towards left.
    Let dV be the decrease in volume and dP be the increase of pressure. 
    The system is isolated from surrounding. Therefore, the change in pressure and volume is according to adiabatic conditions.
    Thus the equation of state can be written as, 
    space space space space space space space space space space PV to the power of straight y equals space constant space

rightwards double arrow space space space space PγV to the power of straight y minus 1 end exponent dV plus straight V to the power of straight y dP equals 0
rightwards double arrow space space space space space dP equals negative yPdV over straight V space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
    The equation (1) gives the change in pressure of gas when the volume of gas is changed by dV.
    Here, change in the volume of gas when the piston is displaced by y is Ay.
    Thus, 
                      dP equals negative yPA over straight V straight y
    Force on piston is,
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#6 {main}</pre>
    The acceleration of piston is,
              straight a equals straight F over straight M equals negative yPA squared over MV straight y equals negative straight omega squared straight y
    So, it is clear from the above equation that the acceleration is proportional to displacement and directed towards the mean position.
    Therefore the motion of piston is simple harmonic motion.
    The time period of oscillation is,
                straight T equals 2 straight pi square root of 1 over straight omega squared end root equals 2 straight pi square root of MV over yPA squared end root
    Question 987
    CBSEENPH11019259

    A body executes simple harmonic motion with time period T1 under the action of force F1 and with time period T2 under the action of force F2. Find the time period of oscillation if both the forces act simultaneously on the body.

    Solution
    Let m be the mass of the body.
    Time period of oscillation of body under the action of force F1 is T1.
    Therefore,
                       straight F subscript 1 equals straight m open parentheses fraction numerator 2 straight pi over denominator straight T subscript 1 end fraction close parentheses squared space straight y
    The time period of oscillation of body under the action of force F2 is T2. 
    Therefore, 
                         space space space straight F subscript 2 equals straight m open parentheses fraction numerator 2 straight pi over denominator straight T subscript 2 end fraction close parentheses squared space straight y
    When both the forces act simultaneously on the body, the net force is F = F1 + F2. 
    Let T be the time period of oscillation.
    Thus, 
    space space space space space space space space space space space space space space space space space space space space space space space space space space space straight F equals straight F subscript 1 plus straight F subscript 2

straight i. straight e. space space space straight m open parentheses fraction numerator 2 straight pi over denominator straight T end fraction close parentheses squared space straight y equals straight m open parentheses fraction numerator 2 straight pi over denominator straight T subscript 1 end fraction close parentheses squared space straight y plus straight m open parentheses fraction numerator 2 straight pi over denominator straight T subscript 1 end fraction close parentheses squared space straight y space

rightwards double arrow space space space space space space space 1 over straight T squared equals fraction numerator 1 over denominator straight T subscript 1 superscript 2 end fraction plus fraction numerator 1 over denominator straight T subscript 2 superscript 2 end fraction space

rightwards double arrow space space space space space space space space space space space space straight T equals fraction numerator straight T subscript 1 straight T subscript 2 over denominator square root of straight T subscript 1 superscript 2 end root plus straight T subscript 2 superscript 2 end fraction
    Question 988
    CBSEENPH11019260

    How does the time of oscillation of simple pendulum change when it is taken from surface of the earth to deep in the mine? 

    Solution
    The time period of oscillation of simple pendulum is given by,
                        straight T equals 2 straight pi square root of straight L over straight g end root
    When a person goes from the surface of the earth to deep in the mine, the acceleration due to gravity decreases. Therefore, time period of oscillation of the simple pendulum increases deep in the mine. 
    Question 989
    CBSEENPH11019261

    The bob of a simple pendulum is made of iron and its time period of oscillation is T. If the bob is replaced by another bob of same size but made of wood, then what will be the effect on the time period of oscillation?

    Solution
    The time period of oscillation is independent of the nature of material and size of the ball. Therefore, the time period of the oscillation of the bob is not affected. 
    Question 990
    CBSEENPH11019262

    A girl is swinging on a swing in sitting position. How will the time, period of swing change if the girl stands up?


    Solution
    The swinging girl can be considered as a simple pendulum.
    The time period of oscillation of simple pendulum is,
                          straight T equals 2 straight pi square root of straight L over straight g end root  
    where,
    L is the length of simple pendulum.
    Length of the simple pendulum is equal to distance from point of suspension to CG of the oscillating body.
    When the girl stands up, the distance of CG from the point of suspension decreases. Therefore, time period of oscillation also decreases. 
    Question 991
    CBSEENPH11019263

    A spring of constant k is divided into two parts of equal length. What is the spring constant of either part?

    Solution
    Let a spring AC of spring constant K be divided into two equal parts AB and BC.
    We have AB = BC, therefore spring constant of each part will be the same and let it be K'.
    Suppose, spring AC consists of two springs AB and BC connected in series. 
     
    The spring constant is given by, 
    space space space space space space 1 over straight K space equals space fraction numerator 1 over denominator K apostrophe end fraction plus fraction numerator 1 over denominator K apostrophe end fraction space equals space 2 over K to the power of apostrophe space

rightwards double arrow space space space K apostrophe space equals space 2 K
    i.e. spring constant of both the parts is 2K.
    Question 992
    CBSEENPH11019264

    Derive an expression for the pressure of a gas.

    Solution
    The molecules of gas are in random motion. So,  they exert pressure on the walls of the container and strike with the walls of the container. Therefore, they exert thrust or force on the wall.  
    The thrust exerted by molecules per unit area of the wall of the container is called pressure. 
     
    Consider a cubical vessel of side  is enclosing an ideal gas. 
    Let m be the mass of gas molecules. 
    Let v1 be the velocity of one of the gas molecules. 
    If v1x , v1y, v1z are the components of v1
    Then, 
    straight v subscript 1 space space equals space straight v subscript 1 straight x end subscript space straight i with hat on top space plus space straight v subscript 1 straight y end subscript space straight j with hat on top space plus straight v subscript 1 straight z end subscript space straight k with hat on top space

straight v subscript 1 to the power of 2 space end exponent equals space space straight v subscript 1 straight x end subscript squared space plus space space straight v subscript 1 straight y end subscript squared space plus space straight v subscript 1 straight z end subscript squared space 
    Let this molecule approaches the face towards L. 
    Before collision with the walls, momentum is given by, 
    straight p subscript 1 space equals space mv subscript 1 space equals space mv subscript 1 straight x end subscript space straight i with hat on top space plus space space mv subscript 1 straight y end subscript straight j with hat on top space plus space mv subscript 1 straight z end subscript space straight k with hat on top space 
    Since molecule collides with face L due to X-component of motion, therefore, due to collision only X-component of momentum will change. 
    Since collision is elastic, therefore momentum after collision is given by, 
    stack straight p subscript 1 apostrophe with rightwards arrow on top space equals space minus mv subscript 1 straight x end subscript straight i with hat on top space plus space mv subscript 1 straight y end subscript straight j with hat on top plus mv subscript 1 straight z end subscript straight k with hat on top 
    Due to collision, the change in momentum of molecule is, 
    space stack straight p subscript 1 apostrophe with rightwards arrow on top minus stack straight p subscript 1 with rightwards arrow on top space equals space minus 2 mv subscript 1 straight x end subscript straight i with hat on top
    Let n be the number of molecules per unit volume.
    Since the molecules approach towards face L with velocity v1x, therefore, the molecules lying in the cylinder of length v1x and area of cross-section A may collide with face L on area in one second. 
    Since molecular motion is random, therefore, it is logical to suppose only half of the molecules that lie in cylinder will strike the area A in one second. 

    Number space of space molecules space that space strike space on space
area space straight A space of space face space straight L space in space one space second space equals space 1 half left parenthesis nAv subscript 1 straight x end subscript right parenthesis space

Change space in space momentum space of space striking space molecules
on space area space straight A space in space one space second space equals space Change space in space momentum
of space one space molecule right parenthesis space straight X space left parenthesis Number space of space molecules space striking space per space sec right parenthesis

space space space equals space minus space 2 straight m space straight v subscript 1 straight x end subscript straight i with hat on top space cross times space 1 half left parenthesis nAv subscript 1 straight x end subscript right parenthesis space equals space minus nm space Av subscript 1 straight x end subscript squared space straight i with hat on top space

Since comma space velocity space of space all space the space molecules space is space not space the space same comma space
therefore comma space straight v subscript 1 straight x end subscript squared space must space be space replaced space by space mean space square space
velocity space straight i. straight e. space straight v with rightwards harpoon with barb upwards on top space subscript straight x squared space

Therefore comma space

Change space in space momentum space of space colliding space molecules
in space one space second space equals space minus nmA straight v with rightwards harpoon with barb upwards on top space subscript straight x squared space straight i with hat on top space

But comma space change space in space momentum space in space one space second space is space force.

Therefore comma space

Force space on space colliding space molecules space equals space minus mA straight v with rightwards harpoon with barb upwards on top space subscript straight x squared space straight i with hat on top space
    According to Newton’s third law of motion, to every action, there is an equal and opposite reaction.
    Therefore, gas molecules exert an equal and opposite force on area A of face L as experienced by gas molecules. 
    Force on area A of face L, straight F subscript straight x equals nmA straight v with rightwards arrow on top subscript straight x squared 
    Therefore pressure on face L is, 
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#6 {main}</pre> 
    Similarly, 
    straight P subscript straight y space equals space nmA space straight v with bar on top subscript straight y squared space

straight P subscript straight z space equals space nmA space space straight v with bar on top subscript straight z squared space

Since comma space gas space exerts space equal space pressure space on space all space the space
walls space of space the space container comma space

straight P subscript straight x space equals space straight P subscript straight y space equals straight P subscript straight z space equals space straight P thin space

rightwards double arrow space straight P space equals space fraction numerator straight P subscript straight x space plus space straight P subscript straight y space plus straight P subscript straight z over denominator 3 end fraction

space space space space space space space space space equals space 1 third space nm space left parenthesis straight v with bar on top subscript straight x squared space plus space space straight v with bar on top subscript straight y squared space plus straight v with bar on top subscript straight z squared space right parenthesis space

space space space space space space space space space space equals space 1 third space nm space straight v with bar on top squared space equals space 1 third space straight rho space straight v subscript straight r squared space space space space space space space left square bracket because space nm space equals space straight rho right square bracket

where comma space straight v subscript straight r squared space is space the space mean space square space velocity space of space
gas space molecules. space
    Question 993
    CBSEENPH11019265

    What is the length of second's pendulum on the surface of the moon?

    Solution
    The pendulum whose time period is 2 seconds is called Secon's pendulum. 
    That is, time period of a pendulum is given by, 
    space space space space space space space straight T equals 2 straight pi square root of straight L over straight a end root
rightwards double arrow space space straight L equals fraction numerator straight T squared over denominator 4 straight pi squared end fraction straight a space

Here space space space space space space space straight T equals 2 straight s space semicolon space space space straight a equals straight g divided by 6 
    [because on the surface of the moon, a=g/6]
    Length space of space the space pendulum comma space space straight L equals fraction numerator straight T squared over denominator 4 straight pi squared end fraction straight a space equals space fraction numerator left parenthesis 2 right parenthesis squared over denominator 4 straight pi squared end fraction straight g over 6
           
                                                        = 0.1655m
                                                        = 16.55cm
    Question 994
    CBSEENPH11019266

    Explain the phenomenon of thermal expansion on the basis of atomic theory.

    Solution
    According to the Atomic theory, when the substance is heated, it requires more space for their motion and hence it expands. Intermolecular force of attraction between the molecules decreases when a substance is heated. 
    Question 995
    CBSEENPH11019267

    What do you mean by isotropic material?

    Solution
    Isotropic material is whose expansion is same in all the directions. Isotropic means having identical values of a property in all direction. 
    Question 996
    CBSEENPH11019268

    What do you mean by anisotropic material?

    Solution
    A material whose property is different in different directions is called anisotropic material.
    Question 997
    CBSEENPH11019269

    A spring pendulum oscillates with frequency f. If we cut a part of spring and then again disturb, how does the frequency of oscillation change?

    Solution
    Frequency of oscillation of spring pendulum is,
                          
straight f equals fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root
    If we cut a part of spring, the spring constant of the rest of the part of the spring will increase.
    Therefore, the frequency of oscillation will increase.
    Question 998
    CBSEENPH11019270

    Do all the bodies expand on heating?

    Solution
    No, all bodies do not expand on heating.

    For e.g. water when heated from 0°C to 4°C contracts. 
    Question 999
    CBSEENPH11019271

    Is there any specific direction of expansion of a solid when heated?

    Solution
    No, solid expands in all directions. But, expansion may be different in different directions.
    Question 1000
    CBSEENPH11019272

    Define temperature coefficient of expansion.

    Solution
    The temperature coefficient of expansion is the ratio of the change in dimensions per unit temperature to original dimensions per unit change in temperature.
    Question 1001
    CBSEENPH11019273

    What is the dimensional formula of temperature coefficient of expansion?

    Solution
    The dimensional formula of temperature coefficient of expansion is [M°L°T°K–1]
    Question 1002
    CBSEENPH11019274

    Two springs of constants k1 and k2 are connected in parallel as shown. What is the frequency of oscillation?


    Solution

    The two springs shown in the figure are parallel.
    Therefore, the spring constant of the combination is k = k1 + k2.

    Therefore, time period of oscillation of system is,
    space space space space space space space space
space straight T equals 2 straight pi square root of fraction numerator straight m over denominator straight k subscript 1 plus straight k subscript 2 end fraction end root 

    Question 1003
    CBSEENPH11019275

    A pendulum is made of a vessel filled with water suspended from a string. The water gradually flows out from a hole at the bottom of the vessel. How will the period of pendulum change as the liquid flows out?

    Solution
    Time period of simple pendulum is,
                              straight T equals 2 straight pi square root of straight ell over straight g end root
    where, 
    Error converting from MathML to accessible text. is the distance of CM of bob from point of suspension.
    As the water flows out, T first increases to a maximum and then starts decreasing.
    When the whole of the water flows out, time period first increases. Then, having a maximum value T starts decreasing and finally has the same time period as initial. 
    Question 1005
    CBSEENPH11019277

    What is the relation between straight alpha comma space straight beta space and space straight gamma ?

    Solution

    Relation between straight alpha comma space straight beta space and space straight gamma space is given by, 
    space space space space space space space space space space space space space space straight gamma space equals space 3 straight alpha

space space space space space space space space space space space space space straight beta space equals space 2 straight alpha space

i. e. comma space space space 6 straight alpha space equals space 3 straight beta space equals space 2 straight alpha 

    Question 1006
    CBSEENPH11019278

    How much length of a rod of length 1 m increases when heated through 1°C?

    Solution
    The increase in the length of rod is,
    increment ell space equals space ell αθ
space space space space space space space equals space 1 cross times straight alpha cross times 1
space space space space space space space equals space straight alpha 
    That is,
    Increase in the length is equal to the temperature coefficient of linear expansion.
    Question 1007
    CBSEENPH11019279

    The radius of a metallic ring is R and coefficient of linear expansion is   straight alpha. What is the increase in area of ring when heated by 1°C?

    Solution

    Radius of the metallic ring = R
    Coefficient of linear expansion = straight alpha 
    The ring is heated by a temperature of 1oC. 
    Therefore, increase in are of the ring = 2 παR squared.

    Question 1008
    CBSEENPH11019280

    Why are the two pieces of rail track separated by a small distance in between? 

    Solution
    A small gap is left between adjacent rails so as to allow expansion due to the change in temperature. Or otherwise the rail may bend and may derail from its tracks. 
    Question 1009
    CBSEENPH11019281

    Two spheres of same material and same mass - one solid and other hollow are heated through same range of temperature. Which will expand more?

    Solution
    Hollow sphere contains air inside it and the coefficient of expansion of air is greater than solids. Therefore, the hollow sphere will expand more than the solid sphere.
    Question 1010
    CBSEENPH11019282

    A simple pendulum oscillates with angular amplitude ϕ . If I is the length of string and m is the mass of pendulum, then what is the tension in string at the mean position? 


    Solution

    At the mean position B, tension is given by, 
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#6 {main}</pre>
    where,
    v is the velocity of bob at lowest point B.
    Let A be the extreme position of the bob during oscillation and is at a height h from the lowest point B.
    From the figure,
    space space space space space space space space space OC equals straight ell space cosϕ space

therefore space space space space BC equals straight h equals straight ell left parenthesis 1 minus cosϕ right parenthesis
    As the bob moves from A to B, the potential energy decreases by mgh and is converted into kinetic energy at B.
    
straight i. straight e. space space space space mgh space equals space mg straight ell open parentheses 1 minus cosϕ close parentheses equals 1 half mv squared
rightwards double arrow space space space space space space space space space space straight v squared equals 2 straight g straight ell open parentheses 1 minus cosϕ close parentheses space

therefore space space space space space space space straight T equals mg plus fraction numerator straight m 2 straight g straight ell left parenthesis 1 minus cosϕ right parenthesis over denominator straight ell end fraction space

space space space space space space space space space space space space space space equals mg left square bracket 1 plus 2 left parenthesis 1 minus cosϕ right parenthesis right square bracket space

space space space space space space space space space space space space space space equals space mg left square bracket left parenthesis 3 minus 2 space cosϕ right parenthesis right square bracket space
  

    Question 1011
    CBSEENPH11019283

    If someone tells you that difference between the lengths of two different rods of different material is constant at all the temperatures would you believe him?

    Solution

    Yes, the length of two rods of different material is constant at all the temperatures. The length does not vary on heating. It remains the same. 

    Question 1012
    CBSEENPH11019284

    Two rods of length L1 and L2 of materials having temperature coefficient of linear expansions α1 and α2, are placed parallel to each other. What is the relation between their lengths, so that the difference in the length of two rods is independent of temnerature?

    Solution

    The length of the rods is independent of the temperature. 
    The relation between the length of the rods is given by, 
                               straight L subscript 1 over straight L subscript 2 equals straight alpha subscript 2 over straight alpha subscript 1

    Question 1013
    CBSEENPH11019285

    A container contains some liquid so that the volume of empty portion of container is independent of the temperature of system. What is quantity of liquid in the container?

    Solution

    The quantity of the liquid in the container is given by, 
                  space space space straight V subscript liq equals left parenthesis straight V subscript cont right parenthesis end subscript straight alpha subscript cont over straight alpha subscript liq 
    where, 
    Vliq = Vol. of liquid, 
    Vcont = Vol. of container
    straight alpha subscript liq  = Temp. coefficient of liquid. 
    straight alpha subscript space container end subscript space = Temp. coefficient of container. 

    Question 1014
    CBSEENPH11019286

    Two identical simple pendulums are suspended from the roof of two lifts at rest and have time period of one second. One lift starts falling freely and second lift descends with constant velocity. How does the time period of the pendulums alter?

    Solution
    Time period of simple pendulum is,
                          straight T equals 2 straight pi square root of straight ell over straight g end root 
    The body in the lift, which falls freely, will be in the state of weightlessness.
    The effective value of free  g = 0.
    Therefore, time period, t = ∞.
    i.e. pendulum will not oscillate.
    The lift is moving with constant velocity, the value of g does not change.
    Therefore, the pendulum continues to beat the seconds.
    Question 1015
    CBSEENPH11019287

    At what point along the path of simple pendulum, the tension in the string is maximum? Explain.

    Solution
    Tension in the string at any point is, 
                     straight T equals mg space coθ plus mv squared over straight ell
    where,
    straight theta is the angle that the string makes with vertical and v is the speed of bob.
    At the lowest point, cos θ = 1 i.e. maximum.
    Also, the speed of bob is maximum at the lowest point.
    Therefore, maximum tension is at the lowest point which is the mean position. 
    Question 1016
    CBSEENPH11019288

    Which of the two: coefficient of real expansion or that of apparent expansion of liquid is greater?

    Solution
    The coefficient of real expansion of a liquid is greater than that of apparent expansion. 
    Question 1017
    CBSEENPH11019289

    What is the relation between coefficient of real expansion and that of apparent expansion of liquid?

    Solution

    Relation between coefficient of real expansion and that of apparent expansion of liquid is given by, 
    straight gamma subscript app space equals space straight gamma subscript real space minus space straight gamma subscript container

    Question 1018
    CBSEENPH11019290

    A copper vessel of volume 100 cc is filled to brim with liquid. On heating the liquid, 3.2cc of liquid overflows. Is the actual expansion of liquid 3.2cc?

    Solution
    No. The actual expansion of the liquid is more than 3.2 cc and that is why the liquid overflows on heating.
    Question 1019
    CBSEENPH11019291

    Why we define only volume expansion of liquids and gases?

    Solution
    Liquids and gases do not have a shape of their own. Hence, we define volume expansion of liquids and gases. 
    Question 1020
    CBSEENPH11019292

    A brass rivet is tightly fitted in a steel plate. Would you cool or heat the system to loose the rivet from the hole?

    Solution
    The coefficient of expansion of brass is greater than steel. Therefore, to loosen the rivet, system should be cooled down. 
    Question 1021
    CBSEENPH11019293

    Why is platinum wire used in a fuse enclosed in glass?

    Solution
    Platinum wire is used in a fuse in a glass because the temperature coefficient of linear expansion of platinum and glass is almost the same. Hence, the wire will not undergo a short circuit easily. 
    Question 1022
    CBSEENPH11019294

    The length of spring of the oscillator is L and beats one second when loaded by mass m. If spring is cut into two equal parts each of length L/2, then by what percentage the loaded mass must be changed so that the time period of oscillation remains unchanged? 

    Solution
    Time period of spring pendulum is given by,
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#6 {main}</pre>
    If the spring is cut into two equal parts, the spring constant of each part becomes equal to 2k.
    Mass of the load is increased to m' to keep the time period same. 
    Thus comma space space space
straight T equals 2 straight pi square root of straight m over straight k end root equals 2 straight pi square root of fraction numerator straight m apostrophe over denominator 2 straight k end fraction end root space

rightwards double arrow space space space space space space space space space space space space space straight m to the power of apostrophe space equals space 2 straight m
 
    Therefore, the percentage increase in mass is 100%. 
    Question 1023
    CBSEENPH11019295

    A metal disc has a hole in it. What happens to the size of hole when disc is heated?

    Solution
    When the metal disc is heated, the size of the hole increases because of thermal expansion. 
    Question 1024
    CBSEENPH11019296

    Two beakers are filled with water to brim at 4°C. The temperature of one beaker is increased and that of second beaker is decreased. What will you observe?

    Solution
    Water in both the beakers will overflow because the density of water is maximum at 4°C. On increasing or decreasing the temperature of the beaker, the water will overflow as it is filled upto the brim. 
    Question 1026
    CBSEENPH11019298

    The top of lake is frozen. Air in contact is at temperature –10°C. What do you expect the temperature of water in contact with lower surface of ice and at the bottom of lake?

    Solution
    The temperature of water in contact with the lower surface of ice is 0°C and at the bottom of the lake is –4°C. This is the anomalous behaviour exhibited by water. The volume of the water molecules decreases as the temperature rises and expands as the pressure increases. 
    Question 1027
    CBSEENPH11019299

    Why invar is used in making pendulum clocks?

    Solution
    Invar is an alloy which has a low value of temperature coefficient of expansion. The length of a pendulum made of invar does not change with temperature and hence the time period of oscillation remains the same. Hence the time shown by the clock is accurate. 
    Thus, invar is used in making the pendulum clocks. 
    Question 1028
    CBSEENPH11019300

    A pendulum clock keeps correct time at 25°C. Will it lose or gain at 35°C?

    Solution
    The pendulum clock will run slow and hence lose time. 
    If the pendulum of the clock is made with invar, it will give the time accurately at any temperature. Invar is an alloy having small temperature coefficient of expansion. 
    Question 1030
    CBSEENPH11019302

    What do you mean by anomalous expansion of liquids?

    Solution
    Anomalous expansion of liquid implies that with the decrease in temperature, the volume of liquid increases and density decreases. 
    Question 1031
    CBSEENPH11019303

    A wire is stretched between two rigid supports. What would happen on cooling the wire?

    Solution
    The wire becomes taut and may break on excessive cooling. 
    Question 1034
    CBSEENPH11019306

    What is the principle of thermostat?

    Solution
    The principle of the thermostat is that different metals have different coefficients of linear expansion. 
    Question 1035
    CBSEENPH11019307

    How would you remove a dent in the tennis ball?

    Solution
    When a tennis ball is put in the hot water, tennis ball encloses the air inside it and undergoes expansion. This will remove the dent from the tennis ball. 
    Question 1036
    CBSEENPH11019308

    A spring of force constant 1200 N/m is mounted on a horizontal table as shown. A mass of 30 kg is attached to the free end of the spring, pulled side ways to a distance of 2-0 cm and released.

    (a) What is the frequency of oscillation of the mass?

    (b) What is the maximum acceleration of mass?

    (c) What is the maximum speed of mass?

    Solution

    We have,
    Spring constant, k=1200 N/m
    Mass of the spring, m=30kg
    Amplitude of vibration = 2.0 cm =0.02m
    (a) We know frequency of oscillation is,
    straight v equals fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root equals fraction numerator 1 over denominator 2 straight pi end fraction square root of 1200 over 3 end root equals fraction numerator 20 over denominator 2 straight pi end fraction equals 10 over straight pi Hz
    (b) We know maximum acceleration of mass is, space space space a equals straight omega squared straight A
    therefore space straight a equals left parenthesis 2 πv right parenthesis squared space straight A equals open parentheses 2 straight pi 10 over straight pi close parentheses squared straight x 0.02 equals 8 space straight m divided by straight s squared 
    (c) Maximum speed is,
    V equals A omega equals A 2 pi v
space space space equals 0.02 cross times 2 pi cross times 10 over pi
space space space equals 0.4 space m divided by s

    Question 1037
    CBSEENPH11019309

    Which phase of matter has maximum value of temperature coefficient of cubical expansion?

    Solution
    Gaseous phase has maximum value of temperature coefficient of cubical expansion.
    Question 1038
    CBSEENPH11019310

    A steel tape is calibrated at 20°C. A piece of wood is being measured by steel tape at 10°C and reading is 30cm on the tape. State whether the real length of wood is more or less than 30cm.

    Solution
    On cooling, the steel tape contracts. Therefore, the real length of wood piece would be greater than 30cm. 
    Question 1039
    CBSEENPH11019311

    Show that if the acceleration due to gravity on any planet is n2 times the acceleration due to gravity on the earth, then time period of simple pendulum on that planet is 1/n times the time period on simple pendulum on the earth.

    Solution
    Time period of oscillation is given by, 
                          straight T equals 2 straight pi square root of straight ell over straight g end root 
    Let, g and g' be the acceleration due to gravity on the earth and planet respectively, and
    T and T’ be their respective time periods.
    Then, 
    T equals 2 pi square root of straight ell over g end root space space space space space a n d space
T apostrophe equals space T equals 2 pi square root of fraction numerator straight ell over denominator g apostrophe end fraction end root space

therefore space space space fraction numerator T apostrophe over denominator T end fraction equals square root of fraction numerator g over denominator g apostrophe end fraction end root equals square root of fraction numerator g over denominator n squared g end fraction end root equals 1 over n

rightwards double arrow space space space space T apostrophe equals 1 over n T 
    Hence, proved. 
    Question 1040
    CBSEENPH11019312
    Question 1041
    CBSEENPH11019313

    Name the two uses of bimetallic strip.

    Solution
    Bimetallic strip is used as thermometer and thermostat.
    Question 1042
    CBSEENPH11019314

    Why do the substances expand on heating? Explain using atomic theory.

    Solution
    i) In matter, interatomic/molecular forces bind the atoms or molecules. The interatomic separation between the atoms determines the potential energy due to binding force. 
    The variation is shown in the figure below. 

    ii) Consider two atoms A1 and A2 at O and P. At a point of minimum energy, the atoms are at a distance r0 apart and temperature is minimum at this point.
    iii) If we increase the temperature, the atoms absorb the energy and energy of each atom increases.
    iv) Let at temperature T1, the energy be U1. There are two positions r1(=OL) and r2(=OK) corresponding to energy U1.
    v) Both the positions r1 and r2 are equally probable. Therefore the atom A2 starts vibrating between positions L and K.
    vi) At temperature T1, the position of atom Ais taken at G (mean position of L and K).
    vii) The potential energy curve is unsymmetrical and G is on the right side of P. Hence, at temperature T1, the average distance between A1 and Ais OG = r0 +∆r > r0.
    viii) Thus with the increase in temperature, the distance between two atoms increases and hence substance expands on heating. 
    Question 1043
    CBSEENPH11019315

    What is temperature coefficient of linear expansion? Find the length of a rod when it is heated through θ°C.

    Solution
    The temperature coefficient of linear expansion is defined as the ratio of the increase in length per unit original length per unit rise in temperature.
    It is represented by straight alpha
    Let us consider a rod of length l0 at 0°C.
    If we increase the temperature of rod by θ°C, the rod will expand.
    The increase in the length of rod is, 
    (i) proportional to the original length 
    (ii) proportional to rise in temperature, 
    i.e.,           increment l proportional to l subscript 0 theta 
    rightwards double arrow           increment straight l space equals space αl subscript 0 straight theta 
    The length of rod at temperature straight theta degree straight C is 
                   straight l space equals straight l subscript 0 plus αl subscript 0 straight theta
space space equals space straight l subscript 0 left parenthesis 1 plus αθ right parenthesis 
                
    Question 1044
    CBSEENPH11019316

    Two bodies P and Q of equal asses are suspended from two separate springs of constants K1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, then what is the ratio of the amplitudes of vibration of P and Q?

    Solution
    Frequency of oscillation of body P is,
                      straight omega subscript 1 equals square root of straight K subscript 1 over straight m end root 
    Frequency of oscillation of body Q is,
                     straight omega subscript 1 over straight w subscript 2 equals square root of straight K subscript 2 over m end root
    The ratio of the frequency of  oscillation of body P to Q is,
                      
straight omega subscript 1 over straight omega subscript 2 equals square root of straight K subscript 1 over straight K subscript 2 end root 
    Let, Aand Abe the amplitude of oscillation of P and Q respectively.
    Since the velocity of P and Q at mean position (maximum velocity) is equal,  therefore
    
space space space space space space space space space straight A subscript 1 straight omega subscript 1 equals straight A subscript 2 straight omega subscript 2 space

rightwards double arrow space space space space straight A subscript 1 over straight A subscript 2 equals straight omega subscript 2 over straight omega subscript 1 equals square root of straight K subscript 2 over straight K subscript 1 end root 
    Question 1045
    CBSEENPH11019317

    What is temperature coefficient of superficial expansion? Find the area of a sheet when it is heated through θ°C.

    Solution

    The temperature coefficient of superficial expansion is defined as the ratio of the increase in area per unit original area per unit rise in temperature.
    It is represented by β. 
    Consider a sheet of area S0 at 0°C.
    If we increase the temperature of sheet by 0°C, it will expand.
    The increase in the area of sheet is,
    (i) proportional to the original area 
    (ii) proportional to rise in temperature, 
    i.e.       increment straight S proportional to straight S subscript 0 straight theta  
    rightwards double arrow      increment straight S space equals space βS subscript 0 straight theta 
    The area of sheet at temperature straight theta degree straight C is 
              straight S equals straight S subscript 0 plus βS subscript 0 straight theta space equals straight S subscript 0 left parenthesis 1 plus βθ right parenthesis

    Question 1046
    CBSEENPH11019318

    A cylinder is floating in a liquid with 20cm of its length immersed in liquid of density 1.2gm/cc. Find the time period of oscillation when disturbed from mean position.

    Solution
    Time period of oscillation of cylinder floating in the liquid is given by,
                      straight T equals 2 straight pi square root of straight ell over straight g end root
    where,
    Error converting from MathML to accessible text. is the length of the cylinder inside the liquid.
    Here, Error converting from MathML to accessible text.= 20 cm.
    Therefore time period of the oscillation is given by, 
    straight T equals 2 straight pi square root of 20 over 980 end root space

space equals fraction numerator 2 straight pi over denominator 7 end fraction equals 0.897 straight s 
    Question 1047
    CBSEENPH11019319

    A vertical U tube of uniform cross-section contains mercury up to height 39-2cm. Find the time period of oscillation when it is disturbed. Also find the time period of oscillation if mercury in the tube is replaced by water to same height.

    Solution
    The time period of oscillation of liquid in U tube is given by,
                           straight T equals 2 straight pi square root of straight L over straight g end root 
    where,
    L is the length of the mercury column in each limb of U- tube. 
    We have,
    Length, L = 39.2 cm
    therefore space T i m e space p e r i o d comma space straight T equals 2 straight pi square root of fraction numerator 39.2 over denominator 980 end fraction end root equals 1.257 straight s
    Since the time period of oscillation does not depend on the nature of liquid in U tube, therefore time period of oscillation will remain same on replacing mercury by water. 
    Question 1048
    CBSEENPH11019320

    What is temperature coefficient of cubical expansion? Find the volume of a cube when it is heated through θ°C.

    Solution

    The ratio of the increase in volume per unit original volume per unit rise in temperature is defined as the temperature coefficient of cubical expansion.
    It is represented by γ. 
    Consider a cube of volume V0 at 0°C.
    If the temperature of cube  is increased by θ°C, it will expand.
    The increase in the volume of cube is, 
    (i) proportional to the original volume 
    (ii) proportional to rise in temperature, 
    i.e.           increment straight V proportional to straight V subscript 0 straight theta 
     
                   increment straight V equals γV subscript 0 straight theta 
    The volume of cube at temperature straight theta degree straight C is, 
                    straight V equals straight V subscript 0 plus γV subscript 0 straight theta space equals space SV subscript 0 left parenthesis 1 plus γθ right parenthesis

    Question 1049
    CBSEENPH11019321

    A spring oscillates with frequency 30 per minute when loaded by M kg. On increasing the load by 0.5 kg, the frequency of oscillation decreases to 24 per minute. Find M.

    Solution

    The frequency of oscillation is given by,
                        straight nu space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root 
    Given, when m = M ; straight nu space equals space 30 space per space min
    therefore space 30 space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

When comma space straight m space equals space straight M space plus space 0.5 comma space then space

space space space space space space space space space space space space space straight nu space equals space 24 space per space min

So comma space space 24 space equals space fraction numerator 1 over denominator 2 straight pi end fraction space square root of fraction numerator straight k over denominator straight M plus 0.5 end fraction end root space space space space space space... space left parenthesis 2 right parenthesis thin space

Dividing space equation space left parenthesis 1 right parenthesis space by space left parenthesis 2 right parenthesis comma space we space get space

space space space space space space 30 over 24 space equals space square root of fraction numerator straight M space plus space 0.5 over denominator straight M end fraction end root space

rightwards double arrow space space fraction numerator straight M space plus space 0.5 over denominator straight M end fraction space equals space 25 over 16

rightwards double arrow space space space 16 space straight M space plus space 8 space equals space 25 space straight M space

rightwards double arrow space space space space 9 space straight M space equals space 8 space

rightwards double arrow space space space space space space straight M space equals space 8 over 9 kg 

    Question 1050
    CBSEENPH11019322

    A block is kept on a horizontal rough table (μ = 0.6) executing simple harmonic motion with amplitude 3cm. Find the maximum frequency at which the block does not slip over table.

    Solution

    Given, 
    Coefficient of friction, μ = 0.6 and
    Amplitude, A = 3 cm
    Let m be the mass of block and ω the maximum frequency of oscillation.
    The limiting friction between block and table is,

                  Flimiting = μmg

    The maximum acceleration of table is,

                      amax = ω2A

    The maximum pseudo force on the block is,

    Fpseudo = mamax = mω2A

    For the block not to slip over the table, the maximum pseudo force should not exceed the limiting friction.
    Therefore,
    space space space space space space space space space space space space space space space space mω squared straight A equals μmg space

rightwards double arrow space space straight omega equals square root of μg over straight A end root equals square root of fraction numerator 0.6 cross times 980 over denominator 3 end fraction end root equals 14 space rad divided by straight s
 
    straight omega is the maximum frequency at which the block does not slip over the table. 

    Question 1051
    CBSEENPH11019323

    A spring of constant 245N/m is loaded by two blocks of mass 4 kg and 5 kg. If 4 kg mass is suddenly removed, then what will be the frequency and amplitude of oscillation of 5kg mass?

    Solution

    We have,
    Spring constant, k=245N/m
    Mass, m= 4 kg
    MAss, m= 5kg 
    When 4 kg mass is removed from the loaded spring, the the amplitude of oscillation is equal to extension produced due to 4 kg load.
    straight i. straight e.
space space space straight A equals fraction numerator straight m subscript 1 straight g over denominator straight k end fraction equals fraction numerator 4 cross times 9.8 over denominator 245 end fraction equals 0.08 straight m equals 8 space cm space
    The spring remains loaded by 5 kg mass. Therefore, the frequency of oscillation is, 
    straight v equals fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m subscript 2 end root equals fraction numerator 1 over denominator 2 straight pi end fraction square root of 245 over 5 end root equals fraction numerator 7 over denominator 2 straight pi end fraction hz 

    Question 1052
    CBSEENPH11019324

    Derive the relation between alpha and β. Assume that coefficient of linear expansions are same in all directions.

    Solution
    Consider a rectangular sheet of sides L0 and B0 at 0°C.
    Therefore the area of sheet at 0°C is S0 = L0B0.
    When the sheet is heated, area of the plate will increase due to increase in the length and breadth. 
    If straight alpha is coefficient of linear expansion and β is coefficient of superficial expansion, then at temperature θ°C, 
                     straight L equals straight L subscript 0 left parenthesis 1 plus αθ right parenthesis comma

straight B equals straight B subscript 0 left parenthesis 1 plus αθ right parenthesis 
    and            straight S equals straight S subscript 0 left parenthesis 1 plus βθ right parenthesis 
    But           straight S equals LB 
    ∴  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                      equals straight L subscript 0 straight B subscript 0 left parenthesis 1 plus αθ right parenthesis squared 
    rightwards double arrow    left parenthesis 1 plus βθ right parenthesis space equals space left parenthesis 1 plus αθ right parenthesis squared 
    Since,  straight alpha less than less than 1 comma therefore using binomial expansion, 
                 space space space 1 plus βθ space equals space left parenthesis 1 plus αθ right parenthesis squared 
    rightwards double arrow                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Question 1053
    CBSEENPH11019325

    Derive the relation between a and γ. Assume that coefficient of linear expansions are same in all directions.

    Solution
    Let a rectangular cuboid be of sides L0, B0 and H0 be at 0°C.
    Therefore the volume of cuboid at 0°C is VQ = L0B0H0
    When cuboid is heated, all of its sides will expand and hence volume.
    If straight alpha is the coefficient of linear expansion and γ is coefficient of cubical expansion, then at temperature θ°C, 
                      straight L space equals space straight L subscript 0 left parenthesis 1 plus αθ right parenthesis comma

straight B equals straight B subscript 0 left parenthesis 1 plus αθ right parenthesis comma

straight H space equals space straight H subscript 0 left parenthesis 1 plus αθ right parenthesis comma 
    and             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    But,            straight V equals LBH 
    Therefore,
     straight V subscript 0 left parenthesis 1 plus γθ right parenthesis space equals space straight L subscript 0 left parenthesis 1 plus αθ right parenthesis straight B subscript 0 left parenthesis 1 plus αθ right parenthesis space straight H subscript 0 left parenthesis 1 plus αθ right parenthesis 
                    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Since a<<1,  therefore using binomial expansion, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow                  straight gamma equals 3 straight alpha
    Question 1054
    CBSEENPH11019326

    A simple pendulum clock is taken to a height 64 km from the surface of the earth. How much will it lose or gain in a day?

    Solution

    Time period of the simple pendulum is given by,  
                space space space space space space space space space space space straight T equals 2 straight pi square root of straight ell over straight g end root 
    At height h, acceleration due to gravity is,
    straight g subscript blank subscript h end subscript equals straight g open parentheses 1 plus straight h over straight R close parentheses to the power of negative 2 end exponent


    Since h<< R, 
    space space space
space space space space space therefore space space space space space straight g subscript straight h equals straight g open parentheses 1 minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses 
    The time period of simple pendulum clock at height h is,
    space space space space space straight T subscript straight h equals 2 straight pi square root of straight ell over straight g subscript straight h end root space

space space space space space space space space equals 2 straight pi square root of fraction numerator straight ell over denominator straight g open parentheses 1 minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses end fraction end root space

space space space space space space space space equals space space 2 straight pi square root of straight ell over straight g subscript straight h end root open parentheses 1 minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses to the power of negative 1 divided by 2 end exponent space

space space space space space space space equals straight T open parentheses 1 plus straight h over straight R close parentheses space

rightwards double arrow space space space space space space space space straight T subscript straight h equals straight T plus straight T straight h over straight R space

rightwards double arrow space space fraction numerator straight T subscript straight h minus straight T over denominator straight T end fraction cross times 100 equals straight h over straight R cross times 100
    That is, percentage increase in time period of oscillation is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    The time period of oscillation increases.
    Therefore, the clock will lose time in a day.
    Percentage increase in time period is 1%, 
    therefore, it will lose 1 over 100 cross times 86400 equals 864 seconds in a day. 

    Question 1056
    CBSEENPH11019328

    Define the apparent and absolute coefficient of expansion of liquids. Find the relation between them.

    Solution

    Liquids cannot hold themselves alone but have to be contained in the vessel. Therefore when liquids are heated, naturally the vessel will also expand along with liquids. Thus when liquids are heated, they have also to fill the increased volume of the vessel. We cannot observe the intermediate state. We can only observe the initial and the final levels. This observed expansion of the liquid is known as the apparent expansion of the liquid and is less than its actual expansion.

    The apparent coefficient of expansion of liquid is defined as the apparent increase in the volume of liquid in the vessel per unit original volume of liquid in the vessel per unit rise in temperature. It is represented by γa.

    The absolute coefficient of expansion of liquid is defined as the actual increase in the volume of liquid (sum of the apparent increase in the volume of liquid in vessel and increase in the volume of the vessel) per unit original volume of liquid per unit rise in temperature. It is represented by γrRelation between γa and γr

    Let us consider a vessel of volume V0 at 0°C filled to brim with a liquid. Let γa and γr be the apparent and absolute coefficient of volume expansion of liquid respectively and γv is coefficient of cubical expansion of vessel.

    Now heat the vessel to 0°C. Both vessel and liquid will expand. As the container is filled to the brim, therefore on heating, the liquid will overflow. Collect the overflow liquid. The liquid that overflows is the apparent increase in volume. Let it be ∆Va. By definition, the apparent coefficient of expansion of liquid is,


                   γa = VaVoθ   or    Va = Voγaθ
    Let ∆Vv be the increase in the volume of the vessel,

    Vv = V0γvθ

    Now the actual increase in the volume of liquid is,

       V= Va+Vv=Voγaθ+Voγvθ                 ...(1)

    Also, the actual increase in the volume of liquid is,

    V=Voγrθ                                                            ...(2)

    From (1) and (2),

                 Voγrθ = Voγaθ+Voγvθ

    or                γr = γa+γv

    This is the required relation between γa and γr.

    Question 1057
    CBSEENPH11019329

    A rod is clamped between two rigid supports so that it neither bends nor expands. Find the thermal compression force and thermal stress set up in the rod when it is heated through 9°C.

    Solution

    Consider a rod of length L, area of cross-section A, clamped between the two rigid supports so that it neither expands nor bends.
    Heat the rod so that its temperature increases by θ°C.
    As the rod is neither allowed to expand nor bend, therefore a compressive force or stress is set up in the rod. This force or stress is called thermal force or thermal stress. 
    Let straight alpha be the coefficient of linear expansion and Y be the Young’s modulus of material.
    If the ends of rod were free, then its length would increase by, 
                            increment straight L equals Lαθ 
    The rod being clamped, does not expand but a strain is produced in the rod.
    The strain in the rod is given by, 
                        fraction numerator increment straight L over denominator straight L end fraction equals Lαθ over straight L equals αθ 
    The stress in the wire is, 
            Thermal stress = Y x strain 
                                   = γαθ 
    Thermal force developed in the rod is, 
    Thermal force = Thermal Stress x Area 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

    Question 1058
    CBSEENPH11019330

    Two metal strips A and B, each of length L0 and thickness d at temperature Tbe fastened together so that their ends coincide. The temperature coefficient of linear expansion of A is αA and that of B is αA> αB). Find the radius of curvature of strip when it is heated.

    Solution
    Consider two strips A and B each of length L0 which are fastened together so that their ends coincide. 
    The bimetallic strip is heated through a temperature of θ° C.
    Given that, the temperature coefficient of linear expansion of A is greater than that of B and the bimetallic strip will bend in the form of an arc of circle of radius R with A on convex side and B on concave side.
     
    If the strip is heated through θ°C, the length of A and B respectively is, 

    straight L subscript straight A space equals space straight L subscript straight o space left parenthesis 1 space plus space straight alpha subscript straight A straight theta right parenthesis space and

straight L subscript straight B space equals space straight L subscript straight o space left parenthesis 1 space plus space straight alpha subscript straight B straight theta right parenthesis space

Therefore comma space

left parenthesis straight L subscript straight A space minus space straight L subscript straight B right parenthesis thin space equals space straight L subscript straight o space left parenthesis straight alpha subscript straight A space minus space straight alpha subscript straight B right parenthesis space straight theta space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

left parenthesis straight L subscript straight A space plus space straight L subscript straight B right parenthesis space equals space straight L subscript straight o space left square bracket space 2 space plus space left parenthesis straight alpha subscript straight A space plus space straight alpha subscript straight beta right parenthesis space straight theta space right square bracket space space space space space space space space space space... space left parenthesis 2 right parenthesis space

therefore space fraction numerator left parenthesis straight L subscript straight A space minus space straight L subscript straight B right parenthesis over denominator left parenthesis straight L subscript straight A space plus space straight L subscript straight B right parenthesis space end fraction space equals space fraction numerator space left parenthesis straight alpha subscript straight A space minus space straight alpha subscript straight B right parenthesis space straight theta space space over denominator space 2 space plus space left parenthesis straight alpha subscript straight A space plus space straight alpha subscript straight beta right parenthesis space straight theta space end fraction space space space space space... space left parenthesis 3 right parenthesis space space space space space

If space straight phi space be space the space angle space of space arc comma space then space also

straight L subscript straight A space equals space straight ϕ space straight R subscript straight A

straight L subscript straight B space equals space ϕR subscript straight B

therefore space straight L subscript straight A space minus space straight L subscript straight B space equals space space straight ϕ space left parenthesis straight R subscript straight A space minus space straight R subscript straight B right parenthesis space equals space straight ϕ space straight d space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis

where comma space

left parenthesis straight R subscript straight A space minus space straight R subscript straight B right parenthesis space space equals space straight d space left parenthesis thickness space of space each space strip right parenthesis space comma space and space
straight L subscript straight A space plus space straight L subscript straight B space equals space straight ϕ space left parenthesis straight R subscript straight A space plus space straight R subscript straight B right parenthesis thin space equals space straight ϕ space 2 straight R space space space space space space space space space space space space space space space space space space... space left parenthesis 5 right parenthesis thin space

where comma space

straight R space is space the space mean space radius space of space curvature space of space strip. space

therefore space fraction numerator straight L subscript straight A space minus space straight L subscript straight B over denominator straight L subscript straight A space plus space straight L subscript straight B end fraction space equals space fraction numerator ϕd over denominator straight ϕ 2 straight R end fraction space equals space fraction numerator straight d over denominator 2 straight R end fraction space space space space space space space space space space space space space space space space space space space space... space left parenthesis 6 right parenthesis

From space equation space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space have

space space space space space space space space fraction numerator straight d over denominator 2 straight R end fraction space equals space fraction numerator open parentheses straight alpha subscript straight A space minus space straight alpha subscript straight B close parentheses straight theta over denominator 2 space plus space left parenthesis straight alpha subscript straight A plus straight alpha subscript straight B right parenthesis straight theta end fraction

rightwards double arrow space space space space space straight R space equals space straight d over 2 open parentheses fraction numerator 2 space plus space left parenthesis straight alpha subscript straight A plus straight alpha subscript straight B right parenthesis straight theta over denominator open parentheses straight alpha subscript straight A space minus space straight alpha subscript straight B close parentheses straight theta end fraction close parentheses
    Question 1059
    CBSEENPH11019331

    A simple pendulum clock keeps correct time at the sea level and loses 0.9 min per day at the top of a mountain. What is the height of the mountain?

    Solution

    The time period of simple pendulum clock at the surface of the earth is,
                space space space space space
space space space space space space space space space space space space space space space straight T equals 2 straight pi square root of straight L over straight g end root space space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
    Let g' be the acceleration due to gravity at the top of mountain.
    Then, time period of simple pendulum clock at the top of mountain is, 
                               straight T apostrophe equals 2 straight pi square root of fraction numerator straight L over denominator straight g apostrophe end fraction end root space space space space space space space space space space space space space space space.... left parenthesis 2 right parenthesis
    The acceleration due to gravity at the surface of the earth is,
                               straight g equals GM over straight R squared
    where,
    M is mass, and
    R is the radius of the earth.
    Let h be the height of the mountain.

    Acceleration due to gravity at the top of the mountain is given by,
    space space space space space space space space space space space space space space space straight g apostrophe equals fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction space

therefore space space space space space space space space fraction numerator straight g over denominator straight g apostrophe end fraction equals fraction numerator open parentheses straight R plus straight h close parentheses squared over denominator left parenthesis straight R right parenthesis squared end fraction space

Now comma space space space fraction numerator straight T apostrophe over denominator straight T end fraction equals square root of fraction numerator straight g over denominator straight g apostrophe end fraction end root equals fraction numerator left parenthesis straight R plus straight h right parenthesis over denominator left parenthesis straight R right parenthesis end fraction equals 1 plus straight h over straight R
    The loss of time of oscillation of simple pendulum clock in one second is, 
    fraction numerator straight T apostrophe minus straight T over denominator straight T end fraction equals straight h over straight R
    Thus loss of time of oscillation of simple pedulum clock in one day is,
    space space fraction numerator straight T apostrophe minus straight T over denominator straight T end fraction cross times 86400 equals straight h over straight R cross times 86400 equals 54
rightwards double arrow space space space space space straight h equals fraction numerator 54 straight R over denominator 86400 end fraction equals fraction numerator 54 cross times 6400 over denominator 86400 end fraction equals 4 space km
    Height of the mountain is 54 km. 

     
    Question 1060
    CBSEENPH11019332

    Why are the two ends of a long bridge generally kept on a trolley?

    Solution
    Bridges expand or contract due to the change in temperature. Therefore, if bridges are placed on trolley as shown, it will allow the bridge to expand or contract easily without setting up any strain.

                      
    Question 1061
    CBSEENPH11019333

    Why some clocks are slow in summer and fast in winter?

    Solution
    Time period of simple pendulum used in clock is, 
                            straight T equals 2 straight pi square root of ell over straight g end root 
    where,
    L is length of simple pendulum.
    Temperature decreases during winters. Therefore, length of simple pendulum will decrease. Since, T ∝ √L; so time period of simple pendulum will decrease and run faster. 
    In summers, temperature increases. Therefore, the length and hence the time period increases and therefore, the clock will run slower. 
    Question 1062
    CBSEENPH11019334

    Why do the telephone wires get slackened in summer and get tightened in winter?

    Solution
    In summers due to excessive heat, the temperature of the wire increases and wires expand. Also, Young’s modulus decreases and hence wires slack down. In winter, the temperature of wires decrease and wires contract. Hence, wires get tightened. 
    Question 1063
    CBSEENPH11019335

    A bimetallic strip is made of two materials A and B whose coefficients of linear expansions are αA and αB respectively (αA > αB). If length of both the rods is same at 0°C, then what will be the shape of bistrip at –10°C and +10°C?

    Solution
    The coefficient of linear expansion, αA > αB. 
    Therefore, for the same change in temperature, A will expand or contract more than B.

    So, when the bi-strip is heated i.e. temperature is 10°C, bi-strip will bend with A on convex side and B on concave side.
    And when the bi-strip is cooled i.e. temperature is –10°C,  length A decreases more than B. Therefore, A is on concave side and B is on convex side.
    Question 1064
    CBSEENPH11019336

    The amplitude of simple harmonic oscillation is doubled. How would this affect: 

    (i) Time period,

    (ii) Maximum velocity,

    (iii) Acceleration at mean position, 

    iv) Kinetic energy at mean position?

    Solution
    (i) Time period of oscillation of simple harmonic oscillator is given by,
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#6 {main}</pre> 

    Time period is independent of the amplitude of vibration. Therefore, time period remains unchanged on doubling the amplitude of oscillation. 

    (ii) For a particle undergoing Simple Harmonic Motion, maximum velocity is given by

                                v = A ω

    where, A is amplitude.
    When amplitude is doubled, the maximum velocity becomes twice. 

    (iii) Acceleration at mean position is zero and is independent of the amplitude of vibration.
    Therefore, acceleration at mean position remains unaffected.

    (iv) Kinetic energy at mean position is, 
                           straight K equals 1 half mω squared space straight A squared
    If the amplitude of oscillation gets doubled, then kinetic energy at mean position increases to four times. 

    Question 1065
    CBSEENPH11019337

    In hilly area, water pipes get burst during winter. Why?

    Solution
    The density of water is maximum at 4°C and freezing point of water is 0°C. Therefore, when water freezes it will expand and due to increase in volume, the pipes burst.
    Question 1066
    CBSEENPH11019338

    Why is invar used in making clock pendulum?

    Solution
    Invar, which is an alloy of nickel and steel has an exceptionally small value of temperature coefficient of expansion. Therefore its length does not change in summer and winter and the clock gives correct time in all the seasons.
    Question 1067
    CBSEENPH11019339

    When a sitar is brought into cold the pitch increases. Why?

    Solution
    Steel has a greater value of expansion coefficient than wood. When sitar is brought in cold, the decrease in length of wire will be greater than wood. Therefore, tension in the string will increase.
    Since v ∝ √T, therefore, frequency or pitch will increase. 
    Question 1068
    CBSEENPH11019340

    Explain how does a fish stay alive in a frozen pond in winter?

    Solution
    When the temperature falls below 0°C, during winters, the water at the surface freezes into ice. But, at the bottom the temperature of water is at 4°C, because the density of water is maximum at 4°C. Therefore, water at 4°C helps the fish to stay alive in a frozen pond. 
    Question 1069
    CBSEENPH11019341

    Two glasses have stuck together into each other.

    How will you loosen them by the use of:

    (i) hot water (ii) cold water?

    Solution

    Place the two glasses in hot water, such that there is no water in the inner glass. The outer glass expands because of hot water and hence a gap is set up between the two glasses and thus we can take the glass out.
    If we have cold water, then put the cold water in the inner glass. The cold water makes the inner glass contract leaving a gap between the two glasses and thus we can take the other glass out.

    Question 1070
    CBSEENPH11019342

    Why does soda water bottle, some times burst in summer?

    Solution
    The coefficient of expansion of gases is large. Therefore, the gas in the bottle expands. Hence, the pressure inside the bottle increases. Thus, the bottle may burst in hot summer.
    Question 1071
    CBSEENPH11019343

    How does the level of liquid in a container change on heating the liquid of cubical expansion in the container having coefficient of linear expansion α/3?

    Solution
    The coefficient of linear expansion of container is straight alpha/3. Therefore, coefficient of cubical expansion of container is equal to coefficient of cubical expansion of liquid.
    Hence, the level of liquid in container will neither rise nor fall. 
    Question 1072
    CBSEENPH11019344

    How deserts are formed?

    Solution
    In hot regions, the rocks expand at day time and contract at night time. Due to this alternate expansion and contraction, the rocks continue to break and form sand. This results in the formation of deserts. 
    Question 1073
    CBSEENPH11019345

    What is free oscillator? Write the differential equation for free oscillator.

    Solution

    The simple harmonic oscillation is executed when an oscillator is displaced from it's equilibrium position.
    The time period of its oscillation depends upon the oscillator's inertia factor and elastic factor.
    If the amplitude of oscillations remains constant in the absence of any external source of energy, then the oscillations are called free oscillations and oscillator is called free oscillator.
    The frequency with which the oscillator oscillates is called natural frequency.

    The differential equation of free oscillator is,
                      fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight k over straight m straight y equals 0
    The solution of the above equation differential equation is,
                      
straight y equals Asin left parenthesis ωt plus straight ϕ right parenthesis
    The energy of oscillator is,
                   straight E equals 1 half mω squared space straight A squared
    As the amplitude of free oscillations is constant, therefore the energy of oscillation is also constant. 

    Question 1075
    CBSEENPH11019347

    Why the temperature coefficient of volume expansion of all the gases is same whereas it is different for different liquids and solid?

    Solution
    The cohesive force between the molecules of all the gases is negligible. When gases expand, the molecules of all types of gases move away under similar conditions, therefore temperature coefficient of volume expansion of all the gases is same.
    The cohesive forces between solids and liquids is different, and the molecules of solid and liquids is bonded by cohesive forces. When solids or liquids expand, the molecules move away against different cohesive forces. Therefore, temperature coefficient of volume expansion of different solids and liquids are different. 
    Question 1076
    CBSEENPH11019348

    Three identical rods A, B and C form an equilateral triangle. The material of rods A and B is same and that of C is different. When the structure will be heated, what would be the new shape of structure?

    Solution
    The temperature coefficient of expansion of different materials is different. The structure is an equilateral triangle at low temperature. When it is heated, the rod A and B will expand by same length but the increase in length of C will be different as that from A and B. Thus the new structure will form an isosceles triangle.
    Question 1077
    CBSEENPH11019349

    Two rods of length L1 and L2 and coefficient of expansions α1 and α2 are heated. What should be the relation between L1 and L2 so that difference in the length of two rods is independent of temperature?

    Solution
    On increasing the temperature, the length of both the rods will increase.
    If the change in length of both the rods is same irrespective of the rise in temperature, then the difference in the lengths of two rods can be independent of temperature.
    Increase in length of the first rod is,
                           increment straight L subscript 1 space equals space straight L subscript 1 straight alpha subscript 1 straight theta 
    Increase in length of the second rod is, 
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#6 {main}</pre> 
    The required condition is, 
                           space space straight L subscript 1 straight alpha subscript 1 straight theta space equals space straight L subscript 2 straight alpha subscript 2 straight theta  
    rightwards double arrow                        straight L subscript 1 over straight L subscript 2 equals straight alpha subscript 2 over straight alpha subscript 1
    Question 1078
    CBSEENPH11019350

    Find the length of steel and copper rod if steel rod remains 5 cm longer than copper at all the temperatures. Given αs=1.1 x 10–5°CT–1 and αc= 1.7x101–5°C–1.

    Solution
    Let Ls and Lc be the lengths of steel and copper rods respectively.
    The difference in the length remains constant if,
                      straight L subscript straight s over straight L subscript straight c equals straight alpha subscript straight c over straight alpha subscript straight s equals 17 over 11               ...(1) 
    Also,               straight L subscript straight s minus straight L subscript straight c space equals space 5 cm               ...(2) 
    From (1) and (2), we get 
    Length of steel rod, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>, and
    Length of the copper rod, straight L subscript straight c equals 9.17 space cm
    Question 1079
    CBSEENPH11019351

    The density of a solid whose temperature coefficient of cubical expansion is γ is ρ0 at 0°C. What will be its density if it is heated to 6°C?

    Solution
    Let, the mass of the solid be m. 
    If V0 is the volume of solid and ρ0 is density at 0°C, then
                    straight m equals straight V subscript straight o straight rho subscript straight o                               ...(1) 
    If we increase the temperature then the volume will increase.  
    Density will decrease because mass remains constant. 
    Let V be the volume and ρ be the density of the same mass of solid at temperature θ°C.
                      straight m equals Vρ                                ...(2) 
    From (1) and (2), we have 
                    straight V subscript straight o straight rho subscript straight o space equals space Vρ 
    If γ is the temperature coefficient of volume expansion of solid then,
                      space space space straight V equals straight V subscript o left parenthesis 1 plus γθ right parenthesis 
    Thus,      space straight V subscript straight o straight rho subscript straight o space equals space straight V subscript straight o left parenthesis 1 plus gamma theta right parenthesis straight rho
    rightwards double arrow                 straight rho equals fraction numerator straight rho subscript straight o over denominator 1 plus gamma theta end fraction 
    Question 1080
    CBSEENPH11019352

    If the volume of block of metal changes by x % when it is heated through θ°C, then what is the coefficient of linear expansion of the metal?

    Solution
    Let V be the volume of the block and straight alpha the temperature coefficient of linear expansion of the block.  
    Therefore coefficient of cubical expansion of block is 3straight alpha.
    If the temperature of block is increased by straight theta, then change in volume of block is given by, 
                          increment straight V space equals space 3 αVθ 
    The percentage change in volume of block is, 
                      fraction numerator increment straight V over denominator straight V end fraction 100 equals 300 αθ 
    But percentage change in volume is x %.
    Therefore, 
                      300 αθ equals straight x 
    rightwards double arrow             straight alpha equals open parentheses fraction numerator straight x over denominator 300 space straight theta end fraction close parentheses straight C presuperscript straight o superscript negative 1 end superscript 
    Question 1081
    CBSEENPH11019353

    A brass disc, originally at 20°C, has diameter 30cm and a hole of radius 5cm at the centre. Find the circumference of hole at temperature 100°C. The temperature coefficient of linear expansion of brass is l.8 x l0–5 °C–1.

    Solution
    Initial radius of hole is, r = 5 cm 
    Increase in temperature of disc is, straight theta equals 100 minus 20 equals 80 degree straight C. 
    Coefficient of linear expansion of brass is, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
    Therefore, radius of hole at 100°C is, 
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#6 {main}</pre>
     
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#6 {main}</pre> 
    The circumference of hole at 100°C is,
    2 πr space equals space 2 straight pi cross times 5.0072 space equals space 31.461 space cm 
    Question 1082
    CBSEENPH11019354

    What is damped oscillation. Write the differential equation for damped oscillations. Obtain an expression for the displacement in the case of damped oscillatory motion. Discuss how the amplitude of oscillation changes with time?

    Solution
    When an oscillator is oscillating in a resistive medium, the energy of oscillation is consequently decreasing. For such an oscillator, the amplitude decreases with time and ultimately the oscillations die out.
    This type of oscillator is known as a damped oscillator.
                             

    Consider a loaded spring oscillating in the vertical direction in a resistive media.
    Let m be the mass of load and k be the spring constant.
    The load is displaced from the equilibrium position. Let at any instant, y be the displacement from equilibrium position and v is the velocity of the load.
    The different forces acting on load are:

    (i) Restoring force which is proportional to displacement and directed towards equilibrium position.
    i.e.,    F1= – ky

    (ii) Resistive force of medium which is proportional to velocity and opposite to the direction of velocity.
    i.e.,     F2 = –bv
    The net force acting on the load is,

    F = Fx1 + F
       = – ky –bv
    Therefore the equation of motion of load is, 
    space space space space space space space space space space space space space space space space space straight m fraction numerator straight d squared straight y over denominator dt squared space end fraction equals negative ky minus straight b dy over dt

space rightwards double arrow space space space space space space space space space space straight m fraction numerator straight d squared straight y over denominator dt squared space end fraction plus straight b dy over dt plus ky equals 0
rightwards double arrow space space space space space space fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight b over straight m dy over dt plus straight k over straight m straight y equals 0
    This is the required differential equation of motion of damped oscillations.
    The solution of the equation is,
    straight y equals straight A subscript 0 straight e to the power of fraction numerator straight b over denominator 2 straight m end fraction straight i end exponent cos open square brackets open parentheses square root of straight k over straight m minus fraction numerator straight b squared over denominator 4 straight m squared end fraction end root close parentheses straight r plus straight ϕ close square brackets 
    Comparing it with space space straight y equals Acos left parenthesis ωt plus straight ϕ right parenthesis, we get,
    straight A equals straight A subscript 0 straight e to the power of fraction numerator straight h over denominator 2 straight m end fraction straight i end exponent                                      .....(1)
    space space space space straight omega equals square root of straight k over straight m minus fraction numerator straight b squared over denominator 4 straight m squared end fraction end root equals square root of straight omega subscript 0 squared minus fraction numerator straight b squared over denominator 4 straight m squared end fraction end root space space space...... left parenthesis 2 right parenthesis
    Equation (1) gives the amplitude of oscillation, which decreases exponentially with time.

    Equation (2) gives the frequency of oscillation which is less than the natural frequency of oscillation ω0

    Question 1083
    CBSEENPH11019355

    What is forced oscillations? Discuss the vibrations of a system executing damped simple harmonic motion when subjected to an external periodic force.

    Solution

    When an oscillator is made to oscillate by an external periodic force, whose frequency is different from the natural frequency of the free oscillator is defined as forced oscillations. 

    In forced oscillations, the oscillator is under the influence of two forces simultaneously:
    i) Restoring force of its own
    ii) Force due to external periodic force.
    Both these forces have a tendency to make the oscillator oscillate their own way.
    Thus, for the first few, oscillator does not exhibit regular oscillations.
    First, the restoring force is more effective and the oscillator oscillates with natural frequency and then external periodic force dominates and makes the oscillator oscillate with a frequency same as that of external periodic force. 

                        
    Consider a mass m attached to a spring of constant k.

    When an external periodic-force is applied on mass, it will oscillate under the influence of following forces:

    (i) The restoring force which is proportional to displacement and directed towards equilibrium position.

    i.e.,                           F1= – ky

    (ii) Resistive force of medium which is proportional to velocity and opposite to the direction of velocity.
    i.e.,                             F2 = – bv

    (iii) The external periodic force F3 = fo cos ωt.
    The net force acting on the load is, 
    space space space space space space straight F subscript 1 plus straight F subscript 2 plus straight F subscript 3 equals negative k y minus b v plus straight f subscript 0 space c o s omega t 
    Therefore the equation of motion of load is,
    space space space space space space space space space space space space space straight m fraction numerator straight d squared straight y over denominator dt squared end fraction minus ky minus straight b dy over dt plus straight f subscript 0 space cosωt space

rightwards double arrow space space space space space space space space space space space straight m fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight b dy over dt plus ky equals straight f subscript 0 space cosωt space

rightwards double arrow space space space space space space fraction numerator straight d squared straight y over denominator dt squared end fraction plus straight b over straight m dy over dt plus straight k over straight m straight y equals straight f subscript 0 over straight m cosωt
    This is the required differential equation of motion of damped oscillations.
    The solution of the equation is,
    straight y equals Acos left square bracket ωt plus straight ϕ right square bracket space

where comma space
space space space straight A equals fraction numerator straight f subscript 0 over denominator straight m square root of begin display style fraction numerator straight b squared straight omega squared over denominator straight m squared end fraction end style plus open parentheses straight omega squared minus straight omega subscript straight omicron squared close parentheses squared end root end fraction
    From the above equation, we can deduce that  the amplitude of oscillation depends on the difference in the frequency of periodic force and natural frequency.
    Smaller is the frequency difference greater is the magnitude of oscillation and vice versa.

    Question 1084
    CBSEENPH11019356

    Write the differential equation for damped forced oscillations and discuss the conditions of resonance.

    Solution
    The differential equation for damped forced oscillation is, 
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#6 {main}</pre> 
    Solution of the above differential equation is,
    space space space space space space space space space space space space space straight y equals straight A space cos open square brackets ωt plus straight ϕ close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

where comma
space space space space space straight A space equals space fraction numerator straight f subscript straight omicron over denominator straight m square root of begin display style fraction numerator straight b squared straight omega squared over denominator straight m squared end fraction end style plus open parentheses straight omega squared plus straight omega subscript straight omicron squared close parentheses squared end root end fraction space space space space space space space space space space... left parenthesis 2 right parenthesis
    Resonance is the state of vibration in which the oscillator oscillates with maximum amplitude.
    Thus, the condition of resonance is that A should be maximum.
    Therefore denominator of equation (2) should be minimum. 
    straight i. straight e. space space space space straight d over dω open parentheses fraction numerator straight b squared straight omega squared over denominator straight m squared end fraction plus left parenthesis straight omega squared minus straight omega subscript 0 squared right parenthesis squared close parentheses equals 0
rightwards double arrow space space space space space space space straight b squared over straight m squared 2 straight omega plus 2 left parenthesis straight omega squared minus straight omega subscript 0 squared right parenthesis 2 straight omega equals 0
rightwards double arrow space space space space space space space straight omega equals square root of straight omega subscript 0 squared minus fraction numerator straight b squared over denominator 2 straight m squared end fraction end root space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
    Substituting space straight omega in equation (2), we get
    straight A subscript max equals fraction numerator straight f subscript 0 over denominator straight b square root of straight omega subscript 0 squared minus begin display style fraction numerator straight b squared over denominator 4 straight m squared end fraction end style end root end fraction space space space space space space space space space space... left parenthesis 4 right parenthesis 
    The oscillator is in the state of resonance if the frequency of external force is given by, 
                space space space square root of straight w subscript 0 squared minus fraction numerator straight b squared over denominator 2 straight m squared end fraction end root

    Question 1085
    CBSEENPH11019357

    A brass rod and a steel rod are both of length 1.2 m at room temperature 30°C. Both the rods are put in boiling water. Find the difference in the length of two rods in boiling water, αb = 1.8 x 10–5°C–1 and αs=l.2 x 10–5°C–1.

    Solution
    The length L of the rod when heated by straight theta°C increases to L' given by, 
                        straight L apostrophe space space equals space straight L left parenthesis 1 plus αθ right parenthesis 
    ∴ straight L apostrophe subscript straight s space equals space straight L subscript straight s left parenthesis 1 plus straight alpha subscript straight s straight theta right parenthesis space space space and space space space straight L apostrophe subscript straight b space equals space straight L apostrophe subscript straight b left parenthesis 1 plus straight alpha subscript straight b straight theta right parenthesis 
    Here lengths of the two rods are equal, 
    ∴      straight L apostrophe subscript straight s space equals space straight L left parenthesis 1 plus straight alpha subscript straight s straight theta right parenthesis space space space space space space space space and space space space straight L apostrophe subscript straight b space equals space straight L subscript straight b left parenthesis 1 plus straight alpha subscript straight b straight theta right parenthesis 
    The difference in length of the two rods, 
    increment straight L space equals space straight L apostrophe subscript straight b minus straight L apostrophe subscript straight s

space space space space space space space equals space straight L left parenthesis straight alpha subscript straight b minus straight alpha subscript straight s right parenthesis straight theta       
       
         equals 1.2 cross times left parenthesis 1.8 cross times 10 to the power of negative 5 end exponent minus 1.2 cross times 10 to the power of negative 5 end exponent right parenthesis cross times 70 space

equals 5.04 cross times 10 to the power of negative 4 end exponent straight m
    Question 1086
    CBSEENPH11019358

    A brass cube has an edge 10 cm long at 15°C. What will be the area of each face and volume of cube at 60°C (α = 19 x 10–6/°C)?

    Solution

    Let L and L' be the length of edge at 15 degree straight C space and space 60 degree straight C
    ∴           straight L apostrophe space equals space straight L left square bracket 1 plus straight alpha left parenthesis 60 minus 15 right parenthesis right square bracket 
                  space space space equals 10 open curly brackets 1 plus 19 cross times 10 to the power of negative 6 end exponent cross times 45 close curly brackets space

space space space equals 10 left square bracket 1.000855 right square bracket space

space space space equals 10.00855 space cm 
    The edge length of the cube at 60°C is 10.00855 cm.
    Therefore area of each face if cube at 60°C is, 
                 straight S apostrophe space equals space left parenthesis straight L apostrophe right parenthesis squared
space space space space space equals space left parenthesis 10.00855 right parenthesis squared space
space space space space space equals 100.17 space cm squared space 
    And volume of cube at <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> is, 
     straight V apostrophe space equals space left parenthesis straight L apostrophe right parenthesis cubed

space space space space space equals space left parenthesis 10.00855 right parenthesis cubed space
space space space space space equals space 1002.57 space cm cubed 

    Question 1087
    CBSEENPH11019359

    Show that motion of liquid in a U-tube is simple harmonic motion. Also find the time period of oscillation. 

    Solution
    Consider a U-tube of uniform cross-section a.
    Let a liquid of density ρ be poured into the tube such that length of the liquid in each limb is L, large as compared to the width of U-tube.
    If liquid on the left side is depressed by a distance y, then the liquid will rise on the right side by y.
                         
    Therefore,
    The height difference of liquid column in two limbs = 2y.
    Restoring force exerted by weight of liquid column of height 2y is given by, 
                           f with rightwards harpoon with barb upwards on top equals negative 2 y with rightwards harpoon with barb upwards on top Aρg
    This force acts on the whole of liquid in U-tube. 
    The mass of liquid in U-tube is,
                            straight M equals 2 space LAρ 
    Therefore, acceleration of liquid in U-tube is,
    a with rightwards harpoon with barb upwards on top equals fraction numerator rightwards arrow for straight F of over denominator straight M end fraction equals fraction numerator 2 y with rightwards harpoon with barb upwards on top Aρg over denominator 2 LAρ end fraction equals negative straight g over straight L y with rightwards harpoon with barb upwards on top 
    Thus, acceleration is directly proportional to the displacement and directed towards the mean position. Hence, the motion is simple harmonic motion.
    The time period of oscillation is given by, 
                               straight T equals 2 straight pi square root of straight L over straight g end root
    Question 1088
    CBSEENPH11019360

    The coefficient of volume expansion of glycerin is 49 x l0–5 per °C. What is the fractional change in its density for 30°C rise in temperature?

    Solution
    Let m be the mass of glycerin occupying the volume V when its density is d.
    When we increase the temperature, the volume of same mass increases (let it be V') and hence density decreases.
    Let it be ρ'. 
    Therefore,
                 space space space space space ρV space equals space straight rho apostrophe straight V apostrophe 
                     straight rho apostrophe space equals space straight rho fraction numerator straight V over denominator straight V apostrophe end fraction  
    If straight theta be the increase in temperature and straight gamma  be coefficient of volume expansion then, 
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#6 {main}</pre>
     
    ∴                 straight rho apostrophe space equals space straight rho fraction numerator straight V over denominator straight V left parenthesis 1 plus γθ right parenthesis end fraction space equals space straight rho left parenthesis 1 minus γθ right parenthesis 
    ∴                 straight rho apostrophe space equals space straight rho minus pγθ 
    Now, fractional change in density is, 
                fraction numerator straight rho apostrophe minus straight rho over denominator straight rho end fraction equals negative γθ 
    Negative sign shows that density decreases.
    Therefore fractional decrease in density is, 
    γθ space equals space 49 cross times 10 to the power of negative 5 end exponent cross times 30 space equals space 0.0147,  for 30o C rise in temperature. 
    Question 1089
    CBSEENPH11019361

    A piece of metal floats in mercury. The temperature coefficients of volume expansion of metal and mercury are γ1 and γ2 respectively. The temperature of the system is increased by θ. Find the factor by which the fraction of volume of metal submerged in mercury.

    Solution
    Using the law of flotation, the fraction of volume of metal submerged in the mercury is, 
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#6 {main}</pre> 
    where,
    ρ0 and σ0 are the densities of metal and mercury at low temperature.
    When the temperature is increased by straight theta, then the fraction of volume of metal submerged in the mercury is,
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#6 {main}</pre> 
                equals straight rho subscript straight o over straight sigma subscript straight o left parenthesis 1 plus straight gamma subscript 2 straight theta right parenthesis left parenthesis 1 plus straight gamma subscript 1 straight theta right parenthesis to the power of negative 1 end exponent 
    space space space space space space space space space space space space space space space space space equals space straight f subscript straight o space left parenthesis 1 space plus space straight gamma subscript 2 straight theta right parenthesis space left parenthesis 1 minus straight gamma subscript 1 straight theta right parenthesis space

space space space space space space space space space space space space space space space space space equals space space straight f subscript straight o space left square bracket space 1 space plus space left parenthesis straight gamma subscript 2 space minus space straight gamma subscript 1 right parenthesis space straight theta right square bracket space

rightwards double arrow space straight f space minus space straight f subscript straight o space equals space straight f subscript straight o space left parenthesis straight gamma subscript 2 space minus space straight gamma subscript 1 right parenthesis space straight theta space

rightwards double arrow space fraction numerator space straight f space minus space straight f subscript straight o over denominator space straight f subscript straight o end fraction space equals space left parenthesis straight gamma subscript 2 space minus space straight gamma subscript 1 right parenthesis space straight theta space
    Question 1090
    CBSEENPH11019362

    Show that motion of cylinder floating in the liquid is simple harmonic motion and find the time period of oscillations.

    Solution
    Consider a cylinder of length L and density ρ floating in a liquid of density σ. 
    Let A be the area of the cross-section of the cylinder.
    The cylinder floats up to depth ‘h’ in the liquid. 
                   
    According to the law of floatation,
    Weight of body = Weight of liquid displaced
     italic space italic space italic space italic space
italic space italic space italic space italic space italic space italic space A L rho g italic equals A h a g space

space space space space space space space space space space italic space italic space italic space L italic equals fraction numerator h a over denominator rho end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic left parenthesis italic 1 italic right parenthesis
    Let the body be pushed into liquid by distance y.
    Then,
    Weight of the liquid displaced by cylinder of length y will exert net force on cylinder in the upward direction.
    This force will provide the restoring force on the body and is given by, 
                    F space equals space minus Ayσg 
    Therefore acceleration produced by restoring force in cylinder, 
    straight a equals negative Ayσg over ALρ equals negative σg over ρL straight y
    Substituting in (1), we have, 
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#6 {main}</pre>
    therefore space space straight a space proportional to space straight y
    Thus the motion is simple harmonic motion.
    Time period of oscillation is,
    straight T equals 2 straight pi square root of displacement over acceleration end root space

space space equals 2 straight pi square root of straight y over straight a end root space

space space equals 2 straight pi square root of straight h over straight g end root

space space equals 2 straight pi square root of ρL over σg end root
    Question 1091
    CBSEENPH11019363

    A glass tube of length 119cm and of uniform cross section at 0°C is to be filled with mercury so that the volume of unoccupied tube remains same at all the temperatures. The cubical expansion of mercury is seven time that of glass. Find the length of mercury column in the tube.

    Solution
    If the change in the volume of mercury and glass is same at all the temperatures, the volume unoccupied tube remains constant.
    i.e.,              increment straight V subscript straight g space equals space increment straight V subscript Hg
    Let, L and l be the length of tube and length of mercury column in the tube at straight theta°C and 'a' be the area of cross section of tube.
    Since, 
    Error converting from MathML to accessible text. 
    Question 1092
    CBSEENPH11019364

    What is degree of freedom?

    Solution
    The number of independent variables required to specify the dynamic position of particle or system of particles completely is known as Degree of freedom. 

    Question 1093
    CBSEENPH11019365

    How many degrees of freedom the atoms of monontomic gas have?

    Solution
    The atoms of monoatomic gas have three translational degrees of freedom. 
    Question 1094
    CBSEENPH11019366

    How many degrees of freedom the molecules of diatomic gas have?

    Solution
    The molecules of diatomic gas have five degrees of freedom: 3 translational and 2 rotational.
    Question 1095
    CBSEENPH11019367

    How many degrees of freedom the molecules of angular triatomic gas have?

    Solution
    The molecules of angular triatomic gas have six degrees of freedom. 
    That is 3 translational and 3 rotational degree of freedom.


    Question 1096
    CBSEENPH11019368

    How many degrees of freedom the molecules of linear triatomic gas have?

    Solution
    The molecules of linear triatomic gas have seven degrees of freedom.
    The molecules have 3 degree of freedom of translation, 2 degree of freedom of rotation and one degree of freedom of vibration. 
    Question 1097
    CBSEENPH11019369

    A particle is constraint to move on the surface of table. How many degrees of freedom are associated with the motion of particle?

    Solution
    Two degrees of freedom are assosciated with the particle moving on the surface of a table. 
    Question 1098
    CBSEENPH11019370

    What are the degrees of freedom of CO2 molecule?

    Solution
    There are 7 degrees of freedom in a COmolecule. CO2 is a liner molecule. It has 3 rotational, 3 translational and 1 vibrational degree of modes for CO2 molecule. 
    Question 1099
    CBSEENPH11019371

    State law of equipartition of energy. 

    Solution
    Law of equipartition of energy states that:

    For any dynamical system in equilibrium, the total energy is distributed equally among all the possible energy modes and average energy associated with each degree of freedom is 1/2 kBT . 
    Question 1100
    CBSEENPH11019372

    What is average kinetic energy of translation of a molecule of an ideal gas at temperature T ?

    Solution
    Average kinetic energy of translation of a molecule of an ideal gas at temperature T is 3/2 kBT. 
    Question 1101
    CBSEENPH11019373

    What is average kinetic energy of one mole of an ideal gas at temperature T having f degree of freedom for each molecule? 

    Solution
    Average Kinetic energy of 1 mole of an ideal gas is given by, 
                  K.E =f2kBNT=f2RT 
    Question 1102
    CBSEENPH11019374

    Define specific heat.

    Solution
    Specific heat is defined as the quantity of heat required to increase the temperature of unit mass of substance through one degree Celsius. 
    Question 1103
    CBSEENPH11019375

    Define molar specific heat.

    Solution
    Molar specific heat is defined as the amount of heat required to increase the temperature of 1 mole of substance through one degree Celsius. 
    Question 1104
    CBSEENPH11019376

    What is the ratio of molar specific heat to principle specific heat?

    Solution
    The ratio of two given specific heats is numerically equal to molecular weight.
    Question 1105
    CBSEENPH11019377

    What are the different units of specific heat?

    Solution

    The SI unit of specific heat is J kg-1 K-1.

    The CGS unit of specific heat is cal gm-1 K-1.

    Question 1106
    CBSEENPH11019378

    A vessel is completely filled with 500cc of water and 800cc of mercury. When the vessel is heated through 40°C, 8cc of water overflows. Find the coefficient of volume expansion of mercury.

    Given:

    straight gamma subscript straight w equals 1.5 cross times 10 to the power of negative 4 end exponent space straight C presuperscript straight o superscript negative 1 end superscript space and space straight gamma subscript ves space equals space 1.26 cross times 10 to the power of negative 5 end exponent space degree straight C to the power of negative 1 end exponent.

    Solution
    Let on heating, ∆Vhg, ∆VW and ∆VV be the increase in volume of mercury, water and vessel respectively.
    Water being lighter than mercury, floats over mercury and on heating only water will overflow.
    The volume of water overflowing is given by, 
    increment straight V space equals space increment straight V subscript hg plus increment straight V subscript straight w minus increment straight V subscript straight v                 ...(1) 
    Increase in volume of water is, 
    space space space space space increment straight V subscript straight w equals 1300 cross times 1.26 cross times 10 to the power of negative 5 end exponent cross times 40
space space space space space space space space space space space space space space equals space 0.6552 space cc 
    Volume of overflow water is, 
                 increment straight V subscript straight w space equals space 8 cc 
    Now from (1),
                     8 equals increment straight V subscript hg plus 3 minus 0.6552  
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#6 {main}</pre> 
    But,    increment straight V subscript hg space equals space 800 cross times straight gamma subscript hg cross times 40 
    Therefore, 
    straight gamma subscript hg equals fraction numerator 5.6552 over denominator 800 cross times 40 end fraction equals 1.767 cross times 10 to the power of negative 4 end exponent straight C presuperscript straight o superscript negative 1 end superscript, is the coefficient of volume expansion of mercury. 
    Question 1107
    CBSEENPH11019379

    What is the dimension formula of specific heat?

    Solution
    Specific heat = EnergyMass×time = M0L2T-2K-1
    Question 1108
    CBSEENPH11019380

    What is the SI unit of molar specific heat?

    Solution
    The SI unit of molar specific heat is J mole-1 K-1
    Question 1109
    CBSEENPH11019381

    The coefficient of apparent expansion of a liquid in a copper vessel is C and in silver vessel is S. The coefficient of linear expansion of copper is A. Find the coefficient of linear expansion of silver.

    Solution
    Let γ be the coefficient of volume of expansion of liquid and B be the coefficient of linear expansion of silver.
    Therefore, the coefficient of apparent expansion of a liquid in a copper vessel is C, given by 
                      straight C equals 3 straight A minus straight gamma                            ...(1) 
    The coefficient of apparent expansion of a liquid in  a silver vessel is, 
                       straight S equals 3 straight B minus straight gamma                             ...(2) 
    From (1) and (2), 
     straight B equals fraction numerator straight S minus straight C plus 3 straight A over denominator 3 end fraction , is the required coefficient of linear expansion of silver. 
    Question 1110
    CBSEENPH11019382

    Write the dimension formula of molar specific heat. Let the dimension formula of mole be [μ].

    Solution

    The dimensional formula for molar specific heat is, 
    space space space space space space space open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent straight K to the power of negative 1 end exponent straight mu to the power of negative 1 end exponent close square brackets

    Question 1111
    CBSEENPH11019383

    Define heat capacity.

    Solution
    Heat capacity is the amount of heat required to increase the temperature of a body by 1°C.
    Question 1112
    CBSEENPH11019384

    What are the units of heat capacity?

    Solution
    SI unit of heat capacity = J/K
    cgs unit of heat capacity = cal/K


    Question 1113
    CBSEENPH11019385

    Write the dimension formula of heat capacity.

    Solution
    Dimensional formula of heat capacity is M1 L2 T-2 K-1
    Question 1114
    CBSEENPH11019386

    A body of mass m has specific heat S. What is the heat capacity of the body?

    Solution
    Mass = m
    Specific Heat = S

    So, Heat capacity of the body = mS. 
    Question 1115
    CBSEENPH11019387

    Define water equivalent of substance.

    Solution
    Water equivalent of substance is defined as the mass of water that will absorb or lose the same quantity of heat as the substance for the same change in temperature.
    Question 1116
    CBSEENPH11019388

    What is relation between water equivalent of substance and thermal capacity?

    Solution
    The water equivalent of substance is numerically equal to thermal capacity of a substance but different dimensionally. 
    Question 1117
    CBSEENPH11019389

    What are the units of water equivalent of substance?

    Solution
    The units of water equivalent of substance is Kg and gm.
    Question 1118
    CBSEENPH11019390

    Which of the two - gram specific heat or molar specific heat has approximately same value for solids?

    Solution
    The molar specific heat for all the solids is approximately same and is equal to 3R where R is the gas constant.

    The value of gas constant, R = 8.314 J/mol.K  
    Question 1119
    CBSEENPH11019391

    Is the specific heat of substance same for all the phases of matter?

    Solution
    The specific heat of substance is not the same for all the phases of matter. 

    It's value is different for different phases of matter. 

    The heat transferred depends on three factors:
    1. Change in temperature,
    2. Mass of the substance, and
    3. Phase of the substance.  
    Question 1120
    CBSEENPH11019392

    A disc has mass M and radius R. If temperature coefficient of linear expansion of material of disc is a, then find the increase in the moment of inertia of disc when heated through θ°C.

    Solution
    The moment of inertia of disc at low temperature is, 
                            straight I subscript straight o space equals space 1 half MR squared 
    When the disc is heated its mass remains same but radius will increase.
    Let R' be the radius of disc at temperature θ. 
    That is, 
                          straight R apostrophe space equals space straight R left parenthesis 1 plus αθ right parenthesis 
    Therefore the moment of inertia of disc at high temperature is, 
       straight I space equals 1 half MR apostrophe squared space
space space equals space 1 half straight M left square bracket straight R left parenthesis 1 plus αθ right parenthesis right square bracket squared space 
        equals straight I subscript straight o left parenthesis 1 plus αθ right parenthesis squared space space

equals space straight I subscript straight o left parenthesis 1 plus 2 αθ right parenthesis 
    Increase in moment of inertia of disc is, 
              straight I minus straight I subscript straight o space equals space 2 straight I subscript straight o αθ 
    Question 1121
    CBSEENPH11019393

    At what temperature specific heat of water is minimum?

    Solution
    As the temperature is increased after a certain level, the thermal conductivity decreases for liquids.
    In liquid water, specific heat of water decreases at a temperature of 130o C and thereafter. 
    Question 1122
    CBSEENPH11019394

    Define Dulong and Pettit’s law of specific heat.

    Solution
    Dulong and Pettit's law states that the molar specific heat of all the monoatomic solids is constant throughout the range of temperature and is equal to 3 R.
    Question 1123
    CBSEENPH11019395

    Who explained the correct variation of specific heat with temperature in the lower range of temperature?

    Solution
    The correct variation of specific heat with temperature in the lower range of temperature was explained by Debye.  

    Question 1124
    CBSEENPH11019396

    Do the gases have only one value of specific heat?

    Solution
    No, the gases have infinite value of specific heat.
    Question 1125
    CBSEENPH11019397

    What are the principle specific heats of gases?

    Solution
    The principle specific heat of gases are CP and Cv.

    Cis the specific heat at constant pressure
    Cis the specific heat at constant volume.

    Question 1126
    CBSEENPH11019398

    Which has greater value – specific heat at constant volume or specific heat at constant pressure?

    Solution
    The specific heat at constant pressure has greater value than specific heat at constant volume.

    We have CP - CV = R 

    i.e., CP = CV + R
    Question 1127
    CBSEENPH11019399

    What is γ value of a gas?

    Solution
    γ value of the gas is the ratio of specific heat at constant pressure to the specific heat at constant volume.
    Question 1128
    CBSEENPH11019400

    On what factor the value of γ of a gas depends?

    Solution
    The value of γ depends on the atomicity of the gas.
    Question 1129
    CBSEENPH11019401

    How does γ change with atomicity of gas?

    Solution
    With increase in atomicity of the gas, γ decreases.


    Question 1130
    CBSEENPH11019402

    What is the significance of γ (the ratio of the specific heats)?

    Solution
    The ratio of specific heat, γ gives us an explanation about the structure of molecules of a gas and the atomicity of the gas.

    Question 1131
    CBSEENPH11019403

    A metal rod A of 25 cm length expands by 0.05 cm when heated through 100°C and metal rod B of length 40cm expands by 0.04cm when heated through same temperature. A third rod of length 50 cm, which is made of pieces of A and B placed end to end, expands by 0.06 cm for the same rise in temperature i.e. 100°C. Find the length of the each portion of the composite rod.

    Solution
    Let α1 and α2 be the temperature coefficients of linear expansion of two rods A and B respectively.  
    Therefore, 
                      increment straight L subscript 1 space equals space straight L subscript 1 straight alpha subscript 1 straight theta 
    rightwards double arrow                straight alpha subscript 1 space equals space fraction numerator increment straight L subscript 1 over denominator straight L subscript 1 straight theta end fraction 
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#6 {main}</pre> 
    rightwards double arrow                straight alpha subscript 2 space equals space fraction numerator increment straight L subscript 2 over denominator straight L subscript 2 straight theta end fraction 
    Let the composite rod be made of <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> length of A and straight L apostrophe subscript 2 length of B. 
    ∴         straight L equals straight L apostrophe subscript 1 plus straight L apostrophe subscript 2 equals 50 cm                       ...(1) 
    When this rod is heated through θ, then increase in the length is, 
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#6 {main}</pre>     
     
       
             space equals open parentheses straight L apostrophe subscript 1 fraction numerator increment straight L subscript 1 over denominator straight L subscript 1 straight theta end fraction plus straight L apostrophe subscript 2 fraction numerator increment straight L subscript 2 over denominator straight L subscript 2 straight theta end fraction close parentheses straight theta    
              equals straight L apostrophe subscript 1 fraction numerator increment straight L subscript 1 over denominator straight L subscript 1 end fraction plus straight L apostrophe subscript 2 fraction numerator increment straight L subscript 2 over denominator straight L subscript 2 end fraction
    Here, we have 
    increment straight L subscript 1 space equals space 0.05 cm comma space space space increment straight L subscript 2 equals 0.04 cm comma space space increment straight L equals 0.06 cm 
       straight L apostrophe subscript 1 equals 25 space space cm comma space space space straight L apostrophe subscript 2 space equals space 40 space cm
    ∴      0.06 space equals space straight L apostrophe subscript 1 fraction numerator 0.05 over denominator 25 end fraction plus straight L apostrophe subscript 2 fraction numerator 0.04 over denominator 40 end fraction 
    rightwards double arrowspace space space space 2 straight L apostrophe subscript 1 plus straight L apostrophe subscript 2 equals 60                          ...(2) 
    From equation (1) and (2), we have
     straight L apostrophe subscript 1 equals 10 cm comma space space space space space space space space space space space space space space space space space space straight L apostrophe subscript 2 equals 40 cm
    L1' and L2' represent the length of each portion of the composite rod. 
    Question 1132
    CBSEENPH11019404

    What is the value of molar specific heat of an ideal gas at constant volume in terms of the degree of freedom of a gas molecule?

    Solution
    Molar specific heat of an ideal gas at constant volume in terms of the degree of freedom of a gas molecule is given by, 
                                       f2R 

    where, f is the degree of freedom. 
    Question 1133
    CBSEENPH11019405

    What is the value of molar specific heat of an ideal gas at constant pressure in terms of the degree of freedom of a gas molecule?

    Solution
    Molar specific heat of an ideal gas at constant pressure in terms of the degree of freedom of a gas molecule is given by, 

                                    f2R

    where, f is the degree of freedom. 
    Question 1134
    CBSEENPH11019406

    What is Mayer’s relation?

    Solution
    The relation between Cp and Cv is known as Mayer’s relation.

    The relation is given by, 

                                 Cp-Cv=R

    Where CCand R are measured in same units.

    Cis the specific heat at constant pressure, and 
    CV is the specific heat at constant volume. 
    Question 1135
    CBSEENPH11019407

    What is the difference in the molar specific heats of an ideal gas in SI unit?

    Solution
    The difference in molar specific heats of an ideal gas is given by the gas constant. 

    We have, 
                   CP - CV = R = 8.314 J / mol. K
    Question 1136
    CBSEENPH11019408

    Is it possible to supply the heat at constant pressure without increasing the temperature?

    Solution
    When the phase of a substance changes, heat at constant pressure can be supplied without increasing the temperature. 



    Question 1137
    CBSEENPH11019409

    The specific heat of an ideal gas at constant volume is Cv and R is the universal gas constant. What are the degrees of freedom of the ideal gas?

    Solution
    Given,
    Specific heat of an ideal gas = CV 
    Gas constant = R

    So, degree of freedom of the ideal gas = 2C
    v/R.
    Question 1138
    CBSEENPH11019410

    At what temperature the total internal energy of substance is zero?

    Solution
    There is no temperature at which the total internal energy of substance is zero.
    Question 1139
    CBSEENPH11019411

    What is latent heat? 

    Solution
    Latent heat is the quantity of heat required to transit the phase of one gram of substance at a constant temperature.
    That is,  
                               L = Q/m

    L is the latent heat. 
    Question 1140
    CBSEENPH11019412

    Define latent heat of fusion.

    Solution
    Latent heat of fusion is the quantity of heat required to transit the phase of one gram of solid into liquid at it's melting point.
    Question 1141
    CBSEENPH11019413

     Define latent heat of vaporisation.

    Solution
    Latent heat of vaporisation is the quantity of heat required to transit the phase of one gram of liquid into gas at it's boiling point.
    Question 1142
    CBSEENPH11019414

    What are different units of latent heat?

    Solution
    The SI unit of latent heat is J/kg.
    The CGS unit of latent heat is cal/gm.
    Question 1143
    CBSEENPH11019415

    A clock with a metallic pendulum gains 5s in a day at 15°C and loses 10s in a day at 30°C. Find the temperature at which the clock will read correct time.

    Solution
    Let L be the length and straight alpha the temperature coefficient of linear expansion of pendulum clock. 
    If the temperature of the clock is increased or decreased by θ, the length of the pendulum will increase or decrease by Lstraight alpha . And the clock runs slow or fast.
    Let the clock keep correct time when the length is L and temperature is t°C.
    We know time period of the pendulum is given by, 
                            straight T equals 2 straight pi square root of straight L over straight g end root 
    The time period of oscillation of pendulum when the temperature is changed by θ is, 

    straight T apostrophe space equals space 2 straight pi square root of fraction numerator straight L left parenthesis 1 plus αθ right parenthesis over denominator straight g end fraction end root space equals space straight T left parenthesis 1 plus 1 half αθ right parenthesis 
    The change in time period of oscillation of pendulum when the temperature is changed by θ is, 
    straight T apostrophe minus straight T equals dT equals 1 half αθT 
    The change in the time in a day, 
                    increment straight T equals 1 half αθT subscript straight o 
    where, straight T subscript straight o is the duration of the day. 
    At space space 15 degree straight C comma the gain is 5s.
    Therefore,
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#6 {main}</pre>               ...(1) 
    At 30 degree straight C comma the loss is 10s.
    Therefore,
               space minus 10 equals 1 half straight alpha left parenthesis straight t minus 30 right parenthesis straight T subscript straight o              ...(2) 
    From (1) and (2), we get 
     space straight t equals 20 degree straight C, is the temperature at which the clock will read correct time. 
    Question 1144
    CBSEENPH11019416
    Question 1145
    CBSEENPH11019417

    What is the SI unit of molar specific heat?

    Solution
    Molar specific heat is the amount of heat energy required to raise the temperature of 1 mole of a substance by 1 K.
    S.I. unit of molar specific heat is J/mole/K.
    Question 1146
    CBSEENPH11019418

    Define heat capacity.

    Solution
    Heat capacity is the quantity of heat required to increase the temperature of body through 1°C.
    Question 1147
    CBSEENPH11019419

    What are the units of heat capacity?

    Solution
    Units of heat capacity are Joule/Kelvin in SI system and cal/0C in CGS system. 
    Question 1148
    CBSEENPH11019420

    Write the dimension formula of heat capacity.

    Solution

    S.I. unit of heat capacity is Joule/Kelvin and the dimensional formula is open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent straight K to the power of negative 1 end exponent close square brackets

    Question 1149
    CBSEENPH11019421

    Define water equivalent of substance.

    Solution
    The mass of water which would absorb same heat as the given substance, to raise the temperature by same amount is known as the water equaivalent of the substance. 
    Question 1150
    CBSEENPH11019422

    What is relation between water equivalent of substance and thermal capacity?

    Solution
    The water equivalent of substance is numerically equal to thermal capacity.
    Question 1151
    CBSEENPH11019423

    A meter scale is truly calibrated at 20°C whose least count is 1 mm. If the coefficient of linear expansion of scale is l.2 x l0–5C–1 then what will be the least count at temperature 120°C and actual length of bar, if measured length of this bar is 40cm?

    Solution
    Least count of scale is 1 mm, which means that the value of 1 division of calibration is 1 mm.
     
    The distance between the divisions will increase, when the rod is heated to temperature 120°.
    At 120°C the length of 1 division becomes, 
    straight y space equals straight x left parenthesis 1 plus αθ right parenthesis space

space space equals space 1 left parenthesis 1 plus 1.2 cross times 10 to the power of negative 5 end exponent cross times 100 right parenthesis space mm space

space space equals space 1 plus 0.0012 space equals space 1.0012 space mm 
    Thus least count of scale at 120°C increases to 1.0012 mm. 
    Measured length of rod with hot scale is, 
    space space space straight L subscript straight m equals 40 space cm
space space space space space space space space equals space 400 space div.           
           
    Least count of hot scale = 1.0012 mm
    Therefore actual length of rod is,
    straight L subscript straight a equals 400 cross times 1.0012 mm
space space space space space equals space 400.48 mm
space space space space space equals space 40.048 space cm 
    Question 1152
    CBSEENPH11019424

    Is the specific heat of substance same for all the phases of matter?

    Solution
    No, the specific heat of substance is different in different phases of matter. 
    Question 1153
    CBSEENPH11019425

    Define Dulong and Pettit’s law of specific heat.

    Solution
    Dulong and Pettit's law states that the molar specific heat of all the monoatomic solids is constant throughout the range of temperature and is equal to 3 R.
    Question 1154
    CBSEENPH11019426

    Who explained the correct variation of specific heat with temperature in the lower range of temperature?

    Solution
    Debye explained the correct variation of specific heat with temperature in the lower range of temperature.
    Question 1155
    CBSEENPH11019427

    Do the gases have only one value of specific heat?

    Solution
    The gases have more than one value of specific heat of gas. 
    The specific heat for gases are given at constant pressure and volume both, whereas solids and liquids have only one single value of specific heat. 
    Question 1156
    CBSEENPH11019428

    Define latent heat of vaporisation.

    Solution
    Latent heat of vaporization is the quantity of heat required to transit the phase of one gram of liquid into gas at boiling point.
    Question 1157
    CBSEENPH11019429

    What are different units of latent heat?

    Solution

    The SI unit of latent heat is J/kg.

    The CGS unit of latent heat is cal/gm.

    Question 1158
    CBSEENPH11019430

    What is the dimension formula of latent heat?

    Solution
     [ML2T–2 K0] is the dimensional formula of latent heat. 
    Question 1159
    CBSEENPH11019431

    A body weighs W1at 0°C and W2 at θ°C in a liquid. If ρand σ0 be the densities of the body and liquid at 0°C and γb and γL be the coefficients of expansions of body and liquid respectively and γb < γL then which is greater: W1 or W2?

    Solution
    Consider a body of volume V0 at 0°C.
    Since ρ0 and σ0 are the densities of body and liquid at zero degree Celsius, then weight of body in liquid at 0°C is, 
                         straight W subscript 1 equals straight V subscript straight o left parenthesis straight rho subscript straight o minus straight sigma subscript straight o right parenthesis straight g
    If the temperature of system is increased, the volume of body increases and densities of body and liquid decrease, given by 
                           straight V equals straight V subscript straight o left parenthesis 1 plus straight gamma subscript straight b straight theta right parenthesis 
    Therefore,
     Density space of space the space body comma space straight rho equals fraction numerator straight rho subscript straight o over denominator 1 plus straight gamma subscript straight b straight theta end fraction
Density space of space the space liquid comma space straight sigma space equals space fraction numerator straight sigma subscript straight o over denominator 1 plus straight gamma subscript straight e straight theta end fraction 
    At temperature θ, the weight of body in liquid is,
    space straight W subscript 2 equals straight V left parenthesis straight rho minus straight sigma right parenthesis straight g
    That is, 
    equals space straight V subscript straight o left parenthesis 1 plus straight gamma subscript straight b straight theta right parenthesis space open square brackets fraction numerator ρθ over denominator left parenthesis 1 plus straight gamma subscript straight b straight theta right parenthesis end fraction space minus space begin inline style fraction numerator straight sigma subscript straight o over denominator 1 plus straight gamma subscript straight c straight theta end fraction end style close square brackets space straight g

equals space straight V subscript straight o space open square brackets straight rho subscript straight o space minus straight sigma subscript straight o space fraction numerator 1 plus straight gamma subscript straight b straight theta over denominator 1 plus straight gamma subscript straight c straight theta end fraction space close square brackets space straight g

equals space straight V subscript straight o space left square bracket straight rho subscript straight o space minus straight sigma subscript straight o left curly bracket space 1 space plus space left parenthesis straight gamma subscript straight b space minus straight gamma subscript straight c space right parenthesis space straight theta right curly bracket right square bracket straight g space

Therefore comma space

straight W subscript 2 space equals space straight V subscript straight o space left square bracket space straight rho subscript straight o space minus space straight sigma subscript straight o left curly bracket space 1 space plus space left parenthesis straight gamma subscript straight b space minus straight gamma subscript straight c space right parenthesis space straight theta right curly bracket space right square bracket straight g space

space space space space space space equals space space straight V subscript straight o space left square bracket space straight rho subscript straight o space minus space straight sigma subscript straight o right parenthesis space minus space straight sigma subscript straight o space left curly bracket space left parenthesis space straight gamma subscript straight b space minus straight gamma subscript straight c space right parenthesis straight theta right curly bracket space right square bracket space straight g space

space space space space space space equals space space straight V subscript straight o space left parenthesis straight rho subscript straight o space minus space straight sigma subscript straight o right parenthesis space straight g space open square brackets 1 space minus space fraction numerator straight sigma subscript straight o space left parenthesis straight gamma subscript straight b space minus straight gamma subscript straight c right parenthesis over denominator left parenthesis straight rho subscript straight o space minus space straight sigma subscript 0 right parenthesis end fraction straight theta close square brackets

space space space space space space equals space straight W subscript 1 space open square brackets 1 space minus space fraction numerator straight sigma subscript straight o space left parenthesis straight gamma subscript straight b space minus straight gamma subscript straight c right parenthesis over denominator left parenthesis straight rho subscript straight o space minus space straight sigma subscript 0 right parenthesis end fraction straight theta close square brackets

Since comma space straight gamma subscript straight b space less than straight gamma subscript straight c space

therefore space straight gamma subscript straight b space minus straight gamma subscript straight c space less than space 0

therefore space open square brackets 1 space minus space fraction numerator straight sigma subscript straight o space left parenthesis straight gamma subscript straight b space minus straight gamma subscript straight c right parenthesis over denominator left parenthesis straight rho subscript straight o space minus space straight sigma subscript 0 right parenthesis end fraction straight theta close square brackets space greater than thin space 1 space

That space is comma space

straight W subscript 2 space greater than thin space straight W subscript 1
              
    Question 1160
    CBSEENPH11019432

    What is thermometry?

    Solution
    Thermometry is that branch of heat and thermodynamics which deals with the measurement of temperature.
    Question 1161
    CBSEENPH11019433

    Define temperature.

    Solution
    Temperature is the physical condition that determines the direction of flow of heat when bodies are placed in contact with each other.
    Question 1162
    CBSEENPH11019434

    What is a thermometer?

    Solution
    Thermometer is an instrument used for measuring the temperature.
    Question 1163
    CBSEENPH11019435

    Is temperature a scalar or a vector quantity?

    Solution
    Temperature is a scalar quantity.
    Question 1164
    CBSEENPH11019436

    What is the fact on which the mercury thermometer based?

    Solution
    The mercury thermometer is based on the fact that substance expands on heating. 
    Question 1165
    CBSEENPH11019437

    What is the principle on which resistance thermometer based?

    Solution
    The resistance thermometer is based on the principle of variation of resistance of any conductor with temperature.
    Question 1166
    CBSEENPH11019438

    Why mercury is used in thermometers?

    Solution
    Mercury has a low value of specific heat and high thermal conductivity. Hence, mercury is used in thermometers. 
    Question 1167
    CBSEENPH11019439

    Should a thermometric substance have large heat capacity or small heat capacity?

    Solution
    The thermometric substance should have low value of heat capacity so that it attains the temperature of any substance rapidly, without absorbing any appreciable amount of heat from the substance. 
    Question 1168
    CBSEENPH11019440

    What is the essential property of thermometric substance?

    Solution
    Thermometric liquid should have a uniform rate of expansion, so that the linear scale can be easily marked. 
    Question 1169
    CBSEENPH11019441

    Why do we prefer to use mercury in a thermometer instead of water?

    Solution
    We prefer using mercury in a thermometer because mercury is non-volatile, has low specific heat, higher melting point and lower freezing point as compared to water.
    Question 1170
    CBSEENPH11019442

    Which of the two thermometers is more sensitive-gas thermometer or liquid thermometer?

    Solution
    Gas thermometer is more sensitive than liquid thermometer because the rate of expansion of gas is more than liquid. 
    Question 1171
    CBSEENPH11019443

    At what temperature the Fahrenheit and Celsius scale read the same?

    Solution
    At –40°C, both Fahrenheit and Celsius scale read same.
    Question 1172
    CBSEENPH11019444
    Question 1173
    CBSEENPH11019445

    Name the thermometer that uses the radiations emitted by hot body to measure its temperature.

    Solution
    Pyrometer is the thermometer that uses radiations emitted by hot body to measure its temperature. 
    Question 1174
    CBSEENPH11019446

    What is value of latent heat of fusion of ice?

    Solution
    The value of latent heat of fusion of ice is 80 cal/gm.
    Question 1175
    CBSEENPH11019447

    What is the value of latent heat of vaporisation of water?

    Solution
    The value of latent heat of vaporisation of water is 540 cal/gm.
    Question 1176
    CBSEENPH11019448

    A gas is heated at constant volume. What fraction of heat is converted into work?

    Solution
    At a constant volume, the amount of heat which is converted into work is zero. 
    Question 1177
    CBSEENPH11019449

    When an ideal gas is heated at constant volume then what fraction of heat is converted into internal energy?

    Solution
    The fraction of heat that is converted into internal energy is one when an ideal gas is heated at a constant volume. 
    Question 1178
    CBSEENPH11019450

    A gas is heated at constant pressure. What fraction of heat is converted into work?

    Solution

    The fraction of heat which is converted into work is given by, 
                            1 minus 1 over straight y comma 
    where y is the ratio  of the two specific heats.

    Question 1179
    CBSEENPH11019451

    When an ideal gas is heated at constant pressure then what fraction of heat is converted into internal energy?

    Solution

    When an ideal gas is heated at a constant pressure, thenspace space 1 over straight y fraction of heat is converted into internal energy.
    where,
     y is the ratio the two specific heats. 

    Question 1180
    CBSEENPH11019452

    Name different pyrometers used to measure temperature.

    Solution
    Total radiation pyrometer and optical pyrometer is used to measure temperature. 
    Question 1181
    CBSEENPH11019453

    What is the value of specific heat at constant pressure of an ideal gas having degree of freedom f ?

    Solution

    For an ideal gas having 2 degree of freedom, the value of specific heat at constant pressure is given by, 
                            fraction numerator straight f plus 2 over denominator 2 end fraction straight R

    Question 1182
    CBSEENPH11019454

    What is the principle on which the pyrometers are based?

    Solution
    Pyrometers are based on the principle of Stefan’s law.
    Question 1183
    CBSEENPH11019455

    In which form of energy, the heat energy is converted during melting process?

    Solution
    Heat energy is converted into potential energy of the molecules, during the process of melting. 
    Question 1184
    CBSEENPH11019456

    Which metal is used in resistance thermometer?

    Solution
    Platinum.
    Question 1185
    CBSEENPH11019457

    What are lower fixed point (LFP) and upper fixed point (UFP) in thermometry?

    Solution
    In a thermometer, lower fixed point is melting point of ice and upper fixed point is boiling point of water.
    Question 1186
    CBSEENPH11019458

    (a) What do you understand by degree of freedom? Give an example.

    (b) On what factors does the degree of freedom of gas depend?

    Solution

    (a) The number of independent coordinates required to specify the position of particle or system of particles is known as degree of freedom.
    For e.g. if a particle moves on the surface of table, it has two degrees of freedom. This is because in order to locate the position of particle two independent coordinates are required.

    (b) Degree of freedom depends upon the following factors:

    (i) Temperature of gas.

    (ii) Atomicity of gas
    i.e. whether gas is monoatomic, diatomic, triatomic or polyatomic.

    Question 1187
    CBSEENPH11019459

    What is the value of LFP and UFP on Fahrenheit scale?

    Solution
    The value of LFP is 32°F and UFP is 212°F on Fahrenheit scale.
    Question 1188
    CBSEENPH11019460

    Name the scales on which LFP are same.

    Solution
    The value of LFP on Celsius scale and Reaumer scale is the same.
    Question 1189
    CBSEENPH11019461

    Name the scales on which value of fundamental interval i.e. difference in UFP and LFP is same.

    Solution
    The value of fundamental interval on Celsius scale and Kelvin scale is same.
    In Kelvin scale, the lower fixed point is taken as 273 K and the Upper Fixed Point is taken as 373 K. The interval between 373 K and 273 K is known as the fundamental interval. 
    Question 1190
    CBSEENPH11019462

    Is temperature 1°C equal to 1.8°F?

    Solution
    No.
    1o C =  33.8 o F
    Question 1191
    CBSEENPH11019463
    Question 1192
    CBSEENPH11019464

    Among the four scales of temperature which is most accurate?

    Solution
    Fahrenheit scale is the most accurate scale among the four scales of temperature.
    Fahrenheit scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water. The interval between the two is divided into 180 parts. Hence, a fahrenheit scale is more accurate. 
    Question 1193
    CBSEENPH11019465

    What is the degree of freedom for the following cases:

    (a) A body constraint to move on the surface of the earth.

    (b) A body constraint to move on the surface of the earth only along the equator of the earth.

    (c) Bob of a simple pendulum.

    (d) Cylinder rotating about its axis only.

    (e) Monoatomic gas molecule.

    Solution

    (a) The earth is three-dimensional but the body is constrained to move on the surface of the earth. Therefore, one degree of freedom will decrease and the body has 2 degrees of freedom. 

    (b) Only one degree of freedom.

    (c) One degree of freedom. 

    (d) One degree of freedom.

    (e) Three degrees of freedom. 

    Question 1194
    CBSEENPH11019466

    Two thermometers are designed in same way except that one has a spherical bulb and other an elongated cylindrical bulb. Which will respond quickly to temperature change?

    Solution
    The flow of heat is proportional to the surface area or area of cross-section. We know that, the surface area of a cylindrical bulb is more than that of spherical surface. Therefore, the thermometer with cylindrical bulb responds more quickly to temperature change than the thermometer with spherical bulb. 
    Question 1195
    CBSEENPH11019467

    Why triple point is taken as reference point in thermometry?

    Solution
    Triple point of water is a unique temperature and does not change with external condition i.e. with pressure or density. Hence, triple point can be taken as the reference point. 
    Question 1196
    CBSEENPH11019468

    What is the value of absolute zero on Celsius scale?

    Solution
    The value of absolute zero on Celsius scale is -273.15 °C.
    Question 1197
    CBSEENPH11019469

    What is the value of reference point in modern thermometry?

    Solution
    Absolute zero and triple point of water are reference point in modern thermometry because they are unique values and do not change with external conditions. 
    Question 1198
    CBSEENPH11019470

    What is the value of triple point of water on absolute scale?

    Solution
    The triple point of water on absolute scale is 273.16 Kelvin. 
    Question 1199
    CBSEENPH11019471

    Why we add mercury in bulb of Joly’s constant volume gas thermometer?

    Solution
    We add mercury in order to nullify the expansion of bulb so that volume of gas remains constant.
    Question 1200
    CBSEENPH11019472

    What is the temperature range in which the gas thermometers can be used to measure temperature?

    Solution
    The temperature range of gas thermometers is  –268°C to 1600°C.
    Question 1201
    CBSEENPH11019473

    What is the temperature range in which the resistance thermometers can be used?

    Solution
    The temperature range of resistance thermometer is –200°C to 1200°C.
    Question 1202
    CBSEENPH11019474

    What is the temperature range in which the pyrometer can be used to measure the temperature?

    Solution
    A temperature greater than 600°C is used in pyrometer. 
    Question 1203
    CBSEENPH11019475

    Which liquid other than mercury is commonly used in liquid thermometers?

    Solution
    Alcohol, because the freezing point of alcohol is below 1000 C and can record very low temperatures. 
    Question 1204
    CBSEENPH11019476

    Name the suitable thermometer used to measure the temperature in the range 800°C.

    Solution
    Platinum resistance thermometer or thermocouple thermometer can be used to measure the temperature in the range of 800°C.
    Question 1205
    CBSEENPH11019477

    Name the suitable thermometer used to measure the temperature in the range -80°C.

    Solution
    Gas thermometer is used to measure the temperature in the range –80°C. The coefficient of expansion of gas is large and hence can record low temperature. 
    Question 1206
    CBSEENPH11019478

    (a) How many degrees of freedom are there due to vibration motion of diatomic gas molecules?

    (b) How many degrees of freedom are there for monoatomic, diatomic and triatomic gas due to translational and rotational motion only?

    Solution

    (a) The degree of freedom is one.
    Reason: Diatomic gas molecule has at the maximum six degrees of freedom (2x3 = 6) out of which three are due to translational motion, two are due to rotational motion.

    (b) Monoatomic gas molecule has only three degrees of freedom and they are only translational.

    Diatomic gas molecule has five degrees of freedom.

    Consider a triatomic linear gas molecule like CO2 which has seven degrees of freedom and triatomic angular has six degrees of freedom. 

    Question 1207
    CBSEENPH11019479

    A system of N particles in three dimensions has C constraints. What is the number of degrees of freedom of such a system?

    Solution
    Given, a system of N particles in three dimensions has 3N degrees of freedom.
    There are C constraints which decrease the C degrees of freedom.
    Therefore, net degrees of freedom are,
                       f=3N-C
    Question 1208
    CBSEENPH11019480

    Name the suitable thermometer used to measure the temperature of the sun.

    Solution
    A radiation pyrometer is used to measure the temperature of the sun.
    Question 1209
    CBSEENPH11019481

    What is law of equipartition of energy and how much energy is associated with each degree of freedom?

    Solution

    Law of equipartition of energy:

    According to this law, the total energy of the dynamical system in thermal equilibrium is equally shared among all the degrees of freedom.
    The energy associated with each degree of freedom is 1/2kT,
    where,
    k is Boltzmann’s constant and T is temperature. 

    Question 1210
    CBSEENPH11019482

    What are the differences between heat and temperature?

    Solution
    Heat Temperature
    i) Heat is energy which produces the sensation of warmthness. i) Temperature is the physical condition that tells us the direction of flow of heat.
    ii)  The flow of heat from one body to another does not depend on the amount of heat energy in the body i.e. heat can flow from a body having low heat energy to another body having high heat energy and vice versa. ii) Heat can flow only from a body at high temperature to another body at low temperature spontaneously.
    iii) When heat flows from one body to another, the amount of heat energy lost by the hot body is equal to the amount of heat energy gained by the cold body. iii) Fall of the temperature of the hot body need not necessarily to be equal to rise in temperature of the cold body.
    iv) Heat energy of the body can increase without increasing the temperature e.g. during phase transition the temperature of the body remains constant. iv) Temperature of the body cannot increase without increasing the heat energy of the body 
    Question 1211
    CBSEENPH11019483

    Give at least four effects produced by heat.

    Solution

    Some of the heating effects are:  
    (i) Heat energy brings about the change in dimensions of the body.
    (ii) Heat energy changes the temperature of the body. 
    (iii)Heat energy may also cause chemical changes.
    (iv) Heat energy can also change the state of matter.

    Question 1212
    CBSEENPH11019484

    What is a thermometer? Name some thermometers and their principles on which they work?

    Solution

    Thermometer is an instrument used for measuring the temperature.
    The different types of thermometers are:
    (i) Mercury thermometer: It is based on the principle that substance expands on heating.
    (ii) Gas thermometer: It is based on Regnault’s law which states that at constant volume, the pressure of gas varies directly with temperature. Specific heat of gas is constant for all values of the pressure of gas.
    (iii) Resistance thermometer: It is based on the fact that resistance of conductor increases with the increase in temperature.
    (iv) Thermoelectric thermometer: It is based on Seeback effect. 
    The Seebeck effect is a phenomenon in which a temperature difference between two dissimilar electrical conductors or semiconductors produces a voltage difference between the two substances.

    Question 1213
    CBSEENPH11019485

    What are the advantages and disadvantages of gas thermometers?

    Solution

    Advantages of gas thermometers: 
    (i) Gas thermometers are very sensitive because the expansion of gases is considerable.
    (ii) The gases have regular expansion.
    (iii) Gas thermometers have wide range of temperature scale.
    (iv) Permanent gases have close resemblance with perfect gas. Therefore, the thermometers filled with permanent gases give the reading close to thermodynamic scale.
    (v) Gases have low thermal capacity.
    Disadvantages of gas thermometers:
    (i) It is not a direct reading thermometer.
    (ii) It is not portable.
    (iii) It cannot measure accurately the rapidly changing temperatures.

    Question 1214
    CBSEENPH11019486

    Give at least five characteristic of a liquid used as thermometric liquid used in thermometers.

    Solution

    The thermometric liquid used in thermometers should have the following properties:
    (i) should have low heat capacity,
    (ii) should be good conductor of heat,
    (iii) should have large temperature coefficient of expansion,
    (iv) should have uniform rate of expansion,
    (v) should not be volatile and sticky, and
    (vi) should have high boiling point and low freezing point.

    Question 1215
    CBSEENPH11019487

    What is the reason for using mercury as a thermometric liquid?

    Solution

    Mercury is used as a thermometric liquid because of the following reasons:
    (i) it has low heat capacity,
    (ii) it has constant value of temperature coefficient of expansion,
    (iii) it has high melting point and low freezing point,
    (iv) it is good conductor of heat,
    (v) it does not stick with glass, and
    (vi) it is less volatile and hence has very low vapour pressure

    Question 1216
    CBSEENPH11019488

    What are the disadvantages of using mercury as a thermometric liquid?

    Solution
    The disadvantages of mercury as thermometric liquid are:
    i) its temperature coefficient of expansion is very low and it cannot be used to measure the temperature below –39°C and above 357°C.
    ii) Mercury thermometer cannot be used in very cold regions like arctic regions because it freezes below a temperature of 39o C. 
    Question 1217
    CBSEENPH11019489

    What are the advantages of alcohol over mercury for using as thermometric liquid?

    Solution
    Advantages of using alcohol over mercury as thermometric liquid are:
    i) Alcohol has greater value of temperature coefficient of expansion than mercury.
    ii) it's freezing point is below –100°C. Thus, alcohol thermometer is more sensitive than mercury thermometer and can be used to measure the temperature below –40°C.
    Question 1218
    CBSEENPH11019490

    What are the disadvantages of alcohol over mercury for using as thermometric liquid?

    Solution
    Disadvantages of alcohol over mercury for using as thermometric liquid are:
    i) alcohol has low boiling point,
    ii) high heat capacity,
    iii) sticks with glass, and
    iv) is highly volatile.
    Question 1219
    CBSEENPH11019491

    Can we use water as thermometric liquid?

    Solution

    No, water cannot be used as a thermometric liquid because of the following reasons:
    (i) has a very high value of specific heat, 
    (ii) has non-uniform expansion, 
    (iii) is transparent and is a wet liquid,
    (iv) is a bad conductor of heat, 
    (v) evaporates under vacuum conditions, and 
    (vi) can only be used in the temperature region 4°C to 100°C.

    Question 1220
    CBSEENPH11019492

    Why gas thermometers are more accurate than liquid thermometers?

    Solution
    Gas thermometers are more accurate than liquid thermometers because thermal expansion of gases is much more than those of liquids. Also, expansion of gases is quite regular over wider range of temperature than in liquids.
    Question 1221
    CBSEENPH11019493

    What are the relations between different scales of temperatures?

    Solution
    Relations between different scales of temperature is given by, 
    fraction numerator straight C minus 0 over denominator 100 end fraction equals fraction numerator straight F minus 32 over denominator 180 end fraction equals fraction numerator straight R minus 0 over denominator 80 end fraction equals fraction numerator straight K minus 273 over denominator 100 end fraction
    Question 1222
    CBSEENPH11019494

    Which is greater– change in 1°C or change in 1.8°F?

    Solution
    The fundamental interval on Celsius scale is 100°C.
    The fundamental interval on Fahrenheit scale is 180°F.
    Thus,
            Change space of space 100 degree straight C space equals space Change space of space 180 degree straight F 
       space space space space space space space Change space of space 1 degree straight C space equals space Change space of space 1.8 degree straight F
    Question 1223
    CBSEENPH11019495

    Two thermometers have same volume of bulbs – one spherical bulb and the other an elongated cylindrical bulb. Which of the two will respond quickly to temperature change?

    Solution
    The rate of heat transfer is directly proportional to the  surface area of the substance. And, for a given volume, sphere has minimum surface area. That is, the area of a cylindrical bulb is greater than a spherical bulb. Therefore, thermometer with cylindrical bulb will respond more quickly than thermometer with spherical bulb.
    Question 1224
    CBSEENPH11019496

    Why the gases have infinite number of specific heats?

    Solution

    The quantity of heat which is required to raise the temperature of one mole of gas through one degree Celsius is called the specific heat of a substance.
    This heat supplied to gas is not only used to increase the temperature of gas but also used to expand it.
    Heat required to increase the temperature of the gas by 1°C is a constant quantity. But the heat required to expand it depends on the manner in which it is heated. That is why gases have infinite number of specific heats depending on the manner in which it is heated up. 

    Question 1225
    CBSEENPH11019497

    (a) Define two principle specific heats of gases.

    (b) Why Cp is greater than Cv?

    Solution

    (a) Specific heat at constant volume:

    The quantity of heat required to raise the temperature of 1 mole of gas through 1°C at constant volume is known as specific heat at a constant volume.

    Specific heat at constant pressure:
    The quantity of heat required to raise the temperature of 1 mole of gas through 1°C at constant pressure is called specific heat at constant pressure. 

    (b) When gas is heated at constant volume, the whole of the heat supplied is used to increase the temperature of gas because gas is not allowed to expand and work done against expansion is zero.
    When gas is heated at a constant pressure, it expands, and gas has to do work against the external pressure.
    Therefore, in addition to the heat which is required to raise the temperature, heat is also required to expand the gas. So, at constant pressure, we require more heat than at constant volume.

    Question 1226
    CBSEENPH11019498

    Does the relation Cp – Cv= R hold good for solids or liquids?

    Solution
    No, for solids or liquids Cp– Cv< R.
    This is because the change in volume of solids or liquids is very small as compared to gases. Therefore, work done against pressure for solids/liquids is very small as compared to gases. 
    Question 1227
    CBSEENPH11019499

    What is Dulong and Petit’s law? Is this law applicable in all the ranges of temperature?

    Solution
    According to the Dulong and Petit’s law, the specific heat of all the solids is constant.
    That is, 3R/M is independent of temperature.
    This law is in agreement with observed heat capacity only in the higher range of temperature, because the specific heat of all the substances is found to approach zero at low temperature.
    Question 1228
    CBSEENPH11019500

    (a) Can Cp < Cv?

    (b) Can a system which is not in thermal equilibrium be in thermal equilibrium at a later time? Give an example.

    Solution

    (a) Yes, when water is heated from 0°C to 4°C then Cp < Cv , because water contracts when heated from 0°C to 4°C. Water exhibits anomalous behaviour.

    (b) Yes, when we add a hot piece of metal into water, the temperature of the metal piece starts decreasing and that of water starts increasing. After some time temperature of both becomes equal and hence thermal equilibrium is attained.

    Question 1229
    CBSEENPH11019501

    (a) Define latent heat.

    (b) What are dimensions and units of latent heat?

    Solution

    (a) Latent heat is the heat required to transit the phase of unit mass of a substance at a constant temperature. 

    Latent heat of fusion: It is the quantity of heat required to transit the phase of 1 g of solid into liquid at constant temperature.

    Latent heat of vaporization: It is the quantity of heat required to convert 1 g of liquid into vapours at constant temperature. 

    (b) Dimensions of latent heat are [M 0L2T–2]

    Unit of latent heat is J/kg in SI system and cal/g in cgs system.

    Question 1230
    CBSEENPH11019502

    Can there be a solid thermometer? Explain.

    Solution
    Yes, solid thermometers is used with thermoelectric thermometer and platinum resistance thermometer.
    The thermoelectric thermometer is based upon Seebeck effect and platinum resistance thermometer is based upon the fact that the resistance of conductor rises with the rise of temperature. 
    Question 1231
    CBSEENPH11019503

    Why triple point of water is taken as reference point in modern thermometry?

    Solution
    The reference point should be invariant. Melting point of ice and boiling point of water change with pressure, therefore melting point of ice and boiling point of water need to be changed as reference point. In modern thermometry, the triple point of water is taken as reference point because it is a unique point. 
    Question 1232
    CBSEENPH11019504

    In Joly’s constant volume gas thermometer, why mercury is added in glass bulb?

    Solution
    The volume changes when the temperature of the bulb is changed. Therefore the volume of gas in the bulb will no longer be constant. To nullify the change in volume of glass bulb due to change in temperature, the mercury is added in the bulb of the thermometer.
    Question 1233
    CBSEENPH11019505

    How much mercury is added in glass bulb of Joly’s constant volume gas thermometer?

    Solution
    Mercury equal to one-seventh the volume of glass bulb is added in the glass bulb of Joly’s constant volume gas thermometer. This will nullify the change in volume of glass bulb due to change in temperature.
    Question 1234
    CBSEENPH11019506

    Using the law of equipartition of energy, calculate the total energy of one mole of monoatomic, diatomic and triatomic gases.

    Solution
    According to the law of equipartition of energy, the energy is equally distributed among all the degrees of freedom.
    Therefore, if the degrees of freedom of gas molecule is f, then internal energy of 1 mole of gas will be equal to 1/2 kT.
    i.e.,         straight U space equals space straight f over 2 kNT space equals space straight f over 2 RT
    where, N is the Avogadro number. 
     
    Question 1235
    CBSEENPH11019507

    Is internal energy of gas a function of pressure?

    Solution
    The internal energy of an ideal gas is a function of temperature only and does not depend on the pressure of the gas.
    But, the internal energy of real gas is a function of both temperature as well as pressure. 
    Question 1236
    CBSEENPH11019508

    Do you think that water equivalent of a substance is equal to heat capacity of body?

    Solution
    No, water equivalent of the substance is equal to the mass of water which requires same heat for rise in temperature as required by the body to raise the temperature to the same extent.
    If m is the mass of substance and S is specific heat, then quantity of heat required to raise the temperature by θ is, 

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#6 {main}</pre> 
    If W is mass of water required, then
             space space space space space space straight Q equals straight W 11. straight theta space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
 
    From (1) and (2),
                         W=mS
    Therefore water equivalent of the substance is numerically equal to heat capacity but not dimensionally.
    Water equivalent has dimensions [M1L0T0] and heat capacity has dimensions [M1L2T–2K–1].
    Question 1237
    CBSEENPH11019509

    Why the zero of mercury thermometer changes with use of it?

    Solution
    The constant use of mercury thermometer in a temperature region below the temperature at which it is calibrated results in permanent compression. Also, the bulb would be in  a temperature region higher than the temperature at which it is calibrated, the bulb undergoes permanent elongation. Owing to the permanent compression or elongation shifts the zero mark on the thermometer.
    Question 1238
    CBSEENPH11019510

    Mercury boils at 360°C though it can be used to measure the temperature upto 500°C. Explain how.

    Solution
    The mercury freezes at -39°C and boils at 360°C at normal pressure.
    When nitrogen gas is filled in the stem of mercury thermometer, the range of mercury can be increased beyond 360°C. 
    With increase in temperature, the boiling point of liquid increases. When the mercury expands, the gas in the stem gets compressed and hence boiling point of mercury in the thermometer increases. 
    Question 1239
    CBSEENPH11019511

    Why platinum metal is preferred in resistance thermometer?

    Solution
    Platinum metal is preferred in resistance thermometer because of the following reasons:
    i) it has high melting point,
    ii) high value of temperature coefficient of resistance, and
    iii) temperature scale of platinum resistance thermometer is very close to absolute scale. 
    Question 1240
    CBSEENPH11019512

    What are the advantages of pyrometer thermometer?

    Solution
    Pyrometer need not be placed in contact with the hot body whose temperature is to be measured. Therefore, it can be used to measure the temperature of distant hot bodies. Pyrometer has an unlimited upper range of temperature it can measure.
    Question 1241
    CBSEENPH11019513

    What are the disadvantages of pyrometers?

    Solution
    Disadvantages of pyrometer are: 
    i) Stefan's law, which is applicable to perfectly black body is used in radiation pyrometer to find the temperature  of the body. In actual practice hot body may not be ideally black body, therefore there will be error in the measurement of temperature.
    ii) The optical pyrometer cannot be used to measure the temperature of hot body below 600°C because the image of hot body is not brilliant for temperature below 600°C. 
    Question 1242
    CBSEENPH11019514

    Define specific heat and heat capacity.

    Solution

    Specific heat:
    Specific heat is the quantity of heat required to raise the unit temperature of unit mass of the substance by 1-degree celsius of temperature. 

    If dH heat is required to raise the temperature of m gram of substance through dT degree Celsius, then
     
                             dH = mS(dT)
    where,
     S is the specific heat of the substance.
    Mathematically it is given by, 
    therefore space space space space space straight S equals 1 over straight m open parentheses dH over dT close parentheses
    Unit of specific heat is cal/g/K or J/kg/K. 
              

    Question 1243
    CBSEENPH11019515

    On the stem of mercury thermometer ice point and steam point are separated by 180 mm. If the two divisions of calibration are separated by 0.5 mm then what is the least count of thermometer?

    Solution

    Given,
    Distance between thermometer ice point and steam point =  180 space mm space equals space 100 degree straight C 
    Separation between two divisions of calibrations = 0.5 mm 
    ∴          space space space 0.5 mm space equals space 100 over 180 cross times 0.5 space equals space left parenthesis 5 divided by 18 right parenthesis degree straight C 
    i.e.  least count of thermometer is left parenthesis 5 divided by 18 right parenthesis degree straight C 

    Question 1244
    CBSEENPH11019516

    Explain that heavier metal has lesser value of gram specific heat.

    Solution
    The molar specific heat of all the metals is approximately constant and is equal to 3R.
    The formula for gram specific heat is given by,
    Gram space spepic space heat space equals fraction numerator Molar space specific space heat over denominator Atomic space weight end fraction
    i.e. gram specific heat is inversely proportional to the atomic weight of the metal.
    Thus heavier is the metal lesser is the value of gram specific heat.
    Question 1245
    CBSEENPH11019517

    Is specific heat of solid constant throughout the range of temperature? Discuss. 

    Solution
    No, the specific heat of solid is not constant throughout the range of temperature.
    Specific heat of solid is constant only in the higher range of temperature. In the lower range, specific heat increases with temperature, which is in accordance with the Debye's law. 
    Debye's law is, 
                              Cv ∝ T
    Question 1246
    CBSEENPH11019518

    At what temperature reading on both Fahrenheit scale and Reaumer scale is same?

    Solution

    Relation between Fahrenheit scale and Reaumur scale is given by,  
                         space space space space space space space fraction numerator straight F minus 32 over denominator 180 end fraction equals straight R over 80 
    Let at temperature T, the reading on both Reaumur scale and Fahrenheit scale be the same. 
    ∴                  fraction numerator straight T minus 32 over denominator 180 end fraction equals straight T over 80  
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    Question 1247
    CBSEENPH11019519

    Why the gases have infinite number of specific heats?

    Solution
    Specific heat is the quantity of heat required to raise the temperature of one mole of gas by one degree Celsius.
    This heat supplied to the gas is used to expand the gas apart from increasing the temperature. The heat required to expand the gas depends on the manner in which heat exchange takes place or heating is done. 
    Therefore, gases have an infinite number of specific heats depending on the manner in which it is heated up.
    Question 1248
    CBSEENPH11019520

    At what temperature do the Fahrenheit scale and Kelvin scale coincide?

    Solution

    The relation between Fahrenheit and Kelvin is given as, 
                      space fraction numerator straight F minus 32 over denominator 180 end fraction equals fraction numerator straight K minus 273 over denominator 100 end fraction 
    Let at temperature T, Fahrenheit scale and Kelvin scale readings coincide. 
    ∴              space space space fraction numerator straight T minus 32 over denominator 180 end fraction equals fraction numerator straight T minus 273 over denominator 100 end fraction 
    rightwards double arrow          5 straight T minus 160 space equals space 9 straight T minus 2457 
    rightwards double arrow                      4 straight T space equals space 2297 
    Thus,                 space straight T equals 574.25 degree 
    So at this temperature, the Fahrenheit and the kelvin scale coincide. 

    Question 1249
    CBSEENPH11019521

    Show that Cp – Cv = R.

    Solution

    Consider one mole of an ideal gas enclosed in a cylinder fitted with the movable frictionless piston.

    Let the gas be heated at constant volume.
    On supplying dQ amount of heat, let the temperature of the gas increase by dT. 

    therefore space dQ space equals space straight m space cross times straight C subscript straight P space cross times space dT space

space space space space space space space space space space equals space 1 cross times space straight C subscript straight P space cross times space dT space

According space to space the space first space law space of space thermodynamics comma space

dQ space equals space dU space plus space dW space

space space space space space equals space dU space plus space straight P. dV space left square bracket space dV space equals space 0 right square bracket space

space space space space space equals space dU space
    therefore space space space space space space space dQ apostrophe equals 1 cross times straight C subscript straight p cross times dT 
    By first law of thermodynamics, 
               dQ = dU + dW = dU + PdV 
    rightwards double arrow    CpdT = Cx dT + RdT 
    At a constant pressure, 
    P.dV = R.dT 

    
therefore space space space space space space space straight C subscript straight p equals straight C subscript straight v space plus space straight R space

rightwards double arrow space space space space space space space straight C subscript straight p minus straight C subscript straight v equals straight R
 
    where  CpCand R are in same units. 
    Hence, proved. 

    Question 1250
    CBSEENPH11019522

    Can at any temperature Centigrade scale and Kelvin scale read same?

    Solution
    Relation between Kelvin scale and Celsius scale is given by,
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#6 {main}</pre> 
    Let at temperature T, both scales read the same. 
    ∴                  straight T over 100 equals fraction numerator straight T minus 273 over denominator 100 end fraction  
                            space space space straight T equals straight T minus 273 
    rightwards double arrow                  273 = 0; which is not possible. 
    Therefore there is no temperature at which Kelvin scale and Centigrade scale can read the same. 
    Question 1251
    CBSEENPH11019523

    If an ideal gas has f degrees of freedom, then what is the internal energy of gas and what will be the value of Cpand Cfor such a gas?

    Solution
    According to equipartition law of energy, the energy is equally distributed among all degrees of freedom.
    Therefore, the internal energy of gas of one mole having f degrees of freedom is given by,
    space space space space space space space space space space space space straight U space equals space straight f over 2 N k T equals straight f over 2 R T space

therefore space space space space space space straight C subscript straight v equals fraction numerator d U over denominator d T end fraction equals straight f over 2 straight R space

a n d space space straight C subscript straight p equals straight C subscript straight v plus straight R space

space space space space space space space space space space space space space space equals fraction numerator straight f plus 2 over denominator 2 end fraction straight R 
    Question 1252
    CBSEENPH11019524

    Can the specific heat of gas be infinity?

    Solution
    Yes, the specific heat of gas will be infinity. 
    In an isothermal process, the specific heat of the gas is infinity. The whole of the heat supplied is used to do work on gas.
    Increase in the temperature of gas is zero, for an isothermal process.
     i.e. ∆T= 0
    So, specific heat of the gas is given as,
                  space straight C equals fraction numerator d Q over denominator d T end fraction equals fraction numerator d Q over denominator 0 end fraction equals infinity
    Question 1253
    CBSEENPH11019525

    A faulty thermometer reads 10°C in melting ice and 140°C in boiling water at normal pressure. What will be the reading on this thermometer when true thermometer reads 40°C?

    Solution

    Let T be the temperature on the faulty thermometer when the true thermometer reads 40°C. 
    ∴               space space space straight C over 100 equals fraction numerator straight T minus 10 over denominator 140 minus 10 end fraction 


    rightwards double arrow               40 over 100 equals fraction numerator straight T minus 10 over denominator 130 end fraction 

    rightwards double arrow                 space space space space space straight T equals 62     
    Therefore, temperature on the faulty scale will be space 62 degree straight C.

    Question 1254
    CBSEENPH11019526

    Calculate the temperature of Celsius scale whose value is half of that on Kelvin scale.

    Solution

    The relation between Celsius and Kelvin is given by, 
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Let straight t degree straight C space equals space 2 straight t space straight K 
    ∴              straight t over 100 equals fraction numerator 2 straight t minus 273 over denominator 100 end fraction 
    rightwards double arrow                   straight t equals space 2 straight t minus 273 
     
    rightwards double arrow                    straight t equals 273 degree straight C 

    Question 1255
    CBSEENPH11019527

    Two thermometers – one Celsius and other Fahrenheit are put in a hot bath. The reading on Fahrenheit is 2.6 times the reading on Celsius scale. Show that this temperature is in the range of temperature of human body.

    Solution

    The relation between Celsius and Fahrenheit is, 
                       space space fraction numerator straight C minus 0 over denominator 100 end fraction equals fraction numerator straight F minus 32 over denominator 180 end fraction 
    Let space space space space straight t degree straight C space equals space 2.6 straight t degree straight F 
    ∴             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre> 
    rightwards double arrow                   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#6 {main}</pre> 
    rightwards double arrow                   4 straight t equals 160 
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#6 {main}</pre> 
    That is, the temperature is 40 degree straight C space and the temperature of human body is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 

    Question 1256
    CBSEENPH11019528

    By using first law of thermodynamics, find the change in internal energy during boiling of liquids.

    Solution

    Let the latent heat of vaporisation of a liquid of mass m at its boiling temperature be L. 
    Let V be the volume of liquid and Vv be the volume of vapours and let evaporation take place at constant pressure P.

    According to the first law of thermodynamics,
    space space space space space space space space space space increment straight Q equals increment straight U plus straight P increment straight V space

rightwards double arrow space space space space space space mL equals increment straight U plus straight P left parenthesis straight V subscript straight v minus straight V subscript straight ell right parenthesis space

rightwards double arrow space space space increment straight U equals mL minus straight P left parenthesis straight V subscript straight v minus straight V subscript straight ell right parenthesis space space 
    This is the required expression for change in internal energy. 

    Question 1257
    CBSEENPH11019529

    When a bulb of constant volume air thermometer is placed in melting ice and boiling water at normal pressure, the pressure is found to be 70 cm and 100 cm of mercury respectively. What will be the temperature of the bath, pressure of the gas being 120 cm of Hg?

    Solution

    We have,
    Pressure space when space thermometer space is space placed space in space melting space ice comma space straight P subscript straight o equals space 70 space cm space of space Hg space

Pressure space when space placed space in space boiling space water comma space straight P subscript 100 space equals space 100 space cm space of space Hg space

Pressure space of space the space gas comma space straight P subscript straight theta space equals space 120 space cm space of space Hg 
    where. 
    straight P subscript straight theta is the pressure of bath whose temperature is to be measured.
    We know,  
        straight theta equals fraction numerator straight P subscript straight theta minus straight P subscript straight o over denominator straight P subscript 100 minus straight P subscript straight o end fraction cross times 100 

          equals fraction numerator 120 minus 70 over denominator 100 minus 70 end fraction cross times 100 space equals space 50 over 30 cross times 100
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#6 {main}</pre>, is the temperature of the bath when pressure is 120 cm of Hg. 

    Question 1258
    CBSEENPH11019530

    A constant volume gas thermometer using helium records a pressure of 20.0 KPa at the triple point of water and pressure of 14.3 KPa at temperature of dry ice (solid CO2). What is the temperature of dry ice?

    Solution

    We know that pressure is directly proportional to the temperature on absolute scale. 
    Temperature of triple point of water, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  
    Let T be the temperature of dry ice on Kelvin scale. 
    So, we have 
                  straight T over straight T subscript straight o equals fraction numerator 14.3 cross times 10 cubed over denominator 20.0 cross times 10 cubed end fraction equals 143 over 200
                      straight T equals 143 over 200 cross times 273.16 straight K
                        equals 195.3 space straight K 
                         equals negative 77.85 degree straight C , is the temperature of the dry ice. 
               

    Question 1259
    CBSEENPH11019531

    n1 and n2 moles of two different gases A and B are mixed together. If Cp1 and Cp2be the specific heats at constant pressure and Cvl and Cv2 be the specific heats at constant volume of gases respectively, then what is the Cp and Cv of mixture and hence find the value of γ.

    Solution

    Heat required to raise the temperature of nmoles of gas A at constant volume through 1°C is given by,

                          straight Q subscript vA equals straight n subscript 1 Cv subscript straight l
    Heat required to raise the temperature of nmoles of gas B at constant pressure through 1°C is, 

                           straight Q subscript vB equals straight n subscript 2 straight C subscript straight v 2 end subscript
    Therefore,
    Total quantity of heat required to raise the temperature of mixture of (n1 + n2) moles through 1°C at constant volume,

                         straight Q subscript straight v equals straight n subscript 1 straight C subscript vl plus straight n subscript 2 straight C subscript straight v 2 end subscript
    Heat required to raise the temperature of 1 mole of mixture through 1°C at constant volume,

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#6 {main}</pre>
    Similarly, heat required to raise the temperature of 1 mole of mixture through 1°C at constant pressure,

    straight C subscript straight P equals fraction numerator straight Q subscript straight v over denominator straight n subscript 1 plus straight n subscript 2 end fraction equals fraction numerator straight n subscript 1 straight C subscript straight P 1 end subscript plus straight n subscript 2 straight C subscript straight p 2 over denominator straight n subscript 1 plus straight n subscript 2 end fraction
    So, the ratio of specific heat is given by, 
    space space space space space space space space space space space space straight y equals fraction numerator straight n subscript 1 straight C subscript straight P 1 end subscript plus straight n subscript 2 straight C subscript straight P 2 end subscript over denominator straight n subscript 1 plus straight C subscript straight v 1 end subscript plus straight n subscript 2 straight C subscript straight v 2 end subscript end fraction
    Question 1260
    CBSEENPH11019532

    What is a resistance thermometer? Give the method of measuring temperature with it.

    Solution

    Resistance thermometer is a thermometer, which determines the temperature by studying the variation in resistance of metal due to change in temperature.
    Resistance of metals approximately increases linearly with temperature.
    By every one-degree rise or fall in temperature, the resistance of conductor increases or decreases by some constant multiple value of resistance of zero degree centigrade. 
    ∴                    space space space space space straight R subscript straight t space equals space straight R subscript straight o plus straight R subscript straight o αt 
                       straight R subscript straight t minus straight R subscript straight o space equals space straight R subscript straight o αt                       ...(1) 
    When t = 100, 
                    straight R subscript 100 minus straight R subscript straight o space equals space straight R subscript straight o straight alpha left parenthesis 100 right parenthesis                 ...(2) 
    Dividing (1) by (2), we get 
             space space space space space space space space space space space space straight t equals fraction numerator straight R subscript straight t minus straight R subscript straight o over denominator straight R subscript 100 minus straight R subscript straight o end fraction cross times 100

    Question 1261
    CBSEENPH11019533

    nmoles of gas A are mixed with nmoles of gas B whose ratio of specific heats are γ1 and γ2. What are the values of specific heats Cp and Cv of the mixture?

    Solution
    The ratio of specific heat of gas A is γ1 and that of gas B is γ2
    The specific heat of gas A and gas B at constant volume is, 
    straight C subscript straight v 1 end subscript equals fraction numerator straight R over denominator straight y subscript 1 minus 1 end fraction space and space straight C subscript straight v 2 end subscript equals fraction numerator straight R over denominator straight y subscript 2 minus 1 end fraction respectively
     
    The specific heat of mixture at constant volume is, 
    space space space space space space space
space space space space space space space space space space space straight C subscript straight v equals fraction numerator straight n subscript 1 straight C subscript straight v 1 end subscript plus straight n subscript 2 straight C subscript straight v 2 end subscript over denominator straight n subscript 1 plus straight n subscript 2 end fraction space space

space space space space space space space space space space space space space equals fraction numerator straight n subscript 1 begin display style fraction numerator straight R over denominator straight y subscript 1 minus 1 end fraction end style plus straight n subscript 2 begin display style fraction numerator straight R over denominator straight y subscript 2 minus 1 end fraction end style over denominator straight n subscript 1 plus straight n subscript 2 end fraction space

rightwards double arrow space space space straight C subscript straight v equals fraction numerator straight R over denominator straight n subscript 1 plus straight n subscript 2 end fraction open parentheses fraction numerator straight n subscript 1 over denominator straight y subscript 1 minus 1 end fraction plus fraction numerator straight n subscript 2 over denominator straight y subscript 2 minus 1 end fraction close parentheses
    Similarly, specific heat of mixture at constant pressure is, 

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#6 {main}</pre>
    Question 1262
    CBSEENPH11019534

    Find the internal energy of V volume of gas at pressure P. The ratio of the specific heats is γ.

    Solution
    The internal energy of n moles of an ideal gas at temperature T is given by,
                           U = n CT

    where,
     Cv is the molar specific heat at constant volume.
    In terms of γ, the value of Cv can be written as


    space space space space space space straight C subscript straight v equals fraction numerator straight R over denominator straight y minus 1 end fraction space

therefore space space space space straight U equals fraction numerator nRT over denominator straight y minus 1 end fraction

space space space space space space space space space space equals fraction numerator PV over denominator straight y minus 1 end fraction

    Question 1263
    CBSEENPH11019535

    The resistance of platinum wire resistance thermometer at ice point is 5.5Ω and 7.5Ω at steam point. The pressure exerted by gas in constant volume gas thermometer at respective points is 100cm and 136cm of mercury. When both the thermometers are put in a bath, the resistance of platinum resistance thermometer is 7.1Ω and the pressure of the gas is 128cm of mercury. Compare the temperature recorded by two thermometers. 

    Solution
    Resistance thermometer: 
    We have,
     straight R subscript straight o space equals space 5.5 space straight capital omega comma space space space space space space space space straight R subscript 100 space equals space 7.5 space straight capital omega comma   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
    Let t be the temperature of bath recorded by resistance thermometer. 
    ∴                straight t subscript 1 space equals space fraction numerator straight R subscript bath minus space straight R subscript straight o over denominator straight R subscript 100 minus straight R subscript straight o end fraction cross times 100
                      space space space space equals fraction numerator 7.1 minus 5.5 over denominator 7.5 minus 5.5 end fraction cross times 100
space space space space equals space 80 degree straight C 
    Gas Thermometer: 
    We have,
    straight P subscript straight o space equals space 100 cm space of space Hg
 
    space space straight P subscript 100 space equals space 136 cm space of space Hg 
    straight P subscript bath space equals 128 space cm space of space Hg
    Let t2 be the temperature of bath recorded by gas thermometer. 
    ∴           straight t subscript 2 equals fraction numerator straight P subscript straight theta minus straight P subscript straight o over denominator straight P subscript 100 minus straight P subscript straight o end fraction cross times 100 

                   equals fraction numerator 128.5 minus 100 over denominator 136 minus 100 end fraction cross times 100
equals space 79.17 degree straight C 
    The temperature of bath on platinum scale is 80°C and that of same bath on gas scale is 79.17°C. 
    Question 1264
    CBSEENPH11019536

    Find the internal energy of air in a room of size 3m x 4m x 6m at pressure 75cm of mercury.

    Solution
    The internal energy of V volume of gas at pressure P is given by, 
                space space space space space space
space space space space space space space space space space space space space U equals fraction numerator P V over denominator y minus 1 end fraction             ... (1)
    we have,
    Volume of the room, V=3m x 4m x 6m = 72m

    Pressure, P=75cm of mercury
                    =75 x 13.6 x 980 dyne/cm2 
                    = 9.996 x 10N/m
    Air is mixture of diatomic gases.
    Therefore,
                     y = 1.4
    Putting values in equation (1), we have
    So, internal energy is given by, 
    
space straight U space equals fraction numerator 9.996 cross times 10 to the power of 4 cross times 72 over denominator 1.4 minus 1 end fraction space space space

space space space space space equals space 1.8 cross times 10 to the power of 7 straight J          
                
    Question 1265
    CBSEENPH11019537

    A constant volume thermometer using helium gas records a pressure of 1.75 x 104 Pa at normal freezing point of water and a pressure of 2.39 x 104 Pa at normal boiling point of water. Obtain from these observations the temperature of absolute zero on Celsius scale.

    Solution
    Let the relation between the temperature on the two scales be, 
                     K = T + b               ...(1) 
    where,
    K is temperature on absolute scale and T on Celsius scale. 
    If,
         K = 0     then T = - b           ...(2) 
    Therefore the absolute zero corresponds to -b on Celsius scale. 
    On Celsius scale, normal freezing and the boiling point of water are 0°C and 100°C respectively.
    From equation (1), the normal freezing and boiling point of water on the absolute scale are b and (100 + b) respectively. 
    We know, 
    Pressure α temperature on absolute scale, 
    ∵        fraction numerator 100 plus straight b over denominator straight b end fraction equals fraction numerator 2.39 cross times 10 to the power of 4 over denominator 1.75 cross times 10 to the power of 4 end fraction equals fraction numerator 2.39 over denominator 1.75 end fraction 

     rightwards double arrow     1 plus 100 over straight b equals 239 over 175 
    rightwards double arrow           100 over straight b equals 239 over 175 minus 1 space equals space 64 over 175
    rightwards double arrow                 straight b equals 175 over 64 cross times 100 space equals space 273.44 
    From equation (2), the value of absolute zero on Celsius scale is, 
    space space space straight T equals negative straight b equals negative 273.44 degree straight C. 
    Question 1266
    CBSEENPH11019538

    Find the quantity of heat required to increase the temperature of 960mg of Ofrom 20°C to 45°C in a rigid container. The molar specific heat of O2 at constant volume is 5cal/mol/K.

    Solution
    Mass of O2 gas = 960mg = 0.96gm

    Therefore the number of moles in 0.96gm of gas are,
    n equals fraction numerator 0.96 over denominator 32 end fraction equals 0.03 space m o l e s 
    The amount of heat required is,
    
space space space space space space space space space space straight Q space equals nC subscript straight v increment straight T

space space space space space space space space space space space space space space equals 0.03 cross times 5 cross times 25 space space

space space space space space space space space space space space space space space equals 3.75 space cal. 
    Question 1267
    CBSEENPH11019539

    The triple points of neon and carbon dioxide are 24.57K and 216.55K respectively. Express these temperatures on Celsius and Fahrenheit scales.

    Solution
    We know that relation between different scales of temperature is, 
             fraction numerator straight K minus 273.15 over denominator stack 100 with underbrace below with straight I below end fraction equals fraction numerator straight C minus 0 over denominator stack 100 with underbrace below with II below end fraction space equals space fraction numerator straight F minus 32 over denominator stack 180 with underbrace below with III below end fraction 
    From (1) and (2), we get 
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#6 {main}</pre>  
    Therefore,
    Triple point of neon on Celsius scale is, 
             straight C subscript Ne space equals space 24.57 minus 273.15 minus 248.58 degree straight C 
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#6 {main}</pre> on Celsius scale is, 
             straight C subscript CO subscript 2 end subscript space equals space 216.55 minus 273.15 space equals space minus 56.60 degree straight C 
    From (2) and (3), we get, 
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#6 {main}</pre> 
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#6 {main}</pre> 
    and  
                  straight F subscript CO subscript 2 end subscript space equals space 9 over 5 left parenthesis negative 56.6 right parenthesis plus 32 space

space space space space space space space space space equals space minus 69.88 degree straight F 

    Question 1268
    CBSEENPH11019540

    For an ideal gas, the value of γ is 1.4. Find the value of specific heats of gas at constant volume and constant pressure, take R = 2 cal/mole/K.

    Solution

    We have,
    Ratio of specific heat, y = 1.4
    R=2 cal/mole/K 
    Now, using the formula,  
    space space space space space space space space space space straight y equals straight C subscript straight P over straight C subscript straight v equals 1.4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space

space space space space straight C subscript straight P minus straight C subscript straight v equals straight R equals 2 space cal divided by mole divided by straight K space space space space space space space space... left parenthesis 2 right parenthesis
space
    Solving(1) and (2), we get
              C= 7 cal/mole/k, is the specific heat at constant volume.
     and    C = 5 cal/mole/K, is the specific heat at constant pressure. 

    Question 1269
    CBSEENPH11019541

    What is the physical condition that determines thermal equilibrium?

    Solution
    Temperature is the physical condition which determines the thermal equilibrium. 
    Question 1270
    CBSEENPH11019542

    When two bodies will be said to be in thermal equilibrium?

    Solution
    When the temperature of two bodies is same they are said to be in thermal equilibrium. 
    Question 1271
    CBSEENPH11019543

    If two bodies are not in thermal equilibrium, then in which direction the heat will flow?

    Solution
    If two bodies are not in thermal equilibrium, then the heat flows from body at high temperature to the body at low temperature. 
    The flow of heat takes place for the bodies which are not in thermal equilibrium. 
    Question 1272
    CBSEENPH11019544

    State Zeroth law in thermo-dynamics.

    Solution
    Zeroth Law in thermodynamic states that, when two bodies are in thermal equilibrium with third body separately, then two bodies are also in thermal equilibrium with each other. 
    Question 1273
    CBSEENPH11019545

    What is indicator diagram?

    Solution
    The graphical representation of the state of system with the help of two thermodynamic variables is known as an indicator diagram. 
    Question 1274
    CBSEENPH11019546

    What do you mean by the internal energy of the system?

    Solution
    The internal energy of a system is the total energy possessed by the system due to molecular motion and molecular configuration. 
    Internal energy in Thermodynamics is the total amount of energy contained within the whole system. The total energy includes the kinetic energy and potential energy. 
    Question 1275
    CBSEENPH11019547

    What are the components of internal energy?

    Solution
    The internal energy of the system is the sum of the internal kinetic energy and internal potential energy of the system. 
    Question 1276
    CBSEENPH11019548

    What do you mean by the internal kinetic energy of system?

    Solution
    The matter is made up of tiny particles, atoms or molecules.
    These molecules of matter are in continuous motion and therefore possess the kinetic energy.
    The sum of the kinetic energy of all the particles is the internal kinetic energy of the system.
    Question 1277
    CBSEENPH11019549

    What do you mean by internal potential energy of system?

    Solution
    The atoms or molecules of matter are bonded by interatomic or intermolecular forces.
    Therefore the molecules have potential energy called internal potential energy.
    Question 1278
    CBSEENPH11019550

    Which phase of matter possesses the minimum internal energy?

    Solution
    Kinetic energy is least in the solids. Hence, solid phase possesses the minimum internal energy. 
    Question 1279
    CBSEENPH11019551

    If we compress the gas, how does the internal potential energy of gas change?

    Solution
    When gas is compressed, work is done on the gas. Therefore, the internal potential energy of the gas will decrease and dW is taken as negative. 
    Question 1280
    CBSEENPH11019552

    What do you mean by internal and external work?

    Solution
    When one part of the system does the work on another part of the same system, then the work is called internal work.
    If a force is exerted on the system and does work on the system then it is called external work.
    Question 1281
    CBSEENPH11019553

    Can internal work change the potential energy of system?

    Solution
    No, internal energy cannot change the P.E of the system.
    The internal energy of an ideal gas consists only of kinetic energy, which depends only on the temperature of the gas. 
    Question 1282
    CBSEENPH11019554

    Can external work change the potential energy of the system? 

    Solution
    Work done by an external agent can change the potential energy of the system. 
    Positive external work leads to an increase in the potential energy of the system. 
    For example, we have an apple that is lying on the ground. The work done by an external agent to pick the apple from the ground or lift the apple is stored as the gravitational potential energy. 
    Question 1283
    CBSEENPH11019555

    For and ideal gas straight R over straight C subscript straight v equals 2 over 3 comma space what is the atomicity of gas?

    Solution

      W e space h a v e comma space

space space space space space straight R over straight C subscript straight v equals 2 over 3 space

rightwards double arrow space space straight C subscript straight v equals 1.5 space straight R space

A l s o comma space space space space straight C subscript straight P minus straight C subscript straight v equals straight R

T h u s comma space space space straight C subscript straight P equals 2.5 straight R space

N o w comma space space straight y equals straight C subscript straight P over straight C subscript straight v equals fraction numerator 2.5 over denominator 1.5 end fraction equals 5 over 3
    Thus the gas is monoatomic.

    Question 1284
    CBSEENPH11019556

    Write the expression for work done on or by the gas during any process.

    Solution

    Work done on or by the gas during any process is given by, 
                      straight W equals integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV
    where, 
    P is the pressure acting on the system, and 
    dV is the change in the volume. 

    Question 1285
    CBSEENPH11019557

    How we can find the work done on/ by the gas from P v/s V graph?

    Solution
    The work done by or on the gas is equal to area under P–V graph. 
    Question 1286
    CBSEENPH11019558

    State first law of thermodynamics.

    Solution
    The first law of thermodynamics states that whenever heat is supplied to a system, a part of it increases the internal energy and remaining heat is used in doing external work. 
    Mathematically, 
                          dQ space equals space dU plus dW
    Question 1287
    CBSEENPH11019559

    Is internal energy an exact differentiable function?

    Solution
    Yes, the internal energy is an exact differentiable function.
    Differential form of internal energy is given by, 
                            dU = dQ + dW
    Question 1288
    CBSEENPH11019560

    Is internal energy a path function?

    Solution
    Internal energy is not a path function. 
    Internal energy is a state function because the value depends only on the current state of the system and not on the path taken or process undergone to prepare it. 
    Question 1289
    CBSEENPH11019561

    Is heat and work a path function?

    Solution
    Yes, heat and work are path functions.
    Rate of heat transfer depends on the path taken. 
    The amount of work done also is path-dependent. 
    Question 1291
    CBSEENPH11019563

    What is process?

    Solution
    A thermodynamic process is a course during which the thermodynamic state of the body changes.
    Question 1292
    CBSEENPH11019564

    What are the thermodynamic variables?

    Solution
    Thermodynamic variables are the parameters that completely define the thermodynamic state of the system.
    For e.g. pressure, volume, temperature, internal energy, entropy, enthalpy etc are thermodynamic variables. 
    Question 1293
    CBSEENPH11019565

    What do you mean by equation of state of system?

    Solution
    The equation of a homogeneous system at any time is described in terms of three thermodynamic parameters namely pressure, volume and temperature. 
    The mathematical relation between these parameters is called the equation of state of the thermodynamic system. 
    Question 1294
    CBSEENPH11019566

    What is an isothermal process?

    Solution
    The process during which the temperature of the system remains constant is called isothermal process.
    Question 1295
    CBSEENPH11019567

    What is the equation of state of isothermal process?

    Solution
    The equation of state of isothermal process is given by,
                      PV = constant
    Temperature is kept constant throughout. 
    Question 1296
    CBSEENPH11019568

    Write the expression for work done by gas when it expands isothermally.

    Solution

    For an isothermal expansion of gas, work done is given by, 
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#6 {main}</pre> 
    where, 
    R is the universal gas constant, 
    T is the temperature of the gas, 
    Vf volume to which the gas expands, and
    Vi initial volume of the gas. 

    Question 1297
    CBSEENPH11019569

    State first law of thermodynamics for isothermal process.

    Solution
    The first law of thermodynamics states that, when some quantity of heat (dQ) is supplied to the sytem, then the quantity of heat absorbed by the system (dQ) is equal to the sum of increase in the internal energy of the system (dU) due to rise in temperature and the external work done by the system (dW) in expansion. 
    i.e.,    dQ = dU + dW

    For isothermal process, 
    dQ = 0 + dW 
    This implies, 
                                 dQ = dW

    Question 1298
    CBSEENPH11019570

    What is the value of isothermal bulk modulus of gas at one atmospheric pressure?

    Solution
    The value of isothermal bulk modulus of gas at one atmospheric pressure is equal to atmospheric pressure.
    i.e. Kiso = 1.0x105 N/m2
    Question 1299
    CBSEENPH11019571

    What is an adiabatic process?

    Solution
    Adiabatic process is that during which the system does not exchange heat with surroundings. 
    Question 1300
    CBSEENPH11019572

    What is the equation of state of adiabatic process?

    Solution
    The equation of state for an adiabatic process is, 
                          PVγ = constant
    where, 

    straight gamma space equals space fraction numerator specific space heat space of space gas space at space constant space pressure over denominator specific space heat space of space the space gas space at space constant space volume end fraction
    Question 1301
    CBSEENPH11019573

    A substance of mass M kg requires an input power P to keep it in molten state. If the latent heat of fusion of substance is L, then how long it will take to solidify it completely?

    Solution
    If the power P is required to keep the substance in molten state, then the radiant power of substance is P.
    Let t be the time taken by substance to completely solidify the substance.
    The energy lost by substance is,
                              E = Pt
    The quantity of heat to be lost to solidify the substance is, 
                      Q = ML 
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#6 {main}</pre>          ML = Pt
    rightwards double arrow space space space space space space space space space space space space space space space space space space straight t space equals space ML over straight P, is the required time to solidify the substance completely. 
    Question 1302
    CBSEENPH11019574

    Write the expression for work done by gas during an adiabatic process.

    Solution

    Work done by gas during adiabatic process is, 
                      space space straight W space equals straight C subscript straight v left parenthesis straight T subscript straight i minus straight T subscript straight f right parenthesis space
space space space space space space equals space fraction numerator 1 over denominator 1 minus straight gamma end fraction left parenthesis straight P subscript straight f straight V subscript straight f minus straight P subscript straight i straight V subscript straight i right parenthesis space 

    Question 1303
    CBSEENPH11019575

    State first law of thermodynamics for adiabatic process.

    Solution
    For an adiabatic process heat transfer, dQ = 0.
    Therefore, 
    First law of thermodynamics becomes,
                     dU + dW = 0
    rightwards double arrow space space space space space space space space dU space equals space minus dW
    Question 1304
    CBSEENPH11019576

    What is isochoric process?

    Solution
    A process that takes place at constant volume is called isochoric process. 
    Question 1305
    CBSEENPH11019577

    What is the equation of state of isochoric process?

    Solution
    Equation of state for an isochoric process is given by,
                        P/T = constant 
    Volume remains constant in an isochoric process. 
    Question 1306
    CBSEENPH11019578

    Write the expression for work done by gas when in isochoric process.

    Solution
    The volume in an isochoric process remains constant. Therefore, the work done by gas is zero.
    Question 1307
    CBSEENPH11019579

    State first law of thermodynamics for isochoric process.

    Solution
    For an isochoric process, the volume of gas remains constant, Hence, work done on the gas is zero. 
    Therefore,
                                dQ = dU 
    Question 1308
    CBSEENPH11019580

    The water at the bottom of the Niagara Falls, which are 50 m high, should be warmer than that at the top. Explain why it is and calculate the temperature rise.

    The heat capacity of 1 mol water (which weighs 0.018 kg) is 80 J/K.

    The acceleration due to gravity is 9.8 m/s2.

    Solution
    Height of Niagara Falls, h = 50m
    Let m be the mass of waterfall.
    Decrease in potential energy of waterfall is,
                              U = mgh 
    U = 0.018 space cross times space 9.8 space cross times 50 space equals space 8.82 space straight J
    This potential energy is converted into heat.
    Let θ be the rise in temperature.
    Therefore,
    m g h space equals space m s theta space

rightwards double arrow space space straight theta equals fraction numerator m g h over denominator ms end fraction
space space space space space space space space space equals open parentheses fraction numerator 8.82 over denominator 0.018 cross times 80 end fraction close parentheses space cross times 0.018 space
space space space space space space space space space equals 0.11025 space to the power of degree straight C
    Question 1309
    CBSEENPH11019581

    What is isobaric process?

    Solution
    A process that takes place at constant pressure is called isobaric process.
    Question 1310
    CBSEENPH11019582

    What is the equation of state of isobaric process?

    Solution
    For an isobaric process, pressure remains constant. 
    Therefore, equation of state is given by, 
                            V/T = constant
    Question 1311
    CBSEENPH11019583

    What is cyclic process?

    Solution
    When a system returns to original thermodynamic state after undergoing a series of processes, then it is known as cyclic process.
    Question 1312
    CBSEENPH11019584

    Write the expression for work done by gas when in isobaric process.

    Solution
    Work done for expansion of gas in an isobaric process is given by,
                         W = P(Vf–Vi).
    Question 1313
    CBSEENPH11019585

    State first law of thermodynamic for isobaric process.

    Solution
    First law of thermodynamics for isobaric process is given by,
                      dQ = dU + P(Vf – Vi).
    Question 1315
    CBSEENPH11019587

    A gas absorbs Q1 heat while going from state A to B and releases Q2 heat while coming back to state A. How much work is done by gas in the process?

    Solution
    Heat absorbed = Q1 
    Heat released = Q2 
    Work done by gas in the process is given by,
                                     Q1– Q2.
    Question 1316
    CBSEENPH11019588

    What is the ratio of the adiabatic bulk modulus to isothermal bulk modulus of a perfect gas?

    Solution
    The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to γ. 
    Question 1317
    CBSEENPH11019589

    In a complete cycle of cyclic process, how the heat absorbed by the gas is used?

    Solution
    The heat absorbed by the gas is used to do work without changing any internal energy, in a complete cycle of the cyclic process. 
    Question 1318
    CBSEENPH11019590

    What is a reversible process? Give an example.

    Solution
    A reversible process is a process which can be made to proceed in two opposite directions with the same ease so that the system and the surroundings pass through exactly the same intermediate state as in the direct process.
     e.g. melting of ice.
    Question 1319
    CBSEENPH11019591

    What is an irreversible process? Give an example.

    Solution
    An irreversible process is a process which cannot be retraced in opposite direction exactly.
    e.g. rusting of iron.
    Question 1320
    CBSEENPH11019592

    Which of the following phenomenon is reversible?

    (a) Waterfall

    (b) Electrolysis.

    Solution
    Electrolysis is a reversible process.
    Question 1321
    CBSEENPH11019593

    The average number of degree of freedom of an ideal gas molecule is six. The gas absorbs 120J of heat at constant pressure. Find the increase in internal energy of gas.

    Solution
    The number of degree of freedom of gas is,
                                 f = 6 
    Therefore the value of y of gas  is,
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#6 {main}</pre>
    When the gas is heated at constant pressure, the fraction of heat converted into energy is space space 1 over y. 
    Thus, 
    space space space space space space space space space space space space space space space fraction numerator increment straight U over denominator straight Q end fraction equals 1 over straight y space

rightwards double arrow space space space space space space space space space space space space space increment straight U equals straight Q over straight y equals fraction numerator 120 over denominator 4 divided by 3 end fraction equals 90 straight J
     
    Question 1322
    CBSEENPH11019594

    What do you mean by free expansion?

    Solution
    Adiabatic expansion against vacuum is known as the free expansion of a gas. 
    Question 1323
    CBSEENPH11019595

    What is the work done by gas during adiabatic expansion?

    Solution
    Zero, because there is no exchange of heat between the system and the surroundings. 
    Question 1324
    CBSEENPH11019596

    What is the change in internal energy of gas during free expansion?

    Solution
     Zero, because there is no transfer of heat between the system and the surroundings. 
    Question 1325
    CBSEENPH11019597

    A piece of copper is hammered. Does the internal energy of copper increase?

    Solution
    Yes, the internal energy of copper increases. when a piece of copper is hammered, the temperature is increased. Therefore, internal energy increases. 
    Question 1326
    CBSEENPH11019598

    An ideal gas is compressed adiabatically. Is there any change in internal energy?

    Solution
    When a gas is compressed, work is done on the gas. Therefore, the internal energy of gas will increase.
    Question 1327
    CBSEENPH11019599

    Three samples of same gas initially having same volume and at same pressure are expanded to double its initial volume. Sample A is expanded isothermally, B adiabatically and C isobarically. In which case maximum work is done?

    Solution
    The maximum work is done in the case of isobaric expansion. This heat has been gained by the gas from the surroundings. Hence, temperature of the gas remains constant. 
    Question 1328
    CBSEENPH11019600

    Three samples of same gas initially having same volume and at same pressure are expanded to double its initial volume, sample A is expanded isothermally, B adiabatically and C isobarically. In which case minimum work is done?

    Solution
    In an adiabatic expansion, the temperature of the gas would drop. therefore, work done is positive. Hence, W is positive. Therefore, the minimum work is done in the case of adiabatic expansion. 
    Question 1329
    CBSEENPH11019601

    Heat is supplied to the gas, but its internal energy does not increase. What is the process involved?

    Solution
    The internal energy of the gas does not increase. Therefore, the process involved is isothermal expansion.
    Question 1330
    CBSEENPH11019602

    In what type of process, the whole of the heat absorbed by system only increases the internal energy?

    Solution
    In an isochoric process, the whole of the heat absorbed by the system only increases the internal energy.
    Volume remains constant is an isochoric process. 
    Question 1331
    CBSEENPH11019603

    Is the gas equation PV=nRT valid for isothermal, adiabatic, isochoric and isobaric change?

    Solution
    Yes, the ideal gas equation is valid for all the processes. 
    Question 1332
    CBSEENPH11019604

    Name the forces in a system, which make the process taking place in it, irreversible in nature?

    Solution
    Frictional force in a system makes the process irreversible. 
    Question 1333
    CBSEENPH11019605

    State Carnot’s theorem.

    Solution
    Carnot theorem states that all reversible engines working between same two temperatures have same efficiency irrespective of the nature of working substance.
    The source and the sink works between the same temperature. 
    Question 1334
    CBSEENPH11019606

    What is Kelvin’s statement of second law of thermodynamic?

    Solution
    Kelvin's states that it is impossible to make a continuous supply of mechanical work by cooling a body to a temperature lower than that of coldest of it's surroundings. 
    Question 1335
    CBSEENPH11019607

    What is Clausius statement of second law of thermodynamic?

    Solution
    Clausius states that it is impossible for a self-acting machine, unaided by any external agency, to transfer heat from a body at a lower temperature to another body at a higher temperature. 
    Question 1336
    CBSEENPH11019608

    The average number of degree of freedom of an ideal gas molecule is five. The gas does 20J of work at constant pressure. Find the increase in internal energy and heat absorbed by the gas.

    Solution
    The number of degree of freedom of gas, f = 5
    Therefore the value of y of gas is,
                         straight y equals fraction numerator straight f plus 2 over denominator straight f end fraction equals 7 over 5 
    When the gas  is heated at constant pressure, the fraction of heat used to do work is  fraction numerator straight y minus 1 over denominator straight y end fraction.
    Thus, 
    space space space space space space space space space space space space space fraction numerator increment straight W over denominator increment straight Q end fraction equals fraction numerator straight y minus 1 over denominator straight y end fraction space

rightwards double arrow space space space space space space space space space space space space increment straight Q equals fraction numerator straight y over denominator straight y minus 1 end fraction increment straight W space

space space space space space space space space space space space space space space space space space space space space space space space equals 7 over 22 cross times 20 equals 70 straight J
 
    The  increase in the internal energy of gas is, 
    
increment straight U equals increment straight O minus increment straight W

space space space space space space equals 70 minus 20 equals 50 straight J       
    Question 1337
    CBSEENPH11019609

    Heat is supplied to one kg of solid material at a constant rate. The temperature of material is changing with heat supplied as shown in the figure below.


    (a) What do the horizontal regions AB and CD represent?

    (b) If CD = 9AB, then what interpretation can you make?

    (c) What does the reciprocal of slope of BC represent? 



    Solution

    (a) The melting of solid is represented by the horizontal portion AB and the magnitude of AB is equal to the latent heat of fusion in kcal/kg.
    The horizontal portion CD represents the boiling of liquid and the magnitude of CD is equal to latent heat of vaporisation in k cal/kg.

    (b) CD = 9AB implies that the latent heat of vaporisation is nine times the latent heat of fusion.

    (c) The curve BC represents the vapour state of substance.

    We know,
                    dQ = msdT
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#6 {main}</pre>
    So, the slope of the graph i.e. dQ over dT represents the heat capacity of the liquid. 

    Question 1338
    CBSEENPH11019610

    Mention three characteristics of ideal heat engine.

    Solution
    The three characteristics of ideal heat engine are it has a source and sink of infinite heat capacity and working substance, an ideal gas. 
    Question 1339
    CBSEENPH11019611

    Can heat engine completely convert heat into mechanical work?

    Solution
    No. The efficiency of the heat engine can never by cent percent. 
    In  a heat engine, thermal efficiency is always less than 100 %. This is because the heat is always rejected to the sink. That is Q2 not equal to space 0
    Question 1340
    CBSEENPH11019612

    Can the mechanical energy be converted completely into heat?

    Solution
    Yes. The mechanical energy can be converted completely into heat. 
    Question 1341
    CBSEENPH11019613

    On what factors the efficiency of an ideal heat engine depends?

    Solution
    The efficiency of an ideal heat engine only depends on the temperature of source and sink. Efficiency is always less than 1 because some heat is always rejected to the sink. 
    Question 1342
    CBSEENPH11019614

    An engine working under isothermal conditions can produce no useful work. Explain.

    Solution
    An engine working under the isothermal conditions can produce no useful work because the efficiency of the engine working between the same temperature is zero. 
    Question 1343
    CBSEENPH11019615

    How much quantity of water at 20°C is required to just melt 800gm of ice at –12°C?

    The specific heat of ice is 0.5cal/gm/K.  

    Solution

    Let m be the mass of water required to just melt the ice.
    By law of calorimetry,
    Heat lost by water = Heat gained by  ice
    Therefore, 
      m x l x 20 = 800 x 0.5 x 12 +800 x 80
    rightwards double arrow      20m = 68800
    rightwards double arrow         m  = 3440gm
                      = 3.44 kg 

    Question 1344
    CBSEENPH11019616

    How efficiency of Carnot’s engine is affected by the nature of working substance?

    Solution
    The efficiency of Carnot’s engine does not depend on the nature of working substance.
    Efficiency depends on the temperature of source T1 and temperature of sink T2
    Question 1345
    CBSEENPH11019617

    What type of process is Carnot’s cycle?

    Solution
    Carnot’s cycle is a reversible cyclic process. 
    Question 1346
    CBSEENPH11019618

    What is heat pump? Give an example.

    Solution
    The heat pump is a device that removes the heat by doing mechanical work. Heat engine converts the heat energy into mechanical work. 
    e.g. refrigerator.
    Question 1347
    CBSEENPH11019619

    Explain how heat engine is different from refrigerator.

    Solution
    Heat engine converts heat energy into mechanical energy.
    Refrigerator transfers heat from cold body to hot body. An ideal refrigerator can be considered as a Carnot's ideal heat engine working in the reverse direction. 
    Question 1348
    CBSEENPH11019620

    In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. For this, steam at 160°C from boiler is passed in the coil dipped in water. The steam condenses and returns back to boiler at 95°C. How many kg of steam are required per hour? Take specific heat of steam 2 cal/gm/K.

    Solution
    Let m gm of steam be required per hour to raise the temperature of water.
    The formula of calorimetry is given by, 
    Q = straight m cross times straight C cross times increment straight t
    where, 
    q is the amount of heat transferred, 
    c is the specific heat of the substance, 
    m is the mass of the substance, and 
    increment straight t is the temperature change. 
    increment straight t = Tf - Ti
    Now, according to the law of calorimetry, we have
    Heat lost by steam = Heat gained by water 
    m x 2 x 60 + m x 540 + m x 1 x 5 = 10,000 x 1 x 60 
    rightwards double arrow      665 m = 60, 000 
    rightwards double arrow             m = 902.26 gm, is the mass of steam required per hour. 
    Question 1349
    CBSEENPH11019621

    A copper ball weighing 400gm is takem from a furnace to 1000gm of water at 20°C. The temperature of water increases to 80°C at equilibrium. Find the temperature of furnace. The specific heat of copper is 0.1 cal/ gm/K and that of water is 1.0 cal/gm/K.

    Solution

    Weight of the copper ball = 400 g
    Weight of water = 1000 gm 
    Temperature = 20o
    Increased temperature = 80o C
    Let T be the temperature of furnace.
    According to law of calorimetry,

    Heat lost by copper = Heat gained by water.
    That is,
    400 x 0.1 x ( T– 80 ) = 1000 x 1 x 60
    rightwards double arrow                   T - 80 = 1580
    rightwards double arrow                          T = 15800

    Question 1350
    CBSEENPH11019622
    Question 1351
    CBSEENPH11019623

    State zeroth law of thermodynamics.

    Solution
    Zeroth law of thermodynamics states that when the thermodynamic systems A and B are separately in thermal equilibrium with a third thermodynamic system C, then the systems A and B are in thermal equilibrium with each other also.
    Question 1352
    CBSEENPH11019624

    A lead bullet just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by obstacle, find the velocity of bullet. Initial temperature of bullet is 27°C, the melting point of lead is 327°C, the specific heat of lead is 0.03 cal/gm/K and latent heat of fusion of lead is 6 cal/gm.

    Solution
    Let m be the mass of bullet and v its velocity.
    So,
    Kinetic energy of bullet is 1 halfmv2.
    On stopping the bullet by obstacle, the kinetic energy of bullet is converted into heat.
    25% of heat is absorbed by obstacle and remaining 75% of heat is absorbed by bullet.
    Thus heat absorbed by bullet is, 
                  straight Q equals 3 over 4 open parentheses 1 half mv squared close parentheses
    This heat first increases the temperature of bullet and then melts it.
    Therefore
    space space space space space space space 3 over 4 open parentheses 1 half mv squared close parentheses equals msθ plus mL
rightwards double arrow space space space space space space space space space space space space space space space space space space space space space straight v space equals space square root of 8 over 3 left parenthesis sθ plus straight L right parenthesis end root
    Here, we have
    Specific heat of lead, s = 0.03 cal/gmK=30 x 4.2 J/kg/K
    Latent heat of fusion, L = 6 cal/gm = 6000 cross times4.2 J/kg
    Therefore, velocity of the bullet is given by, 
               
       straight v equals square root of 8 over 3 left parenthesis 30 cross times 4.2 cross times 300 plus 6000 cross times 4.2 right parenthesis end root 
         = 409.9m/s 
    Question 1353
    CBSEENPH11019625

    An ice cube falls from a height of 1500m. Assuming that whole of potential energy of ice cube converts into heat, find the fraction of mass of ice that melts.

    Take g = 10m/s
    2and J = 4.2cal.

    Solution
    Let M be the mass of ice cube.
    The mass of ice that melts on falling = n
    The potential energy that converts into heat is given by, 
    E= Mgh = M x 1500 x 10 
                 = 15000M Joule 
    Heat required to melt m gram of ice is, 
    Q = mL = m x 80,000 cal 
                 = m x 80,000 x 4.2 Joule 
    Since whole of potential energy is converted into heat and used to melt the ice, therefore
                                    Q = E 
    rightwards double arrow   m x 80, 000 c 4.2 = 15, 000M
    rightwards double arrow space space space space straight m over straight M equals fraction numerator 15 comma 000 over denominator 80 comma 000 cross times 4.2 end fraction equals 0.045
    straight m over straight M is the fraction of ice that melts.
    Question 1354
    CBSEENPH11019626

    80 gm of ice at 0°C is mixed with 80gm of water at 80°C. Find the final temperature of the mixture.

    Solution
    Let T be the final temperature of the mixture.
    According to law of calorimetry,
    Heat gained by  ice = Heat lost by water
    Therefore, 
              space space space space space space space space straight m subscript straight i space straight L space equals space straight m subscript straight w space straight s left parenthesis straight T subscript straight w minus straight T right parenthesis
 
    rightwards double arrow       80 x 80 = 80 x 1 (80-T) 
    rightwards double arrow               80 = 80 - T
    rightwards double arrow                   T=0oC, is the final temperature of the mixture. 
    Question 1355
    CBSEENPH11019627

    Derive an expression for the work done by a gas undergoing expansion from volume V1 to V2.

    Solution
    Consider an ideal gas enclosed in a cylinder fitted with a massless and frictionless piston.
    Let A be area of the cross-section of piston.
    Let V be the volume and P be the pressure exerted by gas on the piston.
    The piston is kept in equilibrium by applying pressure P from outside. 
                      
    Let the applied pressure be decreased by infinitesimally small amount, so that the piston moves by infinitesimal distance dx.
    The amount of work done by the gas in infinitesimal expansion is, 
    dW space equals space straight F. dx space equals space straight P space straight A space dx space

space space space space space space space space equals space straight P space dV space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis     
              
    where,
    dV = Adx, the infinitesimal increase in volume.  
    Total work done by gas in expanding it from volume V1 to V2 can be obtained by integrating equation (1) from V1 to V2.
     i.e.,      straight W equals integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV, is the amount of work done. 
    Question 1356
    CBSEENPH11019628

    What are the sign conventions of heat, work and internal energy?

    Solution
    The sign conventions are as follows:
    i) When heat is supplied to a system, dQ is positive. When heat is drawn from the system, dQ is negative. 
    ii) When the gas expands, work is done by the gas (dW is positive). When gas is compressed, work is done on the gas (dW is negative). 
    iii) When the temperature of gas increases, it's internal energy increases, dU is taken as positive. When the temperature of gas decreases, internal energy decreases and dU is taken as negative. 
    Question 1357
    CBSEENPH11019629

    What is isothermal process? What are essential conditions for the isothermal process to take place?

    Solution
    When a thermodynamic system undergoes a process in such a way that its temperature remains constant, then the process is called isothermal.
    Essential conditions for isothermal process are: 
    i) The container should be perfectly conducting to the surroundings.
    ii) The process must be carried out very slowly so that there is sufficient time for exchange of heat with the surroundings so that temperature remains constant.
    Question 1358
    CBSEENPH11019630

    What is the equation of state for isothermal process? Plot P versus V graph and P versus T graph for isothermal process for an ideal gas. Discuss the isothermal process using first law of thermodynamics. 

    Solution
    An ideal gas equation is given by, 
                       PV = nRT 
    In an isothermal process, temperature T is constant.
    Therefore the equation of state for isothermal process reduces to, 
                        PV = constant 
    The graph of P versus V  and P versus T for an isothermal process is as shown below. 
     
    Now, according to first law of thermodynamics, 
    increment straight Q space equals space increment straight U space plus space straight P increment straight V space

For space an space isothermal space process comma space

space space space space increment straight U space equals space straight C subscript straight v increment straight T space equals space 0 space left square bracket Temperature space remains space constant right square bracket space

therefore space increment straight Q space equals space straight P increment straight V space equals space increment straight W space

In space an space isothermal space expansion comma space

space space space space space space increment straight V space greater than thin space 0 space

rightwards double arrow space increment straight Q space equals space straight P space increment straight V space greater than space 0 space

That space is comma space in space an space isothermal space expansion comma space
work space is space done space by space the space gas space and space heat space is space absorbed
space by space the space gas space equal space to space work space done space by space the space gas. space

In space isothermal space compression comma space

increment straight V space less than space 0 space

rightwards double arrow space increment straight Q space equals space straight P increment straight V space less than 0 space

That space is comma space in space isothermal space compression comma space work space is space done
on space the space gas space and space heat space is space released space by space gas space equal
to space work space done space on space the space gas. space
    Question 1359
    CBSEENPH11019631

    70 calories of heat are required to raise the temperature of two moles of an ideal gas by 5°C at constant pressure. Find the amount of heat required to raise the temperature of same gas through 5°C at constant volume.

    Solution

    Given,
    Heat required to raise the temperature, Q= 70 cal
    Number of moles of gas, n = 2 mole  
    Increase space in space temperature comma space increment straight T equals 5 degree straight C space

Now comma
space space space space space space space space space space space space space space space space space straight Q subscript straight P equals nC subscript straight P increment straight T space

rightwards double arrow space space space space space space space space space space space space space space 70 equals 2 cross times straight C subscript straight P cross times 5 space
           
    rightwards double arrow           CP = 7 cal/mol/K
    Therefore,
                    C= 5 cal/mol/K 
    Therefore, the quantity of heat required to increase the temperature at constant volume is, 
    space space space space space space space straight Q subscript straight v equals nC subscript straight v increment straight T equals 2 space cross times space 5 space cross times space 5 space equals space 50 space cal 

    Question 1360
    CBSEENPH11019632

    The internal energy of ideal gas varies with temperature according to relation U = αT. Find the specific heat of gas at constant pressure.

    Solution
    The specific heat of the gas at constant pressure is given by, 
    space space space space space space space space space space straight C subscript straight P equals dU over dT plus straight R space

Here comma space space straight U equals αT space

therefore space space space dU over dT equals straight alpha space

Thus comma space space space space straight C subscript straight P equals straight alpha space plus space straight R 
    CP is the specific heat of gas at a constant pressure. 
    Question 1361
    CBSEENPH11019633

    Derive the expression for the work done by the gas during isothermal expansion.

    Solution
    Work done by or on the system is given by, 
    straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript straight P. space dV space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

The space ideal space gas space equation space is comma space

space space space space space space space space space space space space space space space PV space equals space nRT space

rightwards double arrow space space space space space space space space space space space space space straight P space equals space nRT over straight V space space space space space space... left parenthesis 2 right parenthesis space

Putting space left parenthesis 2 right parenthesis space in space left parenthesis 1 right parenthesis comma space we space get space

straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript nRT over straight V space dV space

For space an space isothermal space process comma space straight t space is space constant. space

therefore space space space straight W space equals space nRT space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript 1 over straight V. space dV space

space space space space space space space space space space space space equals space nRT space left square bracket space Ln space left parenthesis straight V subscript 2 right parenthesis space minus space Ln left parenthesis straight V subscript 1 right parenthesis right square bracket space

rightwards double arrow space space space space space space straight W subscript isothermal space equals space nRT space open square brackets Ln space straight V subscript 2 over straight V subscript 1 close square brackets
                              
    Question 1362
    CBSEENPH11019634

    What is adiabatic process? What are essential conditions for the adiabatic process to take place?

    Solution
    Any process in which no heat is allowed to be exchanged with surroundings is called an adiabatic process.
    Essential conditions for the adiabatic process to take place are: 
    i) System is perfectly insulated from surrounding.
    ii) The process must be carried out rapidly so that the system hs sufficient time to exchange heat with the surroundings. 
    Question 1363
    CBSEENPH11019635

    What do you mean by phase change of matter?

    Solution
    Phase change is a process accompanied by the emission or absorption of heat without any temperature change.
    Question 1364
    CBSEENPH11019636

    What is fusion curve?

    Solution
    Fusion curve is the locus of the melting point of solid at varying pressure. 
    The curve on the phasor diagram which represents a transition between the solid and the liquid states.  
    Question 1365
    CBSEENPH11019637

    What is vaporisation curve?

    Solution
    Vaporisation curve is the graph between the boiling point of substance in the liquid state at varying pressure. 
    The curve on the phasor diagram which focuses on the point of transition between the liquid and gaseous state. 
    Question 1366
    CBSEENPH11019638

    What is sublimation curve?

    Solution
    Sublimation curve represents the temperature at which solid directly converts into vapours without following the liquid phase at a different pressure.
    Question 1367
    CBSEENPH11019639

    What is the fusion curve, vaporisation curve and the sublimation curve are called for water?

    Solution
    Fusion curve is called ice curve,
    Vaporisation curve is called steam curve, and
    Sublimation curve is called hoarfrost curve respectively. 
    Question 1368
    CBSEENPH11019640

    At melting point, in which phase does the matter exists?

    Solution
    At melting point, the matter may exist in solid phase alone, in liquid phase alone or may coexist in both solid and liquid phases simultaneously. 
    Question 1369
    CBSEENPH11019641

    Is it possible to have liquid and vapours of a substance at same temperature?

    Solution
    Yes, at the boiling point of substance, the liquid phase and vapour phase are at the same temperature.
    Question 1370
    CBSEENPH11019642

    If the ice at unity atmospheric pressure and 0°C temperature is suddenly compressed, then what would happen?

    Solution
    The whole of the ice will melt if it is compressed at unity atmospheric pressure and at a temperature of 0o C.
    Question 1371
    CBSEENPH11019643

    Define triple point.

    Solution
    Triple point is that value of temperature and pressure at which all the three phases of matter coexist simultaneously. 
    Question 1372
    CBSEENPH11019644

    Is triple point a unique point?

    Solution
    Yes, triple point is a unique point where the three phases of matter coexist. 
    Triple point do not vary in accordance with pressure or temperature. 
    Question 1373
    CBSEENPH11019645

     If the pressure on the ice at triple point were decreased, then what would happen?

    Solution
    At triple point, if the pressure on the ice is decreased, then the whole of the ice convert directly into vapours without following water phase. 
    Question 1374
    CBSEENPH11019646

    Is it possible that there is the change in temperature of the system without giving heat to or taking the heat from it?

    Solution
    Yes, in an adiabatic process, there is change in temperature of the system without giving any heat to the surroundings or taking heat away from it. 
    Question 1375
    CBSEENPH11019647

    Is it possible to increase the vapour pressure beyond saturated vapour pressure at a given temperature?

    Solution
    No it is not possible.
    The saturated vapour pressure is the maximum possible pressure at a given temperature.
    Question 1376
    CBSEENPH11019648

    Derive the equation of state for adiabatic process. Plot P versus V graph for the process.

    Solution
    According to the first law of thermodynamics,
                         dQ space equals space dU plus dW 
    In an adiabatic process, no heat is allowed to exchange between the system and surrounding, therefore dQ = 0.
    Thus, 
    space space space space space space space dU space plus space PdV space equals space 0 space

rightwards double arrow space straight C subscript straight P dT space plus space PdV space equals space 0 space
rightwards double arrow space straight C subscript straight V dT space plus space PdV space equals space 0 space

That space is comma space

straight C subscript straight V dT space plus space RT over straight V dV space equals space 0 space left square bracket because space PV space equals space RT right square bracket space

That space is comma space

dT over straight T plus straight R over straight C subscript straight p dV over straight V space equals space 0 space

rightwards double arrow space dT over straight T space plus space left parenthesis straight gamma space minus 1 right parenthesis space dV over straight V space equals space 0 space

On space integrating comma space

space space space space integral dT over straight T space plus space space left parenthesis straight gamma space minus 1 right parenthesis space integral dV over straight V space equals space constant

rightwards double arrow space Ln space straight T space plus space space left parenthesis straight gamma space minus 1 right parenthesis space Ln space straight V space equals space constant

rightwards double arrow space TV to the power of straight gamma space minus space 1 end exponent space equals space constant space comma space which space is space
the space equation space of space state. space

Since comma space PV space equals space space RT space

we space have comma space straight T space equals space PV over straight R space

therefore space TV to the power of straight gamma space minus space 1 end exponent space equals space PV over straight R straight V to the power of straight gamma space minus space 1 end exponent space equals space constant

rightwards double arrow space PV to the power of straight gamma space equals space constant
    P vs. V graph for an adiabatic process is as shown below. 
                          
                     
       
    Question 1377
    CBSEENPH11019649

    What is principle of pressure cooker?

    Solution
    The pressure cooker works on the principle that boiling point of water increases with increase in pressure.
    Question 1378
    CBSEENPH11019650

    Can the vapours be liquefied at any temperature by applying pressure alone?

    Solution
    The vapours can only be liquefied by applying the pressure only if the temperature of vapours is less than the critical temperature.
    Question 1379
    CBSEENPH11019651

    What is the critical temperature of water?

    Solution
    The critical temperature of water is 374.1°C.
    Question 1380
    CBSEENPH11019652

    How does the boiling point of liquids change with increase in pressure?

    Solution
    With the increase in pressure, the boiling point of the liquid increases. 

    Question 1381
    CBSEENPH11019653

    How does the melting point of solids change with increase in pressure?

    Solution
    With increase in pressure, the melting point of solids may increase or decrease. 
    Question 1382
    CBSEENPH11019654

    How does the melting point of ice change with increase in pressure?

    Solution
    When pressure is increased, the melting point of ice decreases. 
    Question 1383
    CBSEENPH11019655

    How does the melting point of solid CO2 change with increase in pressure?

    Solution
    When pressure increases, the melting point of solid COincreases. 
    Question 1385
    CBSEENPH11019657

    Derive an expression for the work done in adiabatic process.

    Solution
    Consider a unit mole of gas contained in a perfectly non-conducting cylinder provided with a non-conducting and frictionless piston.
    Let Cv be the specific heat of gas at constant volume.
    Let at any instant, when the pressure of gas is P, the gas be compressed by small volume dV.
    Then work done on the gas is, 
                              dW = PdV 
    Total work done on gas to compress from volume V1 to V2 is given by 
         integral dW space equals space straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript straight P space dV                   ...(1) 
    According to first law of thermodynamics, 
                  dQ space equals space dU plus PdV 
    For adiabatic process,  dQ space equals space 0 
    ∴           PdV = -dU = negative straight C subscript straight v dT 
    ∴   straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV space equals space integral subscript straight T subscript 1 end subscript superscript straight T subscript 2 end superscript minus straight C subscript straight v dT 
    where,
    T1 is the temperature of gas when volume is V1 and T2 when volume is V2
    Thus, work done is given by, 
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#7 {main}</pre> 
            equals negative straight C subscript straight v left parenthesis straight T subscript 2 minus straight T subscript 1 right parenthesis space equals space straight C subscript straight v left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis
            equals fraction numerator straight R over denominator straight gamma minus 1 end fraction left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis space equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis RT subscript 1 minus RT subscript 2 right parenthesis
equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis straight P subscript 1 straight V subscript 1 minus straight P subscript 2 straight V subscript 2 right parenthesis 
    Therefore,
    space straight W subscript adi space equals space straight C subscript straight v left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis space equals space fraction numerator straight R over denominator straight gamma minus 1 end fraction left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis
space space space space space space space space space space space equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis straight P subscript 1 straight V subscript 1 minus straight P subscript 2 straight V subscript 2 right parenthesis 
    The above expression gives us the amount of work done in adiabatic process. 
    Question 1386
    CBSEENPH11019658

    Show that slope of adiabatic curve is γ times the slope of isothermal process.

    Solution
    The equation of state of an isothermal process is, 
                          PV = constant 
    ∴      PdV plus VdP equals 0 
    rightwards double arrow             dP over dV equals negative straight P over straight V 
    ∴   Slope of isothermal curve is, 
      space space straight S subscript straight i space equals space dP over dV equals negative straight P over straight V                        
    For adiabatic process the equation of state is, 
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    Therefore, 
                    straight P left parenthesis γV to the power of straight gamma minus 1 end exponent dV right parenthesis plus straight V to the power of straight gamma dP equals 0 
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    ∴    Slope of adiabatic curve is straight S subscript straight a equals dP over dV equals negative straight gamma straight P over straight V 
    On comparing the slopes of two curves, we get
                                 straight S subscript straight a equals γS subscript straight i 
    Therefore, the slope of adiabatic curve is straight gamma times the slope of isothermal process. 

    Question 1387
    CBSEENPH11019659

    What is the effect of decrease of pressure on melting point of CO2 and ice?

    Solution
    When pressure decreases, the melting point of CO2 decreases and that of ice increases. 
    Question 1388
    CBSEENPH11019660

    Two gram of solid CO2 is in equilibrium with three gram of CO2 gas. What will be the effect of increase of pressure on the equilibrium?

    Solution
    When the pressure is increased in equilibrium, three gram of CO2 gas will solidify. 
    Question 1389
    CBSEENPH11019661

    What is the difference in the fusion curve of water and CO2?

    Solution
    The slope of fusion curve of water is negative whereas the slope of fusion curve of CO2 is positive. 
    Question 1390
    CBSEENPH11019662

    Distinguish between evaporation and boiling.

    Solution
    Evaporation  Boiling
    1. The process takes place at all temperature.  1. Boiling occurs
    only at boiling temperature.
    2. The process usually occurs on the surface of the matter.  2. Boiling occurs below the surface. that is, it is a bulk phenomenon. 
    Question 1391
    CBSEENPH11019663

    What is absolute humidity?

    Solution
    Absolute humidity is the amount of water that becomes vapours which is actually present in a cubic meter of air.  
    Question 1392
    CBSEENPH11019664

    What is the unit of absolute humidity?

    Solution
    The unit of absolute humidity is gm/m3
    That is gram per m3 is the unit. 
    Question 1393
    CBSEENPH11019665

    Define relative humidity.

    Solution
    The ratio of the mass of water vapours present in a given volume of air to the mass of water vapours required to saturate the same volume of air at the same temperature is known as relative humidity. 
    Question 1394
    CBSEENPH11019666

    Define relative humidity in terms of pressure.

    Solution
    Relative humidity is defined as the ratio of vapour pressure (p) actually present in the air at room temperature to the saturated vapour pressure (P) at same temperature.
    Mathematically, the formula is given by, 
                      bold space bold space
bold space bold space bold space bold space bold space space straight R. straight H. equals straight P subscript actual over straight P subscript saturated
    Question 1395
    CBSEENPH11019667

    What are the dimensions of relative humidity?

    Solution
    Relative humidity is a quantity which is the ratio of pressure.
    Therefore it is a dimensionless quantity. 
    Question 1396
    CBSEENPH11019668

    What is the unit of relative humidity?

    Solution
    Relative humidity has no unit because it is a ratio of pressure. 
    Question 1397
    CBSEENPH11019669

    Define dew point.

    Solution
    Dew point is the temperature at which water vapours present in a certain volume of air is sufficient to saturate it.
    Question 1399
    CBSEENPH11019671

    Comment on the statement:

    Actual vapour pressure at room temperature is equal to saturated vapour pressure at dew point.

    Solution
    The statement is correct as the actual vapour pressure and saturated vapour pressure is same at the dew point. 
    Question 1400
    CBSEENPH11019672

    You have water vapours at temperature 273°C and 1 atmospheric pressure. Can you liquefy the water vapours at constant temperature by applying pressure alone?

    Solution
    Yes, the water vapours can be liquefied at constant temperature by applying pressure alone. This is because water vapours exist at a temperature lower than the critical temperature.
    Question 1401
    CBSEENPH11019673

    Does the heat supplied to the system always increase the internal energy only?

    Solution
    The heat supplied to the system increases the internal energy only during an isochoric process. 
    If the system undergoes isothermal or isobaric changes, then a part of heat is also used to do work. Hence, the increase in internal energy will be less than heat supplied to the system.
    Question 1402
    CBSEENPH11019674

    Air pressure in a car tyre increases during driving. Explain why?

    Solution
    Frictional force acts between the tyres of the car and the road. During driving, the work is done against the frictional force which is converted into heat. Therefore, the temperature and hence the pressure of the air in the tyres increases. 
    Question 1403
    CBSEENPH11019675

    Can ice coexist in equilibrium with water at 1 atmospheric pressure and –4°C?

    Solution
    No, ice cannot coexist in equilibrium with water at 1 atm and at a temperature of -4o C.
    The solid can be in equilibrium with liquid phase only at temperature and pressure corresponding to the melting point. The melting point of ice at 1 atmospheric pressure is 0°C. Therefore, ice at 1 atmospheric pressure and –4°C temperature cannot coexist in equilibrium with water. 
    Question 1404
    CBSEENPH11019676

    Distinguish between gases and vapours.

    Solution
    Gas is the vapours above the critical temperature.
    And, vapours are the gases below the critical temperature. 
    Gases cannot be liquefied by increasing pressure alone while vapours can be liquefied by increasing pressure only.
    Gases nearly follow perfect gas equation while vapours do not follow perfect gas equation. 
    Question 1405
    CBSEENPH11019677

    (a) Why is one’s breath visible in winter but not in hot summer?

    (b) How does water get cooled in earthen pots? 

    Solution

    (a) There are water vapours in the exhaled breath  which condense into small droplets of water in winter and become visible.
    But in summer the condensation does not take place. 

    (b) There are pores in the earthen pots and from these pores, water oozes out and evaporation takes place. Now, since cooling is caused after evaporation, therefore, water gets cooled down.

    Question 1406
    CBSEENPH11019678

    Although the fan throws hot air, even then it produces a sense of coolness in the body. Why?

    Solution
    The sweats that come out from pores of our body gets evaporated slowly. But the wind of the fan makes the process of evaporation faster. The quicker evaporation produces a greater cooling effect than the slower evaporation.
    Question 1407
    CBSEENPH11019679

    Wet clothes usually dry more quickly on a warm day than on a cold day. Why? 

    Solution
    The rate of evaporation depends on the difference between saturated vapour pressure (S.V.P.) and actual vapour pressure(V.P.) at a given temperature.
    More is the difference in the vapour pressure, greater is the rate of evaporation.
    On a hot day, relative humidity is lower than on a cold day and hence the difference of S.V.P. and V.P is more on a hot day as compared to on a cold day. Therefore, the rate of evaporation on a hot day is greater than that on a cold day. Also, when it is hot, the process of evaporation fastens up.
    Therefore, wet clothes dry more quickly on a warm day than on a cold day.
    Question 1408
    CBSEENPH11019680

    We add a piece of ice into water at 0°C. Will the ice melt? 

    Solution
    No, the ice will not melt. Here, both ice and water are at same temperature. Therefore, heat will not flow from water to ice. So, ice will not melt. 
    There should be a gradient in the temperature for the flow of heat to take place. 
    Question 1409
    CBSEENPH11019681

    1 kg of water is at triple point. What amount of water will be in solid, liquid and vapour phase?

    Solution
    There is no definite relation in the amount of water in different phases at the triple point.
    At triple point, the whole of 1 kg water can be in a single phase. The relative amounts of the three phases depend on  internal energy of water.
    For e.g. we can increase the amount of vapour phase of water by adding heat to the water .
    Question 1410
    CBSEENPH11019682

    What is isobaric process? What is its equation of state? Plot P versus V graph. Discuss the isothermal process using first law of thermodynamics.

    Solution

    An isobaric process is in which the pressure remains constant but volume and temperature are allowed to change.
    i.e.                   ∆P = 0
    Ideal gas equation is given by,
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#6 {main}</pre> 
    rightwards double arrow                   straight V equals nR over straight P straight T 
    For an isobaric process, P is constant.
    Therefore the equation of state for isobaric process reduces to,
                           straight V proportional to straight T 
    P versus V graph is as shown below. 
     
    Now, according to first law of thermodynamics, 
                       space space increment straight Q space equals space increment straight U plus straight P increment straight V
    Here, a part of heat absorbed by gas, increases the internal energy and a part of the heat is used to do work for expansion. 

    Question 1411
    CBSEENPH11019683

    What is super-heated water? Does this state lie on PVT surface?

    Solution
    Super-heated water is water in liquid state which is at a temperature higher than the boiling point of water at the given pressure.
    The superheated state of water is an unstable state. Therefore, it does not lie on the PVT surface. 
    Question 1412
    CBSEENPH11019684

    (a) What is an isochoric process? What is its equation of state? Plot P versus V graph. 

    (b) Discuss the isochoric process using first law of thermodynamics. 

    (c) What is the work done by or on the gas in isochoric process? 

    Solution
    (a) Isochoric process is the process in which volume of gas remains constant.
    i.e.,                   ∆V = 0.
    An ideal gas equation is given by, 
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#6 {main}</pre> 
    In an isochoric process, V is constant.
    Therefore the equation of state for isochoric process reduces to,
                           straight P proportional to straight T 
    P versus V graph is as shown below: 
              
    (b) According to first law of thermodynamics,  
                  space space increment straight Q space equals space increment straight U plus straight P increment straight V 
    Here,   increment straight V equals 0. 
    Therefore straight P increment straight V equals 0 
    So, the quantity of heat absorbed or rejected is equal to increase or decrease in energy respectively. 
    (c) The volume of gas is constant in the isochoric process.
    Therefore, the work done by or on the gas is zero. 
    Question 1413
    CBSEENPH11019685

    Derive the expression for the work done by the gas during isobaric expansion.

    Solution
    Work done by or on the system is given by, 
                    straight W equals integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV               ...(1) 
    We have, P is constant.
    Therefore, 
    straight W equals straight P integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript dV
space space space space equals space straight P left parenthesis straight V subscript 2 minus straight V subscript 1 right parenthesis space
    W the work done when the gas is expanded isobarically. 
    Question 1414
    CBSEENPH11019686

    What are super-cooled vapours? Are these vapours stable? Does this state lie on PVT surface?

    Solution
    Super-cooled vapours of a substance are the vapours at a temperature below its boiling point at given pressure. These vapours are not stable and, therefore, the state does not lie on PVT diagram.
    Question 1415
    CBSEENPH11019687

    (a) The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points?

    (b) There were two fixed points in the original Celsius scale i.e. 0°C and 100°C. On the absolute scale one of the fixed point is the triple point of water. What is the other fixed point on Kelvin scale?

    (c) Absolute temperature and temperature on Celsius scale are related as

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#6 {main}</pre>

    Why do we have 273.15 in the relation and not the 273.16?

    Solution

    (a) The triple point of water is a unique temperature and does not change with pressure.
    The melting point and boiling point of water vary with pressure.
    That is, there is no fixed melting and boiling point of water.

    (b) The second fixed point on the absolute scale is zero on the absolute scale.

    (c) We have 273.15 in the relation, because the triple point of water is 0.01 °C.

    Question 1416
    CBSEENPH11019688

    The molar specific heat of hydrogen is 3/2R in lower range of temperature and 5/2 R in the range of temperature about 750K and 7/2 R in higher range of temperature; what do you think is happening?

    Solution
    Given, 
    Molar specific heat of hydrogen in the lower range of temperature = 3/2 R 
    Molar specific heat of hydrogen in the high range of temperature = 5/2 R 
    Molar specific heat of hydrogen in the higher range of temperature = 7/2 R

    We can see here that in the lower range of temperature, the hydrogen gas has only three degrees of freedom corresponding to translational motion.
    As the temperature increases in the range of 250K to 750K, the diatomic gas has both translational and rotational motion. Hence, hydrogen gas has five degree of freedom.
    At further higher temperature degree of freedom further increases by two due to vibration motion. the total degree of freedom at a much higher temperature becomes 7. 
    Question 1417
    CBSEENPH11019689

    Explain why water is produced on rubbing two blocks of ice together?

    Solution
    When two blocks of ice are rubbed together then there arises friction between the two blocks. The mechanical energy is converted into heat energy which melts the ice. Hence, water is produced on rubbing two blocks of ice. 
    Question 1418
    CBSEENPH11019690

    A vessel with a movable piston maintained at a constant temperature by a thermostat contains a certain amount of liquid in equilibrium with its vapours. Do these vapours obey Boyle’ s law? In other words, what happens when volume of vapours is decreased? Does the vapour pressure increase?

    Solution
    Here, the liquid and vapours are in equilibrium.
    Therefore, the vapours are saturated vapours and pressure of vapours is equal to saturated vapour pressure at that temperature.
    Saturated vapours do not obey Boyle’s law. Saturated vapour pressure is the maximum pressure and when volume is decreased at saturation, the vapours condense maintaining the same pressure.
    Question 1419
    CBSEENPH11019691

    Define bulk modulus. Find the bulk modulus of gas for isothermal, adiabatic, isobaric and isochoric processes.

    Solution
    The bulk modulus of gas is defined as the ratio of the change in pressure of the gas to the volume strain in it.
    It is given by, 
    straight K space equals space minus space straight V space dP over dV

For space isothermal space process comma

space PV space equals space constant.

So comma space dP over dV equals space minus straight P over straight V

therefore space straight K space equals space minus space straight V space dP over dV space equals space straight P space

For space adiabatic space process comma space

PV to the power of straight gamma space equals space constant

So comma space

space space space space space space dP over dV space equals space minus straight gamma space straight P over straight V

therefore space straight K space equals space minus straight V space dP over dV space equals space γP space

For space isobaric space process comma space

straight P space equals space constant

That space is comma space

dP over dV space equals space 0

therefore space straight K space equals space minus straight V space dP over dV space equals space 0 space
For space isochoric space process comma space

straight V space equals space constant

rightwards double arrow space dP over dV space equals space infinity space

therefore space straight K space equals space minus straight V space dP over dV space equals space infinity space

               
    Question 1420
    CBSEENPH11019692

    Is it possible to increase the temperature of gas without adding heat to it? Explain.

    Solution
    If gas is compressed adiabatically, the work done on the gas is converted into heat. Hence,  the temperature of gas increases. 
    Question 1421
    CBSEENPH11019693

    If gas loses heat without doing any work, then what type of process it is?

    Solution
    Gas is not doing any work, therefore the volume of gas remains constant. Since the volume remains constant, this is an isochoric process. Therefore, the gases loses heat and so the temperature of the gas will fall.
    Question 1422
    CBSEENPH11019694

    Find the relative humidity on a day when partial vapour pressure of water vapours present in the room is 0.014x105 Pa. The saturated vapour pressure at room temperature is 0.02xl05 Pa.

    Solution
    Here partial vapour pressure of water vapours at room temperature is, 
    p = 0.014 x 10Pa 
    The saturated vapour pressure at room temperature is, P = 0.02 x 10Pa 
    Now, the relative humidity is, 
    P e r c e n t a g e space straight R. straight H. space equals straight P over straight P cross times 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 0.014 cross times 10 to the power of 5 over denominator 0.02 cross times 10 to the power of 5 end fraction cross times 100 space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 70 percent sign
    Question 1423
    CBSEENPH11019695

    A cylinder filled with gas is placed in heatproof jacket. How will the temperature of gas be changed when the volume of gas is decreased slowly?

    Solution
    The slow compression of gas in a heatproof jacket is an example of adiabatic compression. 
    Work done on the gas is converted into internal energy. Hence, the temperature of the gas will increase. 
    Question 1424
    CBSEENPH11019696

    An ideal gas undergoes a process so that PVn = constant. If the molar heat capacity of the gas in the process is zero, then find the value of n.

    Solution
    Molar heat capacity of the process is zero.
    Therefore, no heat is absorbed or released during the process.
    Thus the process is an adiabatic process. 
    Here,      PV to the power of straight n space equals space constant 
    For adiabatic process,
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#6 {main}</pre> 
    Thus,            straight n equals straight gamma.
    Question 1425
    CBSEENPH11019697

    If the gas loses heat without doing any work, then what type of process is it?

    Solution
    Gas is not doing any work therefore the volume of gas remains constant during the process. Hence, the process is an isochoric process.
    Since the gas loses heat, therefore temperature of the gas will fall. 
    Question 1426
    CBSEENPH11019698

    A gas is compressed from volume V1 to V.2  In which case work done will be more-isothermal compression or adiabatic compression?

    Solution
    The P-V curve for isothermal and adiabatic compression is as shown in figure. 
                   
    We know that the area under PV graph is equal to work done on the gas.
    Thus, the work done on gas during adiabatic compression is maximum and isobaric compression is minimum. 
    Question 1427
    CBSEENPH11019699

    A gas expands from volume V1 to V2. In which case work done will be more: isothermal expansion, isobaric expansion or adiabatic compression?

    Solution
    The indicator diagram (i.e., P-V curve) for an isothermal, isobaric and adiabatic process is as shown below. 

                    
    Work done by the gas is represented by the area under the P-V graph.
    Thus, from the above graph, work done by the gas during isobaric expansion is maximum and adiabatic expansion is minimum. 
    Question 1428
    CBSEENPH11019700

    Air in a cylinder, is suddenly compressed to half of its initial volume by a piston which is then maintained at that new position. What happens to the pressure of gas with the passage of time?

    Solution
    Sudden compression is an adiabatic compression.
    Thus, just after compression, the pressure of gas increases to 2γP which is greater than the initial pressure. The temperature increases to 2γ–1 T, which is greater than T. 
    Here, 
    P is the initial pressure, and 
    T is the initial temperature. 
    With the passage of time, the temperature of air starts decreasing due to loss of heat to the surrounding. This results in the decrease of pressure at constant volume. Finally the pressure of air decreases to 2P. 
    Question 1429
    CBSEENPH11019701

    What is the relative humidity of air when the room temperature is 16°C and the dew point is 7.4°C? Given that the saturated vapour pressure at 7°C, 8°C and 16°C is 7.5mm, 8mm and 13.5mm respectively of mercury.

    Solution

    Given,
    Room temperature, T = 16°C

    Dew point temperature, T0 = 7.4°C

    SVP at 7°C = 7.5mm of Hg

    SVP at 8°C = 8mm of Hg

    SVP at 16°C = 13.5mm of Hg

    For a small change in temperature, it can be assumed that saturated vapour pressure changes linearly.
    The saturated vapour pressure increases by 0.5 mm for one degree rise in temperature.
    Therefore increase in saturated vapour pressure for 0.4°C increase in temperature will be 0.2mm of mercury.
    Hence, the saturated vapour pressure at dew point i.e. 7.4°C is 7.7 mm of mercury.
    Now the relative humidity is,
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    Question 1430
    CBSEENPH11019702

    What are the different processes by which heat is transferred?

    Solution
    The three different process by which heat is transferred are:
    i) conduction, 
    ii) convection, and 
    iii) radiation. 
    Question 1431
    CBSEENPH11019703

    How the heat is transferred from one place to another place by the process of conduction?

    Solution
    Conduction is the phenomenon of transfer of heat through one part of the body to another, from particle to particle in the direction of fall of temperature without any actual movement of the particles. 
    Question 1432
    CBSEENPH11019704

    What is the convection process of transferring the heat?

    Solution
    Convection is the phenomenon of transfer of heat with the actual movement of the particles of the body. 
    For example, when the liquid in a flask is heated, the particles of the liquid at the bottom gets heated, become lighter and actually rise up.
    Question 1433
    CBSEENPH11019705

    How the heat is transferred by means of radiations?

    Solution
    Radiation is the phenomenon of transfer of heat from the source to the receiver directly without any actual movement of the source or receiver. Also, heating of the intervening medium is not required. 
    For example, heat from the sun comes to us through through radiation. 
    Question 1434
    CBSEENPH11019706

    Give an example of transfer of heat by means of radiations.

    Solution
    The transfer of heat from the sun to the earth happens by means of radiation process. 
    Question 1435
    CBSEENPH11019707

    Define coefficient of thermal conductivity.

    Solution
    Coefficient of thermal conductivity of a solid is equal to the rate of flow of heat per unit area per unit temperature gradient across the solid. 
    Question 1436
    CBSEENPH11019708

    Write the unit and dimensions of thermal conductivity.

    Solution
    The S.I. unit of thermal conductivity is Wm–1K–1 
    C.G.S unit of thermal conductivity is Cal m–1s–1°C-1 
    The dimensional formula is [M1L1T–3K–1] . 
    Question 1437
    CBSEENPH11019709

    What is thermometric conductivity or diffusivity?

    Solution
    The thermometric conductivity or diffusivity is the ratio of thermal conductivity to the thermal capacity per unit volume of material. 
    Question 1438
    CBSEENPH11019710

    What is thermal resistance?

    Solution
    Thermal resistance is the hindrance offered by a body to the flow of heat. 
    Mathematically, it is given by straight R space equals space fraction numerator increment straight x over denominator KA end fraction
    Thermal resistance is analogous to the electrical resistance. 
    Greater the thermal resistance, better will be the thermal insulation and poorer will be the thermal conduction. 
    Question 1439
    CBSEENPH11019711

    Is it possible to conduct the heat through gases?

    Solution
    Gases are bad conductors of heat. Hence, it is not possible to conduct heat through gases. 
    Question 1440
    CBSEENPH11019712

    In which state of matter the heat conducts most?

    Solution
    Solid phase conducts heat most. 
    Question 1441
    CBSEENPH11019713

    In which phase of matter, the heat is transferred by means of convection process?

    Solution
    Liquid and gas phase, transfer heat by the process of convection. 
    Question 1442
    CBSEENPH11019714

    Why are the bottoms of cooking pans painted black?

    Solution
    The colour black is a good absorber of heat. Therefore, the black painted pans absorb more heat and make cooking faster. 
    Question 1443
    CBSEENPH11019715

    Why are cooking utensils made of metals of high thermal conductivity?

    Solution
    The metals of high thermal conductivity conduct heat at a faster rate. Thus, they help in cooking the food quickly.
    Question 1444
    CBSEENPH11019716

    (a) What are the limitations of first law of thermodynamics? 

    (b) Two identical samples of a gas are expanded so that volume is increased to twice. One is expanded isothermally and second adiabatically. In which sample, pressure is greater? Explain.

    Solution

    Limitations of the First Law of Thermodynamics are: 
    (a) First law of thermodynamics does not tell us about the direction in which a given process can take place and up to what extent it can take place. 
    (b) Let P and V be the pressure and volume of gas.
    Let Pi be the pressure of gas when expanded isothermally.
    Pa be the pressure of gas when expanded adiabatically. 
    For space isothermal space process space space space space space space space space space space space space space space space space Adiabatic space process

space space space space space straight P subscript 1 straight V subscript 1 equals straight P subscript 2 straight V subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight P subscript 1 straight V subscript 1 to the power of straight gamma space equals space straight P subscript 2 straight V subscript 2 to the power of straight gamma 
    ∴     PV equals straight P subscript straight i 2 straight V                          ∴  PV to the power of straight gamma space equals space straight P subscript straight a left parenthesis 2 straight V right parenthesis to the power of straight gamma 
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#6 {main}</pre> 
    Here, we can see, straight gamma greater than 1. space 
    Therefore, straight P subscript straight i greater than straight P subscript straight a
    So, pressure is greater in the isothermal process. 

    Question 1445
    CBSEENPH11019717

    How does the wire gauge help in uniform heating?

    Solution
    The wire gauge made of metal is a good conductor of heat. Therefore, the metal spreads the heat quickly and helps in uniform heating. 
    Question 1446
    CBSEENPH11019718

    Are the woollens themselves the source of heat?

    Solution
    The woollens are bad conductors of heat. Hence, they are not a source of heat themselves. 
    Question 1447
    CBSEENPH11019719

    What are the characteristics of the material of cooking utensils?

    Solution
    The characteristics of the material of cooking utensils are:
    i) should have low specific heat, and
    ii) should have high thermal conductivity.
    Question 1448
    CBSEENPH11019720

    A piece of substance is heated to a high temperature and then cooled quickly. If the substance cracks, then what can be the possible cause of it?

    Solution
    If a substance cracks on heating it to a high temperature and then cooling suddenly, this implies that the substance is having a small value of thermal conductivity.
    Question 1449
    CBSEENPH11019721

    What is free (natural) convection?

    Solution
    Free convection takes place from bottom to top when liquids or gases are heated.
    Gravity is necessary for convection to take place.
    Question 1450
    CBSEENPH11019722

    What is force convection?

    Solution
    Forced convection is a mechanism or type of transport in which fluid motion is generated by an external source (like a pump, fan, suction device, etc.). It should be considered as one of the main methods of useful heat transfer. Using this method, significant amounts of heat energy can be transported very efficiently.
    The force other than gravity is required to carry the process of forced convection. 
    Question 1451
    CBSEENPH11019723

    Does the convection occur in solid?

    Solution

    No, convection only occurs in fluids but not in solids. 

    Question 1452
    CBSEENPH11019724

    In which phase of matter the heat can be transferred by the process of convection?

    Solution
    The heat can be transferred by the process of convection in liquid and gaseous phase of matter. 
    The process of convection does not occur in solids. 
    Question 1453
    CBSEENPH11019725

     In conduction and convection, how heat is passed on?

    Solution
    In conduction and convection, the transfer of heat takes place through the motion of particles. 
    Question 1454
    CBSEENPH11019726

    Which is the fastest mode of transfer of heat?

    Solution
    The rate of transfer of heat is fastest by the radiation process. Radiation process does not include any direct involvement of the source and the receiver. And also, heating of the intervening medium is also not required. Hence, it is the fastest process. 
    Question 1455
    CBSEENPH11019727

    Can free convection take place on heating the liquid in an artificial satellite?

    Solution
    No, because the effective value of gravity in the artificial satellite is zero. 
    Question 1456
    CBSEENPH11019728

    If air is a bad conductor of heat, why do we not feel warm without clothes?

    Solution
    Air is a bad conductor of heat, but it takes away some heat from our body by the process of convection. Hence, we feel cold.
    Question 1457
    CBSEENPH11019729

    Can water be boiled in a vessel made of thin paper?

    Solution
    Yes, water can be boiled in a vessel made of thin paper. The water takes heat from the burner by convection and so the paper is not affected at all.
    Question 1458
    CBSEENPH11019730

    What are thermal radiations?

    Solution
    The energy emitted in the form of radiation by a body due to its temperature is called thermal radiations. 
    Thermal radiations produce a sensation of warmth. 
    Question 1459
    CBSEENPH11019731

    What is the nature of thermal radiation?

    Solution
    Thermal radiations are electromagnetic in nature. The wavelength of thermal radiations ranges from 8 cross times 10 to the power of negative 7 end exponent space straight m space to space 4 cross times 10 to the power of negative 4 end exponent space straight m. They belong to the infrared region of the electromagnetic spectrum. That is why thermal radiations are electromagnetic in nature. 
    Question 1460
    CBSEENPH11019732

    Can heat propagate in vacuum?

    Solution
    Yes. Heat can be propagated in a vacuum. 
    Question 1461
    CBSEENPH11019733

    What is the velocity with which heat radiations propagate in vacuum?

    Solution
    Heat radiation propagates in vacuum with a velocity of 3x108 m/s.
    Question 1462
    CBSEENPH11019734

    What the term diathermancy means?

    Solution
    Diathermancy is a measure of transmitting power of a material. 
    Question 1463
    CBSEENPH11019735

    What are diathermanous substances?

    Solution
    Diathermanous substances allow the thermal radiation to pass through them easily.
    E.g. pure air, glass, rock salt etc. are diathermanous substances. 
    Question 1464
    CBSEENPH11019736

    What are athermanous substances?

    Solution
    Athermanous substances do not allow the thermal radiation to pass through them easily.
    For e.g. water vapours, wood, carbon dioxide etc. are athermanous substances.
    Question 1465
    CBSEENPH11019737

    Define a black body.

    Solution
    A black body is a body that does not reflect or transmit any radiation but absorbs all the radiations of all wavelengths incident on it. 
    Question 1466
    CBSEENPH11019738

    Is it possible to expand the gas at constant temperature without absorbing any heat?

    Solution
    It is not possible to expand the gas at a constant temperature without absorbing any heat. This is because either the gas does the work at the expense of internal energy or heat energy. Since the expansion takes place at a constant temperature, therefore the change in internal energy is zero.
    Thus work is done by gas by absorbing the heat.  
    Alternative answer: 
    According to first law of thermodynamics, 
             space space space space space space space increment straight Q space equals space increment straight U plus increment straight W 
    Since the gas expands at constant temperature, 
    ∴                space space space increment straight U space equals space 0 
    Hence,          increment straight Q space equals space increment straight W 
    The gas expands, therefore 
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#6 {main}</pre> 
    Hence the gas cannot expand at a constant temperature without absorbing any heat. 
    Question 1467
    CBSEENPH11019739

    Give an example of an ideal black body.

    Solution
    Fery’s cavity is an ideal black body. The cavity absorbs radiations of all wavelength. 
    Question 1468
    CBSEENPH11019740

    Is Newton’s law of cooling a special case of Stefan’s law?

    Solution
    Yes. Newton's law of cooling is a special case of Stefan's law. Newton's law of cooling is derived from Stefan's law. 
    Question 1469
    CBSEENPH11019741

    What is the value of solar constant?

    Solution
    The value of the solar constant is 1388 Watt. m-2.  
    Question 1470
    CBSEENPH11019742

    How can we measure high temperature like that of the sun and stars?

    Solution
    Pyrometer can be used to measure high temperature like that of sun and stars. 
    Question 1471
    CBSEENPH11019743

    What is the difference between rate of loss of heat and rate of cooling?

    Solution
    The rate of loss of heat is the quantity of heat lost by a body per unit time.
    The rate of cooling is the fall in temperature per unit time. 
    Question 1474
    CBSEENPH11019746

    A sphere, a cube and a thin circular plate, all made of the same material and having same masses are initially heated to 350°C. Which of these objects will cool fastest and which one slowest, when left in air at room temperature?

    Solution
    The circular plate will cool fastest and the sphere will cool the slowest. The rate of heat transfer is directly proportional to the surface area. And the circular plate has a greater surface area. Hence, circular plate will cool faster. 
    Question 1475
    CBSEENPH11019747

    What is the dimensional formula of Wien’s constant?

    Solution
    According to the Wien's displacement law, 
    straight lambda subscript straight m space equals space straight b over straight T 
    where, b is the constant of proportionality and is the Wien's constant. 
    Therefore, b will have the same dimensional formula as temperature. 
    So, dimensional formula of b is [M0L1T0K1]
    Question 1476
    CBSEENPH11019748

    What does the area under eλ v/s λ graph represent?

    Solution
    The area under eλ v/s λ graph represents the radiant emittance of the body. 
    Question 1477
    CBSEENPH11019749

    What do you mean by emissive power?

    Solution
    The emissive power of a body at certain temperature is defined as the total energy emitted per second per unit area of body at that temperature. 
    Question 1478
    CBSEENPH11019750

    Define emissivity.

    Solution
    The emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body (e) to the total emissive power of black body at the same temperature. 
    Emissivity is represented by straight capital epsilon

    Question 1479
    CBSEENPH11019751

    What is Kirchhoff’s law?

    Solution
    According to Kirchoff's law, the ratio of emissive power to the absorptive power corresponding to a particular wavelength of all the hot bodies at the same temperature is constant and is equal to emissive power of black body at equal temperature, corresponding to the same wavelength.
     i.e. straight e over straight a equals straight e subscript straight b = constant
    Question 1480
    CBSEENPH11019752

    Two samples of an ideal gas are compressed to half of the initial volume. One sample is compressed isothermally and second adiabatically.
    (a) In which sample pressure is high?
    (b) On which sample more work will be done?

    Solution
    (a) The pressure of gas increases to 2P on isothermal compression and that increases to 2yP on adiabatic compression. Thus the pressure of the gas is high when compressed adiabatically. 
                     
    (b) The work done on the gas is equal to the area under P–Vgraph.
    An adiabatic curve is steeper than the isothermal curve, therefore the area under adiabatic curve is greater than that of the isothermal curve.
    Hence work done on gas during adiabatic compression is more than isothermal compression. 
    Question 1481
    CBSEENPH11019753

    Who put forward the theory of heat radiations, that everybody emits heat radiation at every temperature.

    Solution
    The theory of heat radiations was put forward by Prevost in 1792. 
    Question 1482
    CBSEENPH11019754

    One mole of an ideal gas undergoes a cyclic change ABCD shown in the diagram. Find the net work done during the process. Take 1 atmospheric pressure = 105 N/m2.

    Solution
    Work done in a cyclic change is equal to the area of the cycle. 
    i.e. Work done, W = Area ABCDA 
                               = DC X AD 
    This implies, 
    space DC equals 3 space lt space equals space 3 cross times 10 to the power of negative 3 end exponent straight m cubed space

AD space equals space 3 Atm space equals space 3 cross times 10 to the power of 5 straight N divided by straight m squared
 
    Thus,
    Net work done, straight W equals 3 cross times 10 to the power of negative 3 end exponent cross times 3 cross times 10 to the power of 5 space equals space 900 straight J
    Question 1483
    CBSEENPH11019755

    How is heat transmitted from one body to another?

    Solution

    There are three ways by which heat is transmitted from one body to another: 

    (i) Conduction: This is the phenomenon in which thermal energy is transferred from hotter part of the body to colder part or vice versa. There is no contact without any transference of material particles.

    (ii) Convection: In this process, heat is transferred from one point to another point by actual movement of heated particles. 

    (iii) Radiation: In this process of transmission heat travels directly from one place to another place without heating any intervening media.

    Question 1484
    CBSEENPH11019756

    Explain the phenomenon of transfer of heat by means of conduction.

    Solution
    i) Conduction is the phenomenon of the flow of heat through one part of the body to another from particle to particle in the direction of fall of temperature without any actual movement of the particles. 
    ii) According to kinetic theory, the molecules of the body are in continuous oscillatory motion. When one end of the conductor is heated, the atoms/ molecules at the hot end begin to vibrate faster and collide the slow neighbouring atoms.
    iii) On colliding, the fast molecules transfer a part of their energy to slower molecules and put them in quick oscillations and thus show a rise in temperature.
    iv) In this way, the heat travels from atom to atom until it reaches the other end of the conductor.
    v) During the process of conduction, the atoms themselves do not move from one end to the other. 
    For example, all solids are heated through conduction. 
    Question 1485
    CBSEENPH11019757

    State laws of thermal conduction under steady state.

    Solution

    The laws of thermal conduction under steady state the rate of transfer of heat under steady state:

    (i) is directly proportional to temperature difference (T2 – T2) across the ends of conductor,

    (ii) is directly proportional to area of cross section A of the conductor and

    (iii) is inversely proportional to the length £ of the conductor.

    Combining all these laws, we get
    
space space space space space space dQ over dT proportional to fraction numerator open parentheses straight T subscript 1 minus straight T subscript 2 close parentheses straight A over denominator straight ell end fraction space

rightwards double arrow space dQ over dT equals straight K fraction numerator open parentheses straight T subscript 1 minus straight T subscript 2 close parentheses over denominator straight ell end fraction
    where K is constant of proportionality and is called thermal conductivity of conductor.

    Question 1486
    CBSEENPH11019758

    Define coefficient of thermal conductivity.

    Solution
    Thermal conductivity is defined as the quantity of heat that flows in one second through opposite faces of a cubical block of unit side which are maintained at the temperature difference of 1°C. 
    Question 1487
    CBSEENPH11019759

    What is unit and dimension of thermal conductivity?

    Solution

     Rate space of space change space of space heat space is space given space by comma space

space space space space space space space space space space space space space space space space space space space space dQ over dT equals straight K straight A over straight ell straight theta space

therefore space space space space Thermal space conductivity comma space straight K equals dQ over dt straight ell over Aθ
    S.I. unit of thermal conductivity is Watt/m/K or J/s/m/K.

    CGS unit of K is cal/cm/sec/K or ergs/cm/sec/K.

    Dimensional formula of thermal conductivity is [M1L1T–3K–1

    Question 1488
    CBSEENPH11019760

    Explain the terms:

    Transient state,

    Steady state, and

    Temperature gradient.

    Solution
    On heating one end of the bar by keeping the other end at lower fixed temperature, the heat flows along the length of the bar from hot end to cold end. 
                  
    Heat is received equally by all cross - sectional areas. 
    Let Q be the heat received from the adjacent cross-section on the hot side.

    This heat is used in following parts:

    (i) A part of total heat received is retained by the cross-section Qa.

    (ii) A part of heat (Q1) is transferred to next cross-section towards cold end side. As a result, there is the flow of heat. 

    bold italic Q bold italic equals bold italic Q subscript bold italic 1 bold italic plus bold italic Q subscript bold italic A bold italic equals bold italic Q subscript bold italic 1 bold italic plus bold italic Q subscript bold italic 2 bold italic plus bold italic Q subscript bold italic 3

    Since the temperature of various parts of the rod is increasing, the state is called transient state.

    Steady state: As the temperature increases, the rate of loss of heat also increases.
    After some time,
    Q3= Qand Q2 =0.
    At this stage, the temperature of cross section attains a constant value. This stage is called steady state. 

    Temperature gradient: At a steady state, the temperature at a given cross-section is constant.
    But, the temperature actually decreases along the length of the rod in a direction of flow of heat and is different in different directions.
    The rate of change of temperature with distance in the direction of heat flow is called temperature gradient.
    The temperature gradient at any point is represented by dθ/dx. 

    Question 1489
    CBSEENPH11019761

    What is thermometry conductivity or diffusivity? What does it measure?

    Solution

    In transient state, the rate at which the rod gets heated is not the same for all the materials.
    Heating depends not on the thermal conductivity but also depends on the thermal capacity per unit volume of material.

    The rate at which the temperature of any part of rod increases is called thermometric conductivity or diffusivity. Thermal conductivity is equal to the ratio of thermal conductivity to the thermal capacity per unit volume of material. 

    Question 1490
    CBSEENPH11019762

    What is thermal resistance of a conductor? Write its unit and dimensional formula.

    Solution
    Thermal resistance is the hindrance offered by a body to the flow of heat.
    Consider a conductor of length l, the area of cross-section A and thermal conductivity K. 
    When the two ends of conductor are kept at temperature difference straight theta, then the rate of flow of heat is given by,
                           dQ over dt equals KA over straight l straight theta

    i.e., for a given conductor, the rate of flow of heat is directly proportional to the temperature difference across the length of is conductance.
    The reciprocal of thermal conductance i.e. l/KA is called thermal resistance.

    The unit of thermal resistance is J–1 Ks.
    The dimensional formula is [M–1L–2T–3K1]

    Question 1491
    CBSEENPH11019763

    An ideal gas is taken through straight A rightwards arrow straight B rightwards arrow straight C rightwards arrow straight A as shown in figure. Find the work done by gas in the process straight C rightwards arrow straight A. What is the net heat absorbed by the gas in cyclic process?

    Solution
    The magnitude of area under CA curve on PV diagram is, 
    Area = Area of trapezium ACcaA 
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#6 {main}</pre> 
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#6 {main}</pre> 
    In the process straight C rightwards arrow straight A comma the gas gets compressed.
    Therefore, work done by gas is negative.
    Thus,
                     straight W subscript straight C rightwards arrow straight A end subscript space equals space minus 5.5 straight J 
    Now by the first law of thermodynamics for cyclic process, 
           dQ subscript cyclic space equals space dW subscript cyclic 
    rightwards double arrow   dQ subscript cyclic space equals space dW subscript straight A rightwards arrow straight B end subscript plus dW subscript straight B rightwards arrow straight C end subscript plus dW subscript straight C rightwards arrow straight A end subscript 
                       = Area of rectangle AbcaA + 0 + (-5.5)
                       = 10(2 - 1) -5.5  = 4.5J 
    Thus the heat absorbed by gas in the cyclic process is 4.5J. 
    Question 1492
    CBSEENPH11019764

    Why are woollen clothes warmer than cotton clothes?

    Solution
    Woollen clothes are warmer than cotton clothes because the woollen clothes have numerous pores enclosing a large quantity of air. The air being a bad conductor of heat does not allow the heat of body to conduct away.
    Question 1493
    CBSEENPH11019765

    Why do the thick glass tumblers crack on pouring hot liquid into them?

    Solution
    Glass is a bad conductor of heat. Therefore, when hot liquid is poured into thick tumbler, inner side of glass expands while outer side remains the same.
    Therefore, due to unequal expansions of inner and outer sides, the tumbler cracks. 
    Question 1494
    CBSEENPH11019766

    Stainless steel cooking pans are preferred with extra copper bottom. Why?

    Solution
    The conductivity of copper is large as compared to the conductivity of steel.
    Therefore when a cooking pan, fitted with extra copper sheet, is placed over a flame, more heat passes to the food.
    This helps in cooking the food fast. 
    Question 1495
    CBSEENPH11019767

    At what common temperature would a block of wood and a block of metal feel equally cold or equally hot when touched?

    Solution
    When the temperature of an object is equal to the temperature of a human body, no heat is transferred from the object to the body and vice versa.
    Therefore, the object appears equally cold or equally hot when touched. 
    Question 1496
    CBSEENPH11019768

    Explain why a flame from a bunsen burner will not pass through a piece of metal (wire) gauze for quite some time.

    Solution
    When the wire gauze is held over a flame, the gauze conducts a lot of heat from the flame.
    So, the gas reaching above the gauze fails to acquire its ignition temperature and therefore it does not burn above the wire gauze.
    After a sufficient time has passed, the temperature above the gauze will become equal to ignition temperature. Thus, at this temperature, the gas will burn and the flame will appear. 
    Question 1497
    CBSEENPH11019769

    Why are mud-houses colder in summer and warmer in winter?

    Solution
    Clay is a poor conductor of heat as compared to wood and bricks. Therefore, clay neither allows the external heat to enter the house in summer nor it allows the internal heat to escape out in winters.
    Hence, mud houses remain colder in summer and warmer in winter.  
    Question 1498
    CBSEENPH11019770

    In winter why do birds sit with their wings spread out?

    Solution
    Air is a bad conductor of heat. Therefore, by spreading the wings, the birds enclose a lot of air inside them; it does not allow the internal heat of the bird to escape and so they are protected from cold.
    Question 1499
    CBSEENPH11019771

    Why does a metal spoon placed in a glass allow hot water to be poured in without breaking the glass?

    Solution
    When hot water is added in the glass, it breaks due to unequal expansion. This can be avoided by lowering the temperature of liquid.
    Spoon made of metal is a good conductor of heat. Therefore, when hot water is poured into glass, the spoon takes up some heat and the temperature of liquid drops. 
    Question 1500
    CBSEENPH11019772

    A piece of paper wrapped tightly on a wooden rod is found to get burnt when held over a flame, but similar piece of paper when wrapped around a brass rod does not burn, why?

    Solution
    Wood is a bad conductor and brass is a good conductor of heat.
    When a paper is wrapped around wood and is held over a flame, a sufficient quantity of heat supplied to it resides in the paper. Therefore the paper gets burnt.
    On the other hand, if the paper is wrapped around the brass rod, the heat supplied to paper gets conducted into the interior of metal rod and hence the paper wound over metal rod does not get burnt.
    Question 1501
    CBSEENPH11019773

    Two moles of diatomic gas at NTP is expanded to four times its initial volume under adiabatic conditions. Find the final temperature and pressure of gas.

    Solution
    For adiabatic process, 
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#6 {main}</pre>                    ...(1) 
    and   TV to the power of straight gamma minus 1 end exponent space equals space constant        ...(2) 
    From (1),     
             space space space straight P subscript 1 straight V subscript 1 to the power of straight gamma space equals space straight P subscript 2 straight V subscript 2 to the power of straight gamma
    rightwards double arrow            straight P subscript 2 space equals space straight P subscript 1 open parentheses straight V subscript 1 over straight V subscript 2 close parentheses to the power of straight gamma 
    We have,
    straight P subscript 1 space equals space 1 space Atm space space and space space space straight V subscript 1 over straight V subscript 2 equals 1 fourth 
    Therefore,
    straight P subscript 2 equals 1 open parentheses 1 fourth close parentheses to the power of 1.4 end exponent space equals space 0.144 Atm
    From (2),
      straight T subscript 1 straight V subscript 1 to the power of straight gamma minus 1 end exponent space equals space straight T subscript 2 straight V subscript 2 to the power of straight gamma minus 1 end exponent 
     rightwards double arrow     straight T subscript 2 space equals space straight T subscript 1 open parentheses straight V subscript 1 over straight V subscript 2 close parentheses to the power of straight gamma minus 1 end exponent 
    Given here,
    straight T subscript 1 space space equals space 273 straight K space space space space space and space space space space straight V subscript 1 over straight V subscript 2 equals 1 fourth 
    straight T subscript 2 space equals space 273 open parentheses 1 fourth close parentheses to the power of 0.4 end exponent space

space space space space space equals space 156.8 straight K space

space space space space space space equals space minus 116.2 degree straight C         
    Tand P2 is the final temperature and pressure of the gas respectively. 
    Question 1502
    CBSEENPH11019774

    A metal bob and a wooden bob, both are at same temperature higher than our body temperature. If we touch both the bobs, which of the two will appear hotter? Explain your answer.

    Solution
    The metallic bob appears to be hotter than the wooden bob. This is because the thermal conductivity of copper is more than that of wood.
    When we touch the bobs, more heat will conduct to our hand quickly from metallic bob than wooden bob which makes us to feel metallic bob hotter than wooden bob.
    Question 1503
    CBSEENPH11019775

    A gas of heat capacity 1200J/kg absorbs 1500 cal of heat and does 6300 J of work. Find the increase in temperature of system.

    Solution

    Given here,
    Heat  dQ equals 1500 space cal space equals space 6300 straight J 
    Work done, dW equals 6300 straight J 
    According of first law of thermodynamics, 
                  dQ equals dU plus dW 
    rightwards double arrow          dU equals dQ minus dW equals 0 
    Since increase in internal energy of gas is zero, therefore increase in temperature is also zero. 

    Question 1504
    CBSEENPH11019776

    Why does a substance having high electrical conductivity has a high thermal conductivity also?

    Solution
    Electricity and heat are conducted through free electrons. 
    The substance, which is a good conductor of electricity, has very large number of free electrons. These free electrons also conduct the heat and hence also a good conductor of heat.
    Question 1505
    CBSEENPH11019777

    Two rods A and B are of equal length. Each rod has its ends at temperatures T1and T2 What is the condition that will ensure equal rates of flow of heat through both the rods?

    Solution
    Rate of flow of heat through a rod of length l, cross-sectional area A whose ends are maintained at temperatures T1 and T2 (T1 > T2) is,
    dQ over dT equals straight K fraction numerator left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis over denominator straight ell end fraction
    where K is the thermal conductivity of metal.
    Therefore the rate of flow of heat through the above said rods will be same if,
    straight K subscript straight A straight A subscript straight A equals straight K subscript straight B straight A subscript straight B
    The above equation is the required condition.
    Question 1506
    CBSEENPH11019778

    Discuss Weidemann – Franz law.

    Solution
    Weidman-Franz law is based on the experimental observation that, a material, which is good conductor of electricity, is also a good conductor of heat.
    The observation is on the basis of the fact that the free electrons take heat from hot to cold end.
    On the basis of experimental observations, Weidemann and Franz put forward a law which states:
    “ The ratio of the thermal conductivity to electrical conductivity of a metal is proportional to the temperature of metal and is same for all the metals at a given temperature”.
    i.e.
    space space space space space space space space space space space space space straight K over straight sigma proportional to straight T
rightwards double arrow space space space space space space straight K over σT equals constant space space space space
    Question 1507
    CBSEENPH11019779

    An ideal gas has a specific heat C = 2.5R. The gas is kept in a closed vessel of volume 0.008m3 at a temperature 300K and a pressure 1.5x106N/m2. If 6000 cal of heat is supplied to the gas, then find the final temperature and pressure of the gas. R = 2 cal/mole/K.

    Solution
    Volume of the gas is constant. Therefore the work done by gas is zero.
     i.e. W = 0 
    Now, according to first law of thermodynamics, 
                    increment straight Q space equals space increment straight U plus increment straight W equals increment straight U 
    The change in internal energy of an ideal gas is given by, 
                    increment straight U space equals space straight n space straight C subscript straight v increment straight T 
    Also from gas equation,
                         PV equals nRT 
    We have,
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#6 {main}</pre>
    rightwards double arrow      increment straight T space equals space fraction numerator increment straight Q over denominator straight C subscript straight v end fraction RT over PV 
    Here,    
    Heat space change comma space increment straight Q space equals space 6000 space cal comma space

Pressure comma space straight P space equals space 1.5 cross times 10 to the power of 6 straight N divided by straight m squared space

Temperature comma space straight T space equals space 300 straight K comma space

Volume comma space straight V space equals space 0.008 space straight m cubed space

Universal space gas space constant comma space straight R space equals space 2 space cal divided by mole divided by straight K space

Specific space heat comma space space straight C subscript straight v space equals space 2.5 straight R space equals space 5 space cal divided by mole divided by straight K space

therefore space space space increment straight T space equals space 6000 over 5 cross times fraction numerator 8.4 cross times 300 over denominator 1.5 cross times 10 to the power of 6 cross times 0.008 end fraction space

space space space space space space space space space space space space space space equals space 252 straight K 
    Final temperature,
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#6 {main}</pre> 
    From perfect gas equation, at constant volume 
                fraction numerator straight P apostrophe over denominator straight P end fraction equals fraction numerator straight T apostrophe over denominator straight T end fraction 
    rightwards double arrow      space space space straight P apostrophe space equals straight P fraction numerator straight T apostrophe over denominator straight T end fraction equals 1.5 cross times 10 to the power of 6 552 over 300 
                    equals 2.76 cross times 10 to the power of 6 space straight N divided by straight m squared 
    T' and P' are the pressure and temperature of the gas respectively. 

    Question 1508
    CBSEENPH11019780

    Explain the phenomenon of transfer of heat by means of convection.

    Solution
    i) Convection is the transfer of heat by the movement of a heated material.
    ii) When the liquids or gases are heated, they expand and become lighter than colder liquids or gases surrounding them. Therefore, heated liquids or gases rise up, replacing the cooler liquids or gases.
    iii) Then the cooler liquids or gases near the source of heat become warm and rise.
    iv) This movement of heated liquids or gases away from a source of heat and the flow of cooler liquids or gases toward source of heat, transfers the heat.  
    This process is called convection.
    v) During the convection process, the atoms or molecules of media themselves move from one end to the other to transfer heat. 
    Question 1509
    CBSEENPH11019781

    Why felt is used for thermal insulation in preference to air?

    Solution
    Dry air is a bad conductor but heat can be transferred into air by means of convection and radiations.
    Felt is a fine fibrous fluffy material having numerous pores which trap a lot of air.
    But, air molecules cannot move freely and consequently convection is not possible in felt. That is why felt is used for thermal insulation in preference to air. 
    Question 1510
    CBSEENPH11019782

    Explain how convection causes trade winds.

    Solution
    The surface of the earth at the equator is heated more by sun rays than poles. The hot air at equator moves up and sets the region of low pressure at the equator. This pressure gradient between poles and equator causes the cold air to move from poles to the equator. Due to rotation of the earth from west to east, the wind from northern hemispheres appears to come from the northwest and that from southern hemispheres appears to come from the south-west.
    Question 1511
    CBSEENPH11019783

    Two thin blankets are much warmer than one thick blanket. Why?

    Solution
    Two blankets are warmer than single one of double the thickness because two blankets enclose a thin layer of air.
    Air is a bad conductor of heat. Air neither allows the heat of body to escape nor it allows the external heat to sweep in. 
    Question 1512
    CBSEENPH11019784

    Explain why ventilators are provided in the room near the ceiling?

    Solution
    In rooms, the ventilators are provided at the top near the ceiling because the air in the room gets warmer due to respiration of people present in the room.
    Warm air being lighter goes up. It will exhaust out from the room if ventilators are made at the top of the room near the ceiling. 
    Question 1513
    CBSEENPH11019785

    Three moles of an ideal gas at 300K are isothermally expanded to five times its initial volume and then heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 8314kJ heat is required. Find γ of the gas.

    Solution
    Let P and V be the initial pressure and volume.
    P' and V' be the pressure and volume after isothermal expansion. 
    Given,
    Number of poles, n = 3
    Temperature, T = 300K 
    Final volume, straight V apostrophe space equals space 5 straight V space left parenthesis Given right parenthesis 
    Perfect gas equation for isothermal process, 
                 "<pre 
    rightwards double arrow         straight P apostrophe space equals space straight P fraction numerator straight V over denominator straight V apostrophe end fraction equals straight P over 5 
    During isothermal expansion, the change in internal energy of the gas is zero.
    Therefore the work done by the gas is equal to heat absorbed by gas.
    i.e.
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#6 {main}</pre> 
            equals 3 cross times 8.314 cross times 300 left parenthesis Ln 5 right parenthesis 
          space space equals 12042.8 straight J
    Now the gas is heated at constant volume, therefore the process is isochoric.  
    The initial and final values of thermodynamic variables (represented by prime and double prime superscripts) during isobaric process are given as, 
           space straight T apostrophe space equals space 300 straight K comma space space space space space space space space straight P apostrophe space equals space straight P divided by 5 comma space space space space space straight V apostrophe space equals space 5 straight V
straight T " space equals space ? space space space space space space space space space space space space space space space space straight P " space equals space straight P comma space space space space space space space space space straight V " space equals space 5 straight V 
    Gas equation for adiabatic process is, 
                           space fraction numerator straight T " over denominator straight T apostrophe end fraction equals fraction numerator straight P " over denominator straight P apostrophe end fraction 
    rightwards double arrow     straight T " space equals 5 straight T apostrophe space equals space 1500 straight K 
    According to first law of thermodynamics, 
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#6 {main}</pre> 
    For isochoric process, increment straight W equals 0. space 
    Thus, 
                space increment straight Q subscript 2 space equals space nC subscript straight v increment straight T space

space space space space space space space space space space equals space 3 cross times straight C subscript straight v cross times 1200 space

space space space space space space space space space space equals space 3600 straight C subscript straight v 
    Total heat absorbed by gas during the process, 
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#6 {main}</pre>
              83140 equals 12042.8 plus 4500 straight C subscript straight v 
    rightwards double arrow         space space space straight C subscript straight v space equals space 19.75 space straight J divided by mol divided by straight K 
                        equals 4.7 space cal divided by mol divided by straight K 
    Now,
    Ratio of specific heat is, 
    space space straight gamma space equals space straight R over straight C subscript straight v plus 1 equals fraction numerator 2 over denominator 4.7 end fraction plus 1 space equals space 1.426 
    Question 1514
    CBSEENPH11019786

    Explain how the land and sea breezes blow.

    Solution
    During day-time, land absorbs heat and becomes hotter than water in the sea. This is because the heat capacity of water is more than that of land.
    The air in contact with land becomes hot and it moves up. This sets up a region of low pressure on land. As a result, the air above sea moves towards the land. This is called sea breeze. 
    During night, the earth and water in sea lose heat. The heat capacity of water is more than that of land. Therefore, there is a drop in the temperature in the land. Now, the air above water being warmer goes up and creates a region of low pressure above sea water. As a result air above the land moves towards the sea.
    This is called the land breeze.
    Question 1515
    CBSEENPH11019787

    Can we boil water inside the artificial satellite by convection?

    Solution
    The convection currents are only possible in the presence of gravity. When we heat water, the water at the bottom becomes hot and light as compared the water at the surface of the top. The presence of gravity makes heavy cold water molecules on the top to go down. Hence, the lighter water molecules at the bottom will move up. 
    In artificial satellite due to weightlessness the convection current is not possible. Hence, we cannot make water boil on artificial satellite by the process of convection. 
    Question 1516
    CBSEENPH11019788

    What are thermal radiations. State few properties of thermal radiations.

    Solution

    Thermal radiations is the heat energy emitted by a body in the form of radiation.
    Following are the properties of thermal radiations:

    (i) These radiations are electromagnetic in nature.

    (ii) These radiations travel in a straight line with the speed of light.

    (iii) These radiations can travel through vacuum as well as in medium.

    (iv) In a homogenous media, these radiations travel equally in all the direction. 

    (v) These radiations undergo the phenomenon of reflection, refraction, etc. 

    Question 1517
    CBSEENPH11019789

    Define the terms:

    (i) Absorptance,

    (ii) Reflectance

    (iii)Transmittance.

    Solution

    (i) Absorbtance: It is defined as the ratio of the amount of thermal radiation absorbed by the body in a certain time to the total amount of thermal radiation incident on it at the same time.

    (ii) Reflectance: It is the ratio of the amount of thermal radiation reflected from a body in a certain time to the total amount of thermal radiation incident on it in the same time. 

    (iii) Transmittance: It is the ratio of the thermal radiation transmitted by it in a certain time to the total amount of thermal radiation incident on it in the same time.

    Question 1518
    CBSEENPH11019790

    Define spectral absorptive power.

    Solution

    The spectral absorptive power of a body for a wave of wavelength X is defined as the ratio of the heat energy absorbed to the heat energy incident upon per unit time within the unit wavelength range (λ–1/2)to(λ+1/2) around X.

    Specral absorptive power is denoted by aλ.

    Question 1519
    CBSEENPH11019791

    State Kirchhoff’s law. Explain three observations that can be explained using Kirchhoff’s law.

    Solution

    Kirchoff's Law states that the ratio of the emissive power to absorptive power for the radiation of a given wavelength is same for all substances at the same temperature. This constant is equal to the emissive power of perfectly black body of the same wavelength and the same temperature.
    The following observations can be explained using Kirchhoff’s law: 

    (i) In a solar spectrum, a number of dark lines called Fraunhoffer lines are observed. But during solar eclipse, same bright lines are observed as that of Fraunhoffer dark lines. Kirchhoff’s law explains this observation.

    (ii) The outer wall of thermos flask is made shining, so that neither it absorbs more heat from surrounding (contents in the flask) nor it radiates more heat to surrounding (space outside) thereby keeping the temperature of contents in flask constant.

    (iii) Mark an ink dot on a china ware. You will notice the ink dot dark and china ware shines. If it is heated to high temperature and taken to dark room, the dot begins to shine brightly than chine ware. This indicates that good absorber is a good emitter. This application is also explained by the Kirchoff's law. 

    Question 1520
    CBSEENPH11019792

    Three moles of helium gas are initially at 300K and occupy a volume of 25 liters. The gas is first expanded at constant pressure until volume is increased to 60 liters. Then it undergoes an adiabatic expansion until the temperature decreases to 400K. Find the work done by the gas during adiabatic expansion.

    Solution
    Let P, V and T be the initial pressure, volume and temperature.
    Let P', V' and T' be the pressure, volume and temperature after isobaric expansion. 
    We have, 
    Temperature, T = 300K 
    Initial space volume comma space straight V space equals space 25 space lt space equals space 25 cross times 10 to the power of negative 3 end exponent straight m cubed

Final space volume comma space straight V apostrophe space equals space 60 space lt space equals space 60 space cross times space 10 to the power of negative 3 end exponent straight m cubed 
    Using the perfect gas equation for an isochoric process. 
                straight T apostrophe equals straight T fraction numerator straight V apostrophe over denominator straight V end fraction 
    rightwards double arrow        straight T apostrophe equals 300 cross times fraction numerator 60 cross times 10 to the power of negative 3 end exponent over denominator 25 cross times 10 to the power of negative 3 end exponent end fraction

space space space space equals 720 straight K 
    Now the gas is expanded adiabatically.
    Therefore, the work done by gas during adiabatic expansion is, 
    straight W equals nC subscript straight v left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis space equals space fraction numerator nR over denominator straight gamma minus 1 end fraction left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis 
    Given here,
    Number space of space moles comma space straight n equals 3

Ratio space of space specific space heat comma space space straight gamma equals 5 divided by 3

Initial space temperature comma space straight T subscript 1 space equals space 720 straight K

Final space temperature comma space straight T subscript 2 space equals space 400 straight K 
    So, work done by the gas during adiabatic expansion is, 
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#6 {main}</pre>
                    equals 12096 space straight J 
    Question 1521
    CBSEENPH11019793

    Show that in a cyclic process, the area of the loop on the indicator diagram is equal to net work done by or on the gas which in turn is equal to net heat absorbed or rejected by the gas.

    Solution

    When a system undergoes a cyclic process, it is finally brought back to initial state in all respects.
    Suppose, an ideal gas is contained in a cylinder fitted with a massless and frictionless piston.
    Let the gas be initially in the state represented by point A on the P–V graph. Let it be taken from state A to C along path ABC and brought back to A via path CDA.
    Case 1: Work done by gas is equal to area ABCKLA.
    Case 2: Work done on the gas is equal to area CDALKA. 
     
    Now net work done by gas, 
    space dW equals Area left parenthesis ABCKLA right parenthesis space minus space Area left parenthesis CDALKA right parenthesis space

space space space space space space space equals space Area space of space closed space loop left parenthesis ABCDA right parenthesis  
    According to first law of thermodynamics, 
    space increment straight Q space equals space increment straight U plus increment straight W 
    In cyclic process the change in internal energy is zero.
    So, increment straight Q space equals space increment straight W 
    Net heat absorbed during a cyclic process is equal to work done by gas.
    Therefore, the area of the loop on the indicator diagram is equal to both work done and the heat absorbed in the cyclic process. 

    Question 1522
    CBSEENPH11019794

    Two slabs A and B of thermal conductivity Kl and K2, length l2 and l2 and each of cross section area A are placed in scries. The free end of slab A is at temperature T1 and that of slab B is T2 <T1. Find the temperature at the interface of two slabs. Also find the rate of flow of heat through compound slab in steady state.

    Solution

    Let T be the temperature at the interface of two slabs
                        
    The rate of flow of heat through slab A is,
    dQ subscript 1 over dt equals fraction numerator straight K subscript 1 straight A over denominator straight l subscript 1 end fraction left parenthesis straight T subscript 1 minus straight T right parenthesis space space space space space space space space space space space... space left parenthesis 1 right parenthesis
    The rate of flow of heat through slab B is, 
    
dQ subscript 2 over dt equals fraction numerator straight K subscript 2 straight A over denominator straight l subscript 1 end fraction left parenthesis straight T minus straight T subscript 2 right parenthesis space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis
    As two slabs are in series, therefore the rate of flow of heat through them will be same.
    straight i. straight e. space space space space space space space space space space space space space dQ over dt equals dQ subscript 2 over dt space

space space space space space space space straight K subscript 1 over straight l subscript 1 left parenthesis straight T subscript 1 minus straight T right parenthesis equals straight K subscript 2 over straight l subscript 2 left parenthesis straight T minus straight T subscript 2 right parenthesis space

space straight K subscript 1 over straight l subscript 1 straight T subscript 1 plus straight K subscript 2 over straight l subscript 2 straight T subscript 2 equals straight K subscript 1 over straight l subscript 1 straight T plus straight K subscript 2 over straight l subscript 2 straight T space

rightwards double arrow space space space space straight T equals fraction numerator space space space space space space space space space space begin display style straight K subscript 1 over straight l subscript 1 end style straight T subscript 1 plus begin display style straight K subscript 2 over straight l subscript 2 end style straight T subscript 2 space over denominator begin display style straight K subscript 1 over straight l subscript 1 plus straight K subscript 2 over straight l subscript 2 end style end fraction space

space space space space space space space space space space equals fraction numerator straight l subscript 2 straight K subscript 1 straight T subscript 1 plus straight l subscript 1 straight K subscript 2 straight T subscript 2 over denominator straight l subscript 2 straight K subscript 1 plus straight l subscript 1 straight K subscript 2 end fraction
    The rate of flow of heat through composite slab is,
    dQ over dt equals dQ subscript 1 over dt equals fraction numerator straight K subscript 1 straight A over denominator straight l subscript 1 end fraction left parenthesis straight T subscript 1 minus straight T right parenthesis

space space space space space space space space equals fraction numerator straight K subscript 1 straight A over denominator straight l subscript 1 end fraction open parentheses straight T subscript 1 minus fraction numerator straight l subscript 2 straight K subscript 1 straight T subscript 1 plus straight l subscript 1 straight K subscript 2 straight T subscript 2 over denominator straight l subscript 2 straight K subscript 1 plus straight l subscript 1 straight K subscript 2 end fraction close parentheses space

space space space space space space space space equals fraction numerator straight K subscript 1 straight K subscript 2 over denominator straight l subscript 2 straight K subscript 1 plus straight l subscript 1 straight K subscript 2 end fraction straight A left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis 

    Question 1523
    CBSEENPH11019795

    What is heat engine? Obtain a general expression for its efficiency.

    Solution

    A machine that converts heat energy into mechanical energy is called heat engine. 
    Heat engine consists of three essential parts: 
    (i) Source: It is a hot body at high temperature having infinite heat capacity. 
    (ii) Working substance: It is an ideal gas. 
    (iii) Sink: It is cold body at low temperature having infinite heat capacity. 
    The block diagram for the engine is as shown: 
     
    The working substance is carried through a series of cyclic process, in which it absorbs some heat from source. 
    A part of the heat is converted into mechanical work and rest of heat is rejected to sink. 
    In a cyclic process, the working substance absorbs heat from source and rejects Q2 heat to the sink. 
    The change in internal energy of gas is zero.
    Therefore, using first law of thermodynamics, the net amount of work done by an ideal engine is, 
                    straight W equals increment straight Q minus increment straight U equals increment straight Q
 
                    straight W equals straight Q subscript 1 minus straight Q subscript 2 
    The efficiency of heat engine is defined as the ratio of work obtained to the heat taken up from the source. Efficiency is denoted by straight eta.  
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#6 {main}</pre>

    Question 1524
    CBSEENPH11019796

    What is Carnot heat engine? Discuss the essential parts of Carnot heat engine and draw its block diagram.

    Solution
    Carnot heat engine is an ideal, reversible and most efficient heat engine that converts heat energy into mechanical energy.
    The block diagram for the Carnot engine is as shown:

    The essential parts of Carnot heat engine are discussed below: 
    (i) Cylinder: The cylinder of Carnot’s engine has perfectly insulating walls, perfectly insulating base fitted with perfectly insulating, massless and frictionless piston. 
    (ii) Working substance: It is an ideal gas filled in the cylinder. 
    (iii) Source: It is a hot body at high temperature (T1) having infinite heat capacity.
    The working substance draws the heat Q1 once in each cycle of operation. 
    (iv) Sink: It is a cold body at low temperature (T2) having infinite heat capacity. The working substance rejects the heat Q2 once in each cycle of operation. 
    (v) Insulated pad: In one cycle of operation, the gas in the cylinder is to expand and compress adiabatically.
    The cylinder is placed on an insulated pad to make in order to carry out the adiabatic process.
    Question 1525
    CBSEENPH11019797

    Two slabs of thermal conductivity Kand K2, length l1, and l2 of the same cross section are placed in series. Show that the series combination of two slabs behaves as a single slab of thermal conductivity K given by

    space space space space space space space space space straight K equals fraction numerator straight l subscript 1 plus straight l subscript 2 over denominator begin display style straight l subscript 1 over straight K subscript 1 plus straight l subscript 2 over straight K subscript 2 end style end fraction


    Solution
    Let A be the area of the cross-section of each slab.

    Thermal resistance of slab of thermal conductivity K1 is,
    straight R subscript 1 equals fraction numerator straight l subscript 1 over denominator straight K subscript 1 straight A end fraction space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space
    Thermal resistance of slab of thermal conductivity K2 is,
    space space space space space R equals fraction numerator l subscript 1 plus l subscript 2 over denominator K A end fraction space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis space
    For series combination, 
    space space straight R equals straight R subscript 1 plus straight R subscript 2 space space

rightwards double arrow space space space fraction numerator straight l subscript 1 plus straight l subscript 2 over denominator K A end fraction equals fraction numerator straight l subscript 1 over denominator straight K subscript 1 straight A end fraction plus fraction numerator straight l subscript 2 over denominator straight K subscript 2 straight A end fraction space space

rightwards double arrow space space space space space space space space space space space space space straight K equals fraction numerator straight l subscript 1 plus straight l subscript 2 over denominator begin display style straight l subscript 1 over straight K subscript 1 plus straight l subscript 2 over straight K subscript 2 end style end fraction space 
    Question 1526
    CBSEENPH11019798

    Explain tne working of carnot ’s neat engine and find its efficiency?

    Solution

    The working substance is taken through a cycle of four operations known as Carnot’s cycle in a Carnot engine.

    Four operations of Carnot cycle are:
    (i) isothermal expansion,
    (ii) adiabatic expansion,
    (iii) isothermal compression, and
    (iv) adiabatic compression.
    Initially the working substance is heated to a temperature of source T1
    The initial thermodynamic state (P1,V1,T1) of working substance. 
    (i) Isothermal expansion: The cylinder is placed on source and working substance is allowed to expand slowly so that it expands isothermally.
    Let the thermodynamic states of working substance change from (P1,V1,T1) to (P2,V2,T1).
    Temperature remains constant as it expands isothermally. 
    This process is depicted on PV diagram by curve AB.
    Let Q1 be the heat absorbed by the gas and W1 be the work done by it. 
    therefore space straight W subscript 1 space equals space straight Q subscript 1 space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript straight P. dV space equals space RT subscript 1 Ln space open parentheses straight V subscript 2 over straight V subscript 1 close parentheses space

space space space space space space space space space space space space space space space space space space space space space equals space Area space left parenthesis ABbaA right parenthesis thin space space space space space space space... space left parenthesis 1 right parenthesis space
    (ii) Adiabatic expansion: Now, the cylinder is placed on an insulated pad and allowed to expand quickly so that the temperature of working substance falls to the temperature of sink T2.
    The thermodynamic states of working substance in this process changes from (P2,V2,T1) to (P3,V3,T2).
    This process on PV diagram is represented by curve BC.
    In this process, the heat absorbed by gas is zero.
    Work done on the gas is W2
    Therefore comma space

straight W subscript 2 space equals space integral subscript straight V subscript 2 end subscript superscript straight V subscript 3 end superscript straight P. dV space equals space minus integral subscript straight T subscript 1 end subscript superscript straight T subscript 2 end superscript straight C subscript straight P. dT space equals space straight C subscript straight P space left parenthesis straight T subscript 1 space minus space straight T subscript 2 right parenthesis space

space space space space space space equals space Area space left parenthesis BCcbB right parenthesis space space space space space space space space... space left parenthesis 2 right parenthesis thin space
    (iii) Isothermal compression: When an isothermal compression is carried out, the cylinder is placed on the sink and working substance is compressed slowly so that it compresses isothermally.
    The thermodynamic states of working substance change from (P3,V3,T2) to (P4,V4,T2).
    This process is shown on PV diagram by curve CD.
    Let Q2 be the heat evolved and W3 be the work done on the working substance. 
    therefore space minus space straight W subscript 3 space equals space minus straight Q subscript 2 space equals space integral subscript straight V subscript 3 end subscript superscript straight V subscript 4 end superscript straight P. space dV space equals space minus RT subscript 1 space Ln space open parentheses straight V subscript 3 over straight V subscript 4 close parentheses space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space Area space left parenthesis DCcdD right parenthesis space space space space space space space... space left parenthesis 3 right parenthesis thin space
    (iv) Adiabatic compression: Finally, the cylinder is placed on an insulated pad and compressed quickly so that the working substance returns to initial state (P1,V1,T1).
    This process on PV diagram is illustrated by curve DA.
    In this process, the heat evolved by gas is zero.
    Work done on the gas is W4
    negative straight W subscript 4 space equals space integral subscript straight V subscript 4 end subscript superscript straight V subscript 1 end superscript straight P. dV space
space space space space space space space space equals space minus integral subscript straight T subscript 2 end subscript superscript straight T subscript 1 end superscript straight C subscript straight P space dT space
space space space space space space space space equals space minus space straight C subscript straight P space left parenthesis straight T subscript 1 space minus space straight T subscript 2 right parenthesis space

space space space space space space space space equals space Area space left parenthesis ADdaA right parenthesis space space space space space space space space space space... space left parenthesis 4 right parenthesis 
    Net amount of work done by the working substance in 1 complete cycle. 
    In isothermal and adiabatic expansion, the work W1 and W2 is done by the gas.
    Using sign convention W1 and W2 are positive.
    In isothermal and adiabatic compression, the work W3 and W4 is done on the gas.
    Therefore W3 and W4 are negative.
    Thus net work done by working substance in one complete cycle is, 
    W = W1 + W2 + (-W3) + (-W4)
    From equation (2) and (4), 
    W2 = W4
    Therefore, 
    W = W1 - W3 = Q1 - Q2            ... (5) 
    rightwards double arrow space straight W space equals space RT subscript 1 space Ln space open parentheses straight V subscript 2 over straight V subscript 1 close parentheses space minus space RT subscript 1 space Ln space open parentheses straight V subscript 3 over straight V subscript 4 close parentheses space space space... space left parenthesis 6 right parenthesis thin space

From space Pv space graph comma space

straight W space equals space straight W subscript 1 space plus space straight W subscript 2 space plus space left parenthesis negative straight W subscript 3 right parenthesis space plus space left parenthesis negative straight W subscript 4 right parenthesis space

space space space space equals space Area space left parenthesis ABbaA right parenthesis space plus space space Area left parenthesis BCcbB right parenthesis space
space space space space space space space space space minus space Area space left parenthesis DCcdD right parenthesis space minus space Area space left parenthesis ADdaA right parenthesis thin space

space space space space equals space Area space of space loop space left parenthesis ABCDA right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 7 right parenthesis thin space
    That is in a Carnot's heat engine, the mechanical work obtained during each cycle is equal to the area of the loop on PV diagram.

    Efficiency of Carnot’s cycle.
    The efficiency of Carnot heat engine is given by, 
    straight eta equals straight W over straight Q subscript 1 equals fraction numerator straight Q subscript 1 minus straight Q subscript 2 over denominator straight Q subscript 1 end fraction equals 1 minus straight Q subscript 2 over straight Q subscript 1 

    Question 1527
    CBSEENPH11019799

    What is an ideal reversible engine?

    Solution
    Heat engine continuously converts heat energy into mechanical energy.
    The working substance of engine is carried through a cycle of process in which it takes heat from source. Some part of this heat is converted into work. The rest of the heat is rejected to sink and finally the working substance returns to its initial state. After a complete cycle, the condition of hot source, sink and surrounding changes.
    Now, on reversing the cyclic process, all the components of engine resume exactly their initial conditions and no change is left in the surrounding, then engine is called reversible heat engine.
    Question 1528
    CBSEENPH11019800

    Discuss Searle’s method to find the value of thermal conductivity.

    Solution
    Searle's method:
    i) Searle's apparatus consists of a rod XY whose thermal conductivity is to be determined.
    ii) One end X is enclosed in a steam chamber and steam is continuously passed through it. The second end Y is embedded in a cooling pipe C. A constant flow of water runs through pipe C.
    iii) Two thermometers T1 and T2 are fitted in the holes drilled in the rod at P and Q separated by distance d.
    iv) Two other thermometers T3 and T4 are used to measure the temperature of incoming and outgoing water through pipe C.
    v) The rod is thoroughly insulated to stop any heat losses by radiation.
    vi) Heat is conducted through the rod and when steady state is reached, the thermometers T1 T2, T3 and T4 record constant temperature.
    vi) The mass of water flowing through a pipe C is collected in a given interval of time and weighed.
     

    Now, it has been observed that at a steady state, the quantity of heat flowing through every section is same.
    i.e., the rate of flow of heat at point P, Q and Y is same.

    Quantity of heat that flows from section P to Q in one second is given by, 


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#6 {main}</pre> 
    where,
     θ1 and θ2 be the reading of thermometers T1 and T2 respectively.
    Amount of heat gained by water in one second at end Y is,
                  dH over dt equals ms fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight l end fraction space space space space space space space space space space space... left parenthesis 2 right parenthesis
    where,
    m is the mass of water collected in time t,
    s be the specific heat and
    straight theta3 and θ4 be the reading of thermometers T3 and T4. 
    Now, equating (1) and (2), we get
    space space fraction numerator K A over denominator straight d end fraction left parenthesis straight theta subscript 1 minus straight theta subscript 2 right parenthesis equals m s fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight l end fraction
space
rightwards double arrow space space space space space space space space space space space space space space space space space space straight K equals fraction numerator m s d over denominator A t end fraction open parentheses fraction numerator straight theta subscript 4 minus straight theta subscript 3 over denominator straight theta subscript 1 minus straight theta subscript 2 end fraction close parentheses 

    Question 1529
    CBSEENPH11019801

    Why the heat engine does not work without a cold sink?

    Solution
    When a part of heat absorbed from the source is rejected, this heat energy is converted into mechanical energy by the heat engine. The heat absorbed from the source can only be rejected if there is a cold body at a temperature lower than the temperature of the source. 
    Question 1530
    CBSEENPH11019802

    If you were asked to increase the efficiency of heat engine by increasing the temperature of source or decreasing the temperature of sink, which would you prefer? Explain.

    Solution
    On decreasing the temperature of the sink, efficiency of the heat engine can be increased. This way is preferred because on increasing the temperature of the source, there wont be much increase in the efficiency. 
    Question 1531
    CBSEENPH11019803

    No real engine can be more efficient than Carnot engine working between the same temperatures. Explain.

    Solution
    Carnot engine is the most efficient engine because it uses an ideal gas as a working substance. Also, the force of friction between different parts of engine is zero.
    Question 1532
    CBSEENPH11019804

    Is the transfer of heat from cold space inside the refrigerator to warm surrounding against the second law of thermodynamics? Explain.

    Solution
    According to the second law of thermodynamics, the heat cannot flow spontaneously from a body at low temperature to a body at high temperature. An external force is required for the transfer of heat energy.
    In case of the refrigerator, the external agency is the compressor that does the work to transfer the heat.
    Question 1533
    CBSEENPH11019805

    What is refrigeration? Name the device used to produce refrigeration.

    Solution
    The process of cooling bodies to a temperature lower than the temperature of their surrounding by taking out heat from them is called refrigeration.
    The machine used to produce refrigeration is called refrigerator.
    Question 1534
    CBSEENPH11019806

    Name the processes which involve in the working of refrigerator.

    Solution

    The various processes which take place in the working of the refrigerator are:
    (i) Adiabatic compression of working substance (ideal gas) outside the room of refrigerator. 
    (ii) Cooling of adiabatically compressed working substance. 
    (iii) Adiabatic expansion of cooled working substance inside the room of refrigerator.

    Question 1535
    CBSEENPH11019807

    What is refrigerator? Draw its block diagram. Discuss its theory.

    Solution
    Refrigerator is a device used to cool the bodies to a temperature lower than the temperature of their surrounding. 
    According to second law of thermodynamics, it is impossible to have a purely automated machine that can transfer the heat energy from a body at low temperature to a body at high temperature.
    Therefore, in order to cool a cold body, it is necessary to expend some energy.
    The refrigeration involves the use of a working substance (called refrigerant) going about a cycle. In te process, there is the removal of heat from a cold body.
    Work is done on the body and a large amount of heat is rejected to the hot body. 
     
    The above diagram shows the block diagram of a refrigerator. 
    Let the refrigerant extract Q2 heat from cold body.
    Let W be the work required to be done on the refrigerant to reject this heat to hot body and Q1 be the net heat rejected to hot body.
    As the refrigerant undergoes a cycle, there is no change in internal energy.  
    Using first law of thermodynamics, 
    straight Q subscript 2 minus straight Q subscript 1 equals negative straight W 
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#6 {main}</pre> 
    Thus the heat rejected to hot body is greater than the heat absorbed from cold body.
    The best quality of refrigerator is one which extracts maximum heat and expends minimum work to reject it.
    The quality of refrigerator is measured in terms of coefficient of performance and is defined as the ratio of the heat removed from the cold body to the work spent in rejecting the heat to hot body.
    Mathematically, coefficient of performance is given by, 
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#6 {main}</pre>
    ∴             straight beta equals fraction numerator straight Q subscript 2 over denominator straight Q subscript 1 minus straight Q subscript 2 end fraction equals fraction numerator straight T subscript 2 over denominator straight T subscript 1 minus straight T subscript 2 end fraction
    Question 1536
    CBSEENPH11019808

    The temperature of surface of the sun is approximately 6000K. Is it possible to produce a temperature of 7000K by focussing the rays using a big lens?

    Solution
    This case is not possible because it  is violating the Clausius statement of second law of thermodynamic.
    Clausius stated that we cannot transfer heat from a body of low-temperature (6000K) to a body of high-temperature (8000 K), merely by using a lens. 
    Question 1537
    CBSEENPH11019809

    Can the efficiency of Carnot engine be 100%?

    Solution
    The efficiency of Carnot engine is given by, 
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#6 {main}</pre> 
    The efficiency of Carnot engine depends on  temperature of source and sink i.e. T1 and T2.
    The efficiency of the heat engine can be 100% if T2 = 0 i.e. temperature of the sink is absolute zero.
    Since absolute zero cannot be obtained, therefore, efficiency cannot be 100%. 
    Question 1538
    CBSEENPH11019810

    Why Carnot engine cannot be realised in practice?

    Solution

    Carnot engine cannot be realised in practice because of the following reasons: 
    (i) source and sink of infinite thermal capacity cannot be realised in actual practice,
    (ii) we cannot have perfect insulators and perfect conductors in actual practice. 
    iii) compression and expansion cannot be carried out very slowly for the cycle to be reversible. 

    Question 1539
    CBSEENPH11019811

    Can heat energy be converted into work without a temperature difference?

    Solution
    No, without a temperature difference, heat energy cannot be converted into work.
    According to the second law of thermodynamics, heat energy cannot be converted into work till some heat is rejected into the surroundings.
    In order to reject the heat we require a body at a lower temperature and not at the same temperature. 
    Alternatively:
    We know, 
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#6 {main}</pre> 
    If T1 – T2 = 0, then W = 0
    i.e. no heat can be converted into work without a temperature difference. 
    Question 1540
    CBSEENPH11019812

    Why is it impossible for a ship to use the internal energy of the seawater to run?

    Solution
    A ship cannot use the internal energy of sea water to run because there is no suitable sink of lower temperature so that internal energy can be converted into mechanical work. Hence, it is impossible. 
    Question 1541
    CBSEENPH11019813

    Define coefficient of performance of refrigerator.

    Solution
    The coefficient of performance is the ratio of heat absorbed from sink to the work done on the working substance to reject the heat. 
    Mathematically, it is given by 
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#6 {main}</pre>
    Question 1542
    CBSEENPH11019814

    If the door of refrigerator is kept open will it cool the room?

    Solution
    If the door of the refrigerator is kept open, the room will not be cooled but the temperature of the room will rather increase because refrigerator takes less heat from the space inside the refrigerator and rejects more cold atmosphere to the room. 
    Question 1543
    CBSEENPH11019815

    The temperatures of source and sink of a Carnot engine are 300°C and 300K. Find the efficiency of the engine.

    Solution

    Temperature of source, 
                 straight T subscript 1 space equals space 300 degree straight C space equals space 573 straight K 
    Temperature of sink, 
                  straight T subscript 2 space equals space 300 straight K 
    The efficiency of Carnot engine is given by, 
    straight eta space equals 1 minus straight T subscript 2 over straight T subscript 1
space space space equals 1 minus 300 over 573
space space space equals space 273 over 573

space space space equals 0.476

    Question 1544
    CBSEENPH11019816

    The air above the water surface in a pond is at temperature –θ°C as a result of which the water in the pond starts freezing. Find the time taken by the pond to freeze from thickness x1 to x2. Assume the required data.

    Solution
    Let us consider a pond of water of surface area A.
    Let L be the latent heat, K be the thermal conductivity and ρ be the density of ice.
                 
    Let at any instant, a layer of thickness x has been already formed. 
    Let it take 'dt' time to further increase the thickness of ice by 'dx'.
    Heat given up from water to freeze 'dx' layer of ice is, 
    dH equals Adxρ space straight L space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    This amount of heat is conducted from water to air through ice layer of thickness dx in time dt.
    Thus heat conducted through ice is,
    dQ equals KA fraction numerator 0 minus left parenthesis negative straight theta right parenthesis over denominator straight x end fraction dt equals Ka straight theta over straight x dt space space space space space space space space space space space space space space space.... left parenthesis 2 right parenthesis
    Equating (1) and (2)
        space space space space space space space space space KA straight theta over straight x dt equals Adx space straight rho space straight L
rightwards double arrow space space space space space space space space space space space space space space space dt equals ρL over Kθ xdx space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis 
    Time taken to freeze the ice from thickness x1 to x2 can be obtained by integrating the equation (3).
     i.e., 
    integral subscript 0 superscript straight t dt space equals space integral subscript straight x subscript 1 end subscript superscript straight x subscript 2 end superscript ρL over Kθ space straight x space dx space

space space space space right enclose straight t space subscript 0 superscript straight t space equals space ρL over Kθ 1 half straight x squared vertical line subscript straight x subscript 1 end subscript superscript straight x subscript 2 end superscript space

rightwards double arrow space space straight t space equals space fraction numerator ρL over denominator 2 Kθ end fraction left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis space
 
    Question 1545
    CBSEENPH11019817

    An ideal Carnot engine is working between 227°C and 77°C. This engine delivers power of lOkW. Find the rate at which engine rejects the heat to the sink.

    Solution

    Here,
    Lower temperature, straight T subscript 1 space equals space 227 degree straight C space equals space 500 straight K
    Temperature, straight T subscript 2 space equals space 102 degree straight C space equals space 375 straight K 
    The efficiency of engine is, 
              straight eta space equals space 1 minus straight T subscript 2 over straight T subscript 1 equals 1 minus 375 over 500 equals 0.25 
    Since,     straight eta equals straight W over straight Q subscript 1 
    Therefore, the quantity of heat received by engine from source per second is, 
    space space space space space space straight Q subscript 1 equals straight W over straight eta equals fraction numerator 10 comma 000 over denominator 0.25 end fraction equals 40 comma 000 space straight J divided by straight s 
    Now the heat rejected to sink per second, 
    straight Q subscript 2 space equals space straight Q subscript 1 minus straight W 
        equals 40000 minus 10000
equals space 30000 straight J divided by straight s 

    Question 1546
    CBSEENPH11019818
    Question 1547
    CBSEENPH11019819

    A Carnot engine P works between 827°C and T and another engine Q is working between T and 227°C. Find the value of T at which the efficiency of both the engines is same.

    Solution

    The efficiency of engine P is, 
                space space space straight eta subscript straight P equals 1 minus straight T subscript 2 straight P end subscript over straight T subscript 1 straight P end subscript 
                     equals 1 minus fraction numerator straight T over denominator 827 plus 273 end fraction
equals 1 minus straight T over 1100 
    The efficiency of engine Q is, 
               straight eta subscript straight Q equals 1 minus straight T subscript 2 straight Q end subscript over straight T subscript 1 straight Q end subscript equals 1 minus fraction numerator 227 plus 273 over denominator straight T end fraction
space space space space space equals space 1 minus 500 over straight T 
    Here,         straight eta subscript straight P equals straight eta subscript straight Q
    ∴   1 minus straight T over 1100 equals 1 minus 500 over straight T
    rightwards double arrow    space space space straight T over 1100 equals 500 over straight T
    rightwards double arrow              straight T equals square root of 550000 space

space space equals space 741.62 straight K 
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#6 {main}</pre> 
    This is the required temperature at which the efficiency of both engines is same. 

    Question 1548
    CBSEENPH11019820

    The efficiency of heat engine is 0.3. On increasing the temperature of source by 80°C, the efficiency of heat engine increases to 0.4. Find the initial temperature of source and sink in which heat engine is working.

    Solution
    Let T1 and T2 be the initial temperatures of source and sink when the efficiency of the heat engine is 0.3.
    Therefore, 
                    0.3 equals 1 minus straight T subscript 2 over straight T subscript 1
 
    rightwards double arrow            straight T subscript 2 space equals space 0.7 straight T subscript 1            ...(1) 
    On increasing the temperature of the source by 80°C, the efficiency becomes 0.4.
    Therefore, 
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#6 {main}</pre> 
    rightwards double arrow         straight T subscript 2 space equals space 0.6 left parenthesis straight T subscript 1 plus 80 right parenthesis        ...(2) 
    From (1) and (2), we get 
                  0.7 straight T subscript 1 space equals space 0.6 left parenthesis straight T subscript 1 plus 80 right parenthesis 
    rightwards double arrow          0.1 straight T subscript 1 space equals space 48 
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#6 {main}</pre>, is the temperature of the source. 

    and  
      straight T subscript 2 space equals space 0.7 straight T subscript 1 space

space space space space space space equals space 0.7 cross times 480 space

space space space space space space equals 336 straight K space equals space 63 degree straight C
    T2 is the temperature of the sink. 
    Question 1549
    CBSEENPH11019821

    Carnot engine takes 90 cal of heat from the reservoir at temperature 540K. Engine performs 126J of work. To increase the engine performance to 150J, to what temperature the hot reservoir should be raised?

    Solution

    Given,
    Heat space taken space from space the space reservoir comma space straight Q subscript 1 space equals space 90 space cal space equals space 378 straight J space

Work space done space by space the space engine comma space straight W space equals space 126 straight J
    Heat released, straight Q subscript 2 space equals space 378 minus 126 space equals space 252 straight J         

                           straight T subscript 1 space equals space 840 space straight K
    Since,
            straight Q subscript 1 over straight Q subscript 2 equals straight T subscript 1 over straight T subscript 2 
    ∴      space space straight T subscript 2 space equals space straight Q subscript 2 over straight Q subscript 1 straight T subscript 1 space equals space 252 over 378 cross times 540 space equals space 360 straight K 
    Let at temperature space straight T subscript 1 apostrophe space comma  the engine's performance be 150J.
    Thus, 
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#6 {main}</pre> 
    and    space space straight Q subscript 2 apostrophe space equals space 378 minus 150 space equals space 238 straight J 
    Now     straight T subscript 1 apostrophe space equals space fraction numerator straight Q subscript 1 apostrophe over denominator straight Q subscript 2 apostrophe end fraction straight T subscript 2 equals 378 over 228 cross times 360 space equals space 596.84 straight K
    The hot reservoir should be raised to a temperature of 596.84 K. 

    Question 1550
    CBSEENPH11019822

    A man weighing 60kg starts climbing up after eating 5 bread slices of food value 1000 calories. The efficiency of human body is 0-28. Find the height to which a man can climb up.

    Solution

    Given here,
    Heat released, Q = 1000cal = 4200J 
    Efficiency, straight eta equals 0.28           
    Therefore, 
    Work done,straight W equals eta Q space equals space 0.28 cross times 4200 straight J space equals space 1176 straight J      

    Let the man climb up to height h.
    Therefore, 
                          space space space mgh equals straight W 
    rightwards double arrow         60 cross times 9.8 cross times straight h equals 1176 
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#6 {main}</pre> 

    Question 1551
    CBSEENPH11019823

    An engine consumes 30 kg of coal per hour and does the work at the rate of 45H.P. Find the efficiency of heat engine. The heat of combustion of coal is 2 x l06cal per kg.

    Solution

    Power developed = 45 HP = 33570 J/s 
    Coal consumed per hour = 30kg 
    Heat produced per hour = 30 cross times 2 cross times 10 to the power of 6 space cal divided by hr 
    Heat produced per sec = space 30 cross times 2 cross times 10 to the power of 6 fraction numerator 4.2 over denominator 3600 end fraction straight J divided by straight s 
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#6 {main}</pre> 
    Efficiency,   space space space straight eta equals 33570 over 70000 cross times 100 almost equal to 48 % 

    Question 1552
    CBSEENPH11019824
    Question 1554
    CBSEENPH11019826

    It takes 3 hours to freeze the lake from 2.1 cm to 2.4cm. How long will it take to freeze it from 3.6cm to 3.9cm.

    Solution

    Time taken to freeze the lake from thickness xto x2 is given by, 

    straight t space equals space fraction numerator ρL over denominator 2 Kθ end fraction space left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis thin space

straight t space proportional to space space left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis thin space

fraction numerator straight t apostrophe over denominator straight t end fraction space equals space fraction numerator left parenthesis straight x apostrophe subscript 2 squared space minus space straight x apostrophe subscript 1 squared right parenthesis over denominator straight x subscript 2 squared space minus space straight x subscript 1 squared end fraction
fraction numerator straight t apostrophe over denominator 3 end fraction space equals space fraction numerator left parenthesis 3.9 right parenthesis squared space minus space left parenthesis 3.6 right parenthesis squared over denominator left parenthesis 2.4 right parenthesis squared space minus space left parenthesis 2.1 right parenthesis squared end fraction space equals space 5 over 3

straight t apostrophe space equals space 5 space hours

    Question 1555
    CBSEENPH11019827

    How much work has to be done by the compressor of refrigerator to remove 200k cal of heat from the freezer? The temperature of freezer is –8°C and that of room is 27°C. Also find the coefficient of performance.

    Solution

    Given that,
    straight T subscript 1 space equals space 27 degree straight C space equals space 300 straight K
straight T subscript 2 space equals space minus 8 degree straight C space equals space 265 straight K
Heat space to space be space removed comma space straight Q subscript 2 space equals space 200 kcal 
    The coefficient of performance of refrigerator is, 
    space space space straight beta equals fraction numerator straight T subscript 2 over denominator straight T subscript 1 minus straight T subscript 2 end fraction equals fraction numerator 265 over denominator 300 minus 265 end fraction equals 7.57 
    Also, straight beta equals straight Q subscript 2 over straight W 
    ∴ Work done by the compressor of refrigerator to remove 200k cal of heat is given by, 
    space space space space straight W equals straight Q subscript 2 over beta equals fraction numerator 200 over denominator 7.57 end fraction equals 26.420 kcal        

    Question 1556
    CBSEENPH11019828

    Find the time in which a layer of ice 3cm thick on the surface of pond will increase by 2mm when the temperature of air is –25°C. Take density of ice 900kg/m3and thermal conductivity 0.005cal/msK.

    Solution

    Initial thickness of layer of ice, x1 = 3 cm 
    Final thickness of ice, x2 = 3.2 cm
    Temperature of air, straight theta= -25o
    Density of ice, straight rho= 900 kg/m
    Thermal conductivity, K = 0.005 cal/msK 
    Time taken to freeze the lake from thickness to x1 is given by, 
    straight t space equals space fraction numerator ρL over denominator 2 Kθ end fraction left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis thin space

straight t space equals space fraction numerator 0.9 cross times 80 over denominator 2 cross times 0.005 cross times 25 end fraction open square brackets left parenthesis 3.2 right parenthesis squared minus left parenthesis 3 right parenthesis squared close square brackets space

space space equals fraction numerator 90 cross times 80 over denominator 25 end fraction open square brackets left parenthesis 3.2 right parenthesis squared minus left parenthesis 3 right parenthesis squared close square brackets space

space space space equals space 357.12 space straight s 

    Question 1557
    CBSEENPH11019829

    Find the quantity of heat rejected in the room per minute by a refrigerator fitted with compressor of power 240W. The temperature of space inside the refrigerator is -3°C and that of room is 27°C.

    Solution

    Given,
    Initial temperature, straight T subscript 1 space equals space 27 degree straight C space equals space 300 straight K
    Final temperature, straight T subscript 2 space equals space minus 3 degree straight C space equals space 270 straight K
    Power, P = 240 W
    The coefficient of performance of refrigerator, 
                   straight beta equals straight Q subscript 2 over straight W equals fraction numerator straight T subscript 2 over denominator straight T subscript 1 minus straight T subscript 2 end fraction 
    Therefore, 
    Quantity of heat removed from refrigerator per second is, 
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#6 {main}</pre> 
    Quantity of heat rejected to room per second,
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#6 {main}</pre> 
    Quantity of heat rejected to room per minute, 
    straight Q space equals space straight Q subscript 1 cross times 60 space equals space 1 comma 44 comma 000 space straight J

    Question 1558
    CBSEENPH11019830

    State and prove Kirchhoff’s law.

    Solution

    Kirchoff's law states that the ratio of the emissive power to absorptive power for the radiation of a given wavelength is constant for all substances, at the same temperature. This constant is equal to the emissive power of a perfectly black body at the same temperature and corresponding to the same wavelength. 

    Proof:
    Let eλ and aλ be the monochromatic emittance and absorptance of the body for wavelength straight lambda at temperature T.
    Let Q be the amount of radiation incident per unit area on the body in the wavelength range (λ–1/2) to (λ + 1/2 ).

    Therefore,
    Amount of radiation absorbed by body per second per unit area = aλ Q.
    The body is in thermal equilibrium with the enclosure.
    Therefore, the amount of radiations absorbed by body per sec per unit area in unit wavelength region is equal to the amount of radiations emitted by body per sec per unit area in the same unit wavelength region.
    straight i. straight e. space space space space space space space space straight a subscript straight lambda straight Q equals straight e subscript straight lambda
space space space space space space space space space space space space space space space space Q space equals space straight e subscript straight lambda over straight a subscript straight lambda 
    Therefore, for black body
    space space space space space space space space
space space space space space space space space space space space space space space straight a subscript straight lambda equals 1
space space therefore space space space space space space space space straight E subscript straight lambda over 1 equals straight Q space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 2 right parenthesis

    where,
    straight E subscript straight lambda is the emissive power of  black body. 
    From (1) and (2), we have,
     straight e subscript straight lambda over straight alpha subscript straight lambda equals straight E subscript straight lambda equals constant space space space space space space space... left parenthesis 3 right parenthesis, which is Kirchhoff's law.
    Also from (3), we have,
                 space space space space space space space space straight e subscript straight lambda equals straight E subscript straight lambda straight a subscript straight lambda space

rightwards double arrow space space space space straight e subscript straight lambda proportional to space straight a subscript straight lambda space 
    i.e. good absorbers of radiation are also good emitters of radiation. 
    Question 1559
    CBSEENPH11019831

    Discuss Ferry’s construction for a prefect black body.

    Solution
    Ferry’s black body consists of a hollow double-walled sphere which is blackened inside. The sphere has a small hole on its surface.
    Directly opposite to hole, there is a wedge shaped structure inside the sphere. 
    Ferry's black body is shown below, 
                            
    When the radiations of any wavelength enter the hole, they hit the wedge and suffer a number of reflection at the inner wall of sphere until the radiations are completely absorbed.
    Thus, the small hole acts as a black body absorber. To make it black body radiator, the cavity is heated.
    It is observed that radiations come out from the hole.
    Question 1560
    CBSEENPH11019832

    Plot a graph for the monochromatic emittance versus wavelength for the radiation emitted by black body. Discuss the interpretations made from graph.

    Solution
    The graph for emissive power eλ versus wavelength for the radiations from a black body at different temperature is shown below:

    Interpretations from this graph is as follows: 

    (i) At a given temperature, the energy distribution is not uniform in spectral radiations of a hot black body. 

    (ii) At a given temperature, emissive power increases first with increases in wavelength. Emissive power is maximum, at a particular wavelength and then it decreases with further increase in wavelength. 

    (iii) At every temperature, there exists one wavelength at which emissive power of the black body is maximum. This wavelength is called wavelength of maximum emission.

    (iv) With increase in temperature, the emissive power corresponding to each wavelength increases.

    (v) The wavelength of maximum emission decreases with the increase in temperature.

    (vi) The area under each curve represents the total energy emitted by black body per second per unit area.
    It is found that the area under the curve varies directly with fourth power of absolute temperature of black body.
    i.e. E ∝ T4.

    Question 1561
    CBSEENPH11019833

    The wavelength of maximum emission for a certain star in our galaxy is 1400Å. Find the temperature of star.

    Solution

    Here,
    Maximum wavelength,  straight lambda subscript max space equals space 140 straight Å 
    space straight lambda space equals space 1.4 space cross times space 10 to the power of negative 5 end exponent space cm squared space
    According to Wien's displacement law, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#6 {main}</pre> 
    rightwards double arrowstraight T space equals space straight b over straight lambda subscript max equals fraction numerator 0.2898 over denominator 1.4 space cross times space 10 to the power of negative 5 end exponent end fraction space equals space 20700 straight K, is the temperature of the star. 

    Question 1562
    CBSEENPH11019834

    State Stefen-Boltzmann’s law.

    Solution
    Stefan-Boltzmann law states that the total amount of heat radiated by a perfectly black body per second per unit area is directly proportional to the fourth power of absolute temperature.
    That is, 
    straight i. straight e. space space space space space space space space space space space space straight E space proportional to space straight T to the power of 4 space

rightwards double arrow space space space space space space space space space space space space space straight E equals σT to the power of 4 
    Question 1563
    CBSEENPH11019835

    State Wien’s displacement law.

    Solution
    Wien’s displacement law states that the product of wavelength of maximum emission and the absolute temperature of the body is constant.
    straight lambda subscript straight m straight T equals constant equals 0.2892 space cm space straight K 
    According to this law, the wavelength of maximum emission decreases with the increase in temperature of the body. 
    Question 1564
    CBSEENPH11019836

    State Newton’s law of cooling.

    Solution
    Newton's Law of Cooling states that the rate of loss of heat from a hot body is directly proportional to the temperature difference between the body and surrounding when the temperature difference is small. 
    That is, 
    i. e. space space space fraction numerator d T over denominator d t end fraction proportional to space left parenthesis T minus T subscript degree right parenthesis
    Question 1565
    CBSEENPH11019837

    A body of surface area A and emissivity e is heated to a temperature T and placed in an enclosure maintained at temperature T0 (T0<T). Find the net rate of loss of heat by the body by radiation.

    Solution
    According to Stefan’s law, at temperature T, the energy emitted per second per unit area by black body is,
                                  straight E equals σT to the power of 4
    Therefore heat energy emitted by any body at temperature T per second is given by, 
                              straight E subscript c equals eAσT to the power of 4 
    According to Prevost’s theory, due to temperature of enclosure, body absorbs the thermal radiation.
    Therefore, due to temperature T0 of enclosures, the heat energy absorbed by body per second is,
                        space space space space space straight E subscript straight a equals eAσT subscript 0 superscript 4
    Net heat lost by body at T placed in enclosure at temperature T is,
    negative dQ over dt equals straight E subscript straight c minus straight E subscript straight a equals eAσT to the power of 4 minus eAσT subscript 0 superscript 4 
             equals eAσ left parenthesis straight T to the power of 4 minus straight T subscript 0 superscript 4 right parenthesis 
    Negative sign mplies heat lost by the body.
    Question 1566
    CBSEENPH11019838

    Find the rate of energy radiated by a black body of radius 2.1 cm when heated at temperature 600K.
    Take space space space space straight sigma equals 5.67 cross times 10 to the power of negative 8 end exponent Wm to the power of negative 2 end exponent straight K to the power of negative 4 end exponent
                
                     
                
             

    Solution
    Energy radiated per second by black body is given by, 
                           space space space space space space space space straight E equals σAT to the power of 4 
    Given that,
    Stefan apostrophe straight s space constant comma space straight sigma space equals space 5.67 cross times 10 to the power of negative 8 end exponent Wm to the power of negative 2 end exponent straight K to the power of negative 4 end exponentArea space of space the space balck space body comma space straight A equals 4 πr squared equals 4 straight pi left parenthesis 2.1 right parenthesis squared cross times 10 to the power of negative 4 end exponent straight m squared space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 5.544 cross times 10 to the power of negative 3 end exponent straight m squared
Temperature comma space straight T space equals space 600 straight K
    Therefore, 
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#6 {main}</pre>       
                   equals 40.74 space straight J divided by straight s 
    This is the rate of energy radiated by a black body when heated at a temperature of 600 K. 

    Question 1567
    CBSEENPH11019839

    A spherical black body of 5cm radius is maintained at 227°C. At what wavelength the energy radiated is maximum?

    Solution
    According to Wien’s displacement law, 
    space space space space space space space straight lambda subscript max straight T space equals space 0.2898 space cm space straight K
    Given, 
    Temperature, straight T equals 227 degree straight C space equals space 500 straight K
    Therefore, the wavelength at which the energy radiated is given by, 
    straight lambda subscript max equals straight b over straight T equals fraction numerator 0.2898 over denominator 500 end fraction equals 5.8 cross times 10 to the power of negative 4 end exponent cm
space space space space space space space space space equals space 5.8 space micron
    Question 1568
    CBSEENPH11019840

    Two identical bodies A and B are heated to temperature 400K and 350K respectively. The room temperature is 27°C. Compare their rate of loss of heat.

    Solution

    Given that,
    Temperature space of space first space body space straight A comma space straight T subscript 1 space equals space 400 straight K

Temperature space of space body space straight B comma space straight T subscript 2 space equals space 350 straight K
    Initial space temperature comma straight T subscript straight o space equals space 27 degree straight C space equals space 300 straight K 
    According to Stefan's law, 
                     straight E proportional to left parenthesis straight T to the power of 4 minus straight T subscript straight o to the power of 4 right parenthesis 
    ∴          straight E subscript 1 over straight E subscript 2 equals fraction numerator straight T subscript 1 to the power of 4 minus straight T subscript straight o to the power of 4 over denominator straight T subscript 2 to the power of 4 minus straight T subscript straight o to the power of 4 end fraction 
                    equals fraction numerator left parenthesis 400 right parenthesis to the power of 4 minus left parenthesis 300 right parenthesis to the power of 4 over denominator left parenthesis 350 right parenthesis to the power of 4 minus left parenthesis 300 right parenthesis to the power of 4 end fraction equals 2.326
    This is the ratio of loss of heat. 
               

    Question 1569
    CBSEENPH11019841

    State and prove Newton’s law of cooling.

    Solution
    Newton's law of cooling states that the rate of fall of temperature of a hot body is directly proportional to the temperature difference between the body and surrounding when temperature difference is small. 
                             dT over dt proportional to left parenthesis straight T minus straight T subscript 0 right parenthesis 
    Proof:
    Let a body of mass m, surface area A and emissivity e be heated to a temperature T and placed in an enclosure maintained at temperature T0.
    The rate of loss of heat by radiation from body is,
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#6 {main}</pre>
    If dT is the change in temperature of the body due to loss of heat, then
    dQ space equals space msdT space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis
    where, 
    s is the specific heat of the body. 
    From equation (1) and (2), we have
    negative msdT over dt space equals space eAσ space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space
space space space space space space space space dT over dt space equals space eAσ over ms space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space space space... space left parenthesis 3 right parenthesis thin space

If space temperature space difference space left parenthesis straight T minus straight T subscript straight o right parenthesis space between space body space
and space surrounding space is space small. space

Then comma space

straight T to the power of 4 space minus space straight T subscript 0 to the power of 4 space equals space left parenthesis straight T squared space plus space straight T subscript straight o squared right parenthesis space left parenthesis straight T space plus space straight T subscript straight o right parenthesis space left parenthesis straight T minus straight T subscript straight o right parenthesis thin space

space space space space space space space space space space space space space space space space equals space left parenthesis 2 straight T subscript straight o squared right parenthesis space left parenthesis 2 space straight T subscript straight o right parenthesis space left parenthesis straight T minus straight T subscript straight o right parenthesis thin space

space space space space space space space space space space space space space space space space equals space 4 straight T subscript straight o cubed space left parenthesis straight T minus straight T subscript straight o right parenthesis thin space

Therefore comma space equation space left parenthesis 3 right parenthesis space reduces space to space

dT over dt space equals space minus space open parentheses fraction numerator 4 eAσ over denominator ms end fraction straight T subscript straight o cubed close parentheses space left parenthesis straight T minus straight T subscript straight o right parenthesis

dT over dt proportional to space left parenthesis straight T minus straight T subscript straight o right parenthesis space

Hence comma space the space result. space
    Question 1570
    CBSEENPH11019842

    The temperature of furnace is 2327°C and its intensity of emission is maximum at l.2μm. If from a distant star the radiations of wavelength 0.48μm are received, then what is the temperature of the star?

    Solution

    Given,
    Wavelength at which intensity of emission is maximum,  straight lambda subscript 1 space equals space 1.2 μm
    Wavelength of the distant star, straight lambda subscript 2 space equals space 0.48 space μm
     Temperature space of space the space furnace comma space straight T subscript 1 equals 2327 degree straight C space equals space 2600 straight K

Temperature space of space the space star comma space straight T subscript 2 equals ?
    According to Wien's displacement law, 
                      straight lambda subscript max straight T equals constant 
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#6 {main}</pre> 
    rightwards double arrow       1.2 cross times 2600 equals 0.48 straight T subscript 2 
    That is,
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#6 {main}</pre>

    Question 1571
    CBSEENPH11019843

    A body is found to cool from 100°C to 90°C in 5 minutes and from 90°C to 80°C in 6 minutes. Find the temperature of surrounding.

    Solution
    According to Newton’s law of cooling, 
                         dT over dt proportional to straight T subscript straight a minus straight T subscript straight o 
    where,
    Ta is the average temperature of body, and
    T0 is the temperature of surrounding.
    When the body cools from 100°C to 90°C, the fall in temperature is 10°C.
    Average temperature of the body is 95°C.  
    Therefore, 
                  10 over 5 proportional to 95 minus straight T subscript straight o              ...(1) 
    When the temperature of the body falls from 90°C to 80°C the average temperature of body is 85°C.
    Thus, 
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#6 {main}</pre>                ...(2) 
    Dividing (1) by (2), we get 
                     6 over 5 equals fraction numerator 95 minus straight T subscript straight o over denominator 85 minus straight T subscript straight o end fraction
    rightwards double arrow 510 minus 6 straight T subscript straight o space equals space 475 minus 5 straight T subscript straight o
 
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#6 {main}</pre>, is the temperature of the surrounding. 
    Question 1572
    CBSEENPH11019844

    A body initially at temperature 80°C takes 10 minutes to cool to temperature 72°C when the room temperature is 30°C. How long will it take to cool from 65°C to 60°C?

    Solution
    Time taken by the body to cool from temperature T1 to T2 when surrounding temperature is To is given by, 
              space space space space space space space space space space straight t space equals space kLn open parentheses fraction numerator straight T subscript 1 minus straight T subscript straight o over denominator straight T subscript 2 minus straight T subscript straight o end fraction close parentheses 
    Time taken to cool from 80°C to 72°C is 10 min is, 
    ∴    space space 10 equals kLn open parentheses fraction numerator 80 minus 30 over denominator 72 minus 30 end fraction close parentheses space equals straight k left parenthesis 0.17435 right parenthesis        ...(1) 
    Let t be the time taken to cool from 65°C to 60°C. 
    ∴   straight t equals kLn open parentheses fraction numerator 65 minus 30 over denominator 60 minus 30 end fraction close parentheses space equals space straight k left parenthesis 0.15415 right parenthesis         ...(2) 
    From (1) and (2),
             straight t over 10 equals fraction numerator 0.15415 over denominator 0.17435 end fraction

 
    rightwards double arrow          straight t equals 8.84 space min 
    Therefore, time taken to cool from 65o C to 60o C is 8.84 minutes. 
    Question 1573
    CBSEENPH11019845

    A solid sphere of radius R, made from a material of density ρ and specific heat s is heated to temperature T and placed in an enclosure at temperature T0 Show that the:

    (i) rate of loss of heat varies directly with square of radius and independent of specific heat and density of the material,

    (ii) rate of cooling varies inversely with radius, specific heat and density.

    Solution

    i) According to Stefan’s law, the rate of loss of heat by a body is, 
    negative space space space space dQ over dt equals space eAσ space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space
space space space space space space dQ over dt space equals space minus straight e 4 πR squared straight sigma space space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis thin space
rightwards double arrow space space dQ over dt equals space minus space left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis space straight R squared
    From the above relation, it is clear that the rate of loss of heat varies directly with the square of radius and independent of specific heat of material and density of material. 
    ii) Now, the rate of cooling varies inversely with radius, specific heat, and density. 
    space space space space space space space dQ over dt space equals space msdT over dt

rightwards double arrow space space dT over dt space equals space minus left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket straight R squared over ms

space space space space space space space space space space space space space space space equals space minus space left square bracket 4 πeσ left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket space fraction numerator 3 straight R squared over denominator 4 πR cubed ρs end fraction

space space space space space space space space space space space space space space equals space minus space left square bracket 3 eσ space left parenthesis straight T to the power of 4 space minus space straight T subscript straight o to the power of 4 right parenthesis right square bracket 1 over Rρs

rightwards double arrow space dT over dt space proportional to space 1 over Rρs
    Hence, the result. 

    Question 1574
    CBSEENPH11019846

    From the data given below calculate the black body temperature of the sun:
    straight sigma equals 5.67 cross times 10 to the power of negative 8 end exponent straight J divided by straight m squared sK to the power of 4 space space space space space space space space space space space space space space space straight S equals 1340 space straight J divided by straight m squared straight s
    Radius of the sun = 7 cross times 10 to the power of 8 straight m
    Radius of the earth's obrit equals 1.5 cross times 10 to the power of 11 straight m

    Solution

    The surface temperature of sun is given by, 
                      straight T equals fourth root of fraction numerator straight R squared straight S over denominator σr squared end fraction end root 
    Putting the values, we get 
                    Temperature comma space straight T equals fourth root of fraction numerator left parenthesis 1.5 cross times 10 to the power of 11 right parenthesis squared 1340 over denominator left parenthesis 5.67 cross times 10 to the power of negative 8 end exponent right parenthesis left parenthesis 7 cross times 10 to the power of 8 right parenthesis squared end fraction end root space

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 570 straight K

    Question 1575
    CBSEENPH11019847

    At what temperature, pressure remaining same, will the r.m.s velocity of H2 gas molecules be double it's value at N.T.P?

    Solution

    Let CHt and CHO be the r.m.s velocity of H2 gas at a different temperature.
    That is, 
    fraction numerator straight C subscript straight H subscript straight t end subscript space over denominator straight C subscript straight H subscript straight O end subscript end fraction equals square root of fraction numerator straight t space plus space 273 over denominator 273 end fraction end root

rightwards double arrow space 2 space equals space square root of fraction numerator straight t space plus space 273 over denominator 273 end fraction end root

On space squaring space both space sides comma space

4 space equals space fraction numerator straight t space plus space 273 over denominator 273 end fraction

That space is comma space

1092 space minus space 273 space equals space straight t

rightwards double arrow space straight t space equals space 819 to the power of straight o space straight C space

    Question 1576
    CBSEENPH11019848

    Derive an expression for the time taken to cool a body from temperature T1 to T2placed in an enclosure at temperature T

    Solution
    According to Newton’s law of cooling,
    space space space space fraction numerator space dT over denominator dt end fraction equals negative open parentheses 4 eAσ over ms straight T subscript 0 superscript 3 close parentheses left parenthesis straight T minus straight T subscript 0 right parenthesis

rightwards double arrow space space space space space dt equals negative open parentheses 4 eAσ over ms straight T subscript 0 superscript 3 close parentheses fraction numerator dT over denominator straight T minus straight T subscript 0 end fraction

    Let it take time 't' for the temperature to fall from T1 to T2.
    On integrating the above equation, 
    integral subscript 0 superscript straight t dT space equals space minus open parentheses 4 eAσ over ms straight T subscript straight o cubed close parentheses space integral subscript straight T subscript 1 end subscript superscript straight T subscript 2 end superscript space fraction numerator dT over denominator straight T space minus space straight T subscript straight o end fraction space

rightwards double arrow space straight t space equals space open parentheses 4 eAσ over ms straight T subscript straight o cubed close parentheses space Ln space open parentheses fraction numerator straight T subscript 1 space minus space straight T subscript straight o over denominator straight T subscript 2 space minus space straight T subscript straight o end fraction close parentheses

rightwards double arrow space straight t space proportional to space Ln space open parentheses fraction numerator straight T subscript 1 space minus space straight T subscript straight o over denominator straight T subscript 2 space minus space straight T subscript straight o end fraction close parentheses
     
    Question 1577
    CBSEENPH11019849

    Derive the expression for the surface temperature, of sun using solar constant.

    Solution
    Consider the sun to be a perfectly black body of radius r at temperature T.
    According to Stefan’s law, total radiant power of the sun i.e. total energy emitted by the sun in one second is,
    straight P subscript sun equals AσT to the power of 4 equals 4 πσr squared straight T to the power of 4 space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
    Let R be the mean distance between the sun and the earth,
    S be the mean value of solar constant.
    Then total radiant power of the sun is,
    space space space space space space space space space space space straight P subscript sun equals 4 πR squared straight S space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    From (1) and (2),
         4 πσr squared straight T to the power of 4 equals 4 πR squared straight S space

rightwards double arrow space space space space space space space straight T equals fourth root of fraction numerator straight R squared straight S over denominator σr squared end fraction end root
    Question 1578
    CBSEENPH11019850

    In an electric heater, the heat is continuously produced but its temperature becomes constant after some time. Why?

    Solution
    Heat is produced when current is passed in the electric heater. This increases the temperature of the coil. 
    Also, hot bodies radiate the heat and rate of loss of heat increases with the increase in temperature.
    i.e.,     dQ over dt proportional to open parentheses straight T subscript 2 superscript 4 minus straight T subscript 1 superscript 4 close parentheses
    Thus, heat is produced in the heater coil by current.
    At the same time it is radiating the heat by means of radiations.
    At some temperature when the rate of heat produced in electric heater due to current is just equal to the rate of loss of heat by radiation, at that temperature equilibrium is set up and temperature becomes constant. 
    Question 1579
    CBSEENPH11019851

    A solid sphere of copper of radius R and a hollow sphere of the same material of inner radius r and outer radius R are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster? Explain your answer.

    Solution
    We space know space that space rate space of space change space of space heat space is space given space by comma space

space space space space space space space space space space space space space space space space space dQ over dt equals ms dT over dt space

rightwards double arrow space space space space space space space space space space space space dT over dt equals 1 over ms dQ over dt
    In this case, the external radius of both the spheres is same.
    The surface areas area of both spheres are same.
    If the two spheres are heated to the same temperature and allowed to cool in the same environment, then the rate of loss of heat dQ/dt of both the spheres is same.
    Therefore the rate of cooling dT/dt is inversely proportional to mass of the sphere.
    The mass of the hollow sphere is less than that of solid sphere. Therefore, the rate of cooling of hollow sphere is faster than that of solid sphere.
    Question 1580
    CBSEENPH11019852

    Why the ice is packed in gunny bags?

    Solution

    Gunny bags have pores in them which will stop the flow of heat by the process of convection. Also,  gunny bags are bad conductors of heat. Therefore, ice is packed in gunny bags. 

    Question 1581
    CBSEENPH11019853

    A new quilt is warmer than an old one. Why?

    Solution

    A new quilt is warmer because the air which is trapped inside the cotton or woollen clothing acts as an insulator and does not allow the transfer of heat. However, when the quilt gets older, the air spacing of the cotton or wool gets compressed. It no longer remains a good insulator. Hence, the person does not feel warm. 

    Question 1582
    CBSEENPH11019854

    Why do Eskimos build double walled house of ice?

    Solution

    Eskimos make igloos which are double walled houses. Two walls are used so that the air which is trapped in between, prevent conduction of heat from inside of the house to outside of the house. Hence, people inside would feel warm. 

    Question 1583
    CBSEENPH11019855

    Pieces of iron and glass are heated to same temperature. Why does the piece of iron appear hotter than glass on touching?

    Solution

    The thermal conductivity of copper is about 400 times greater than that of glass. That is, the heat flowing through the copper moves about 400 times faster than it moves through the glass. 
    And the glass holds it's heat longer resulting in the slow transfer of heat to the skin. 
    Therefore, the piece of iron appear hotter than glass on touching. 

    Question 1584
    CBSEENPH11019856

    Calculate the molecular kinetic energy of 1 gram of helium( molecular weight 4) at 127°C.

    R= 8.31 J 

    Solution

    Here, we have
    M = 4 
    T = 127 + 273 
    R =8.31 J mole-1 K-1
    Average K.E per mole of helium = 3 over 2 space R space T

    Average K.E of 1 g of helium = 3 over 2 fraction numerator R T over denominator M end fraction 
                   = fraction numerator 3 cross times 8.31 cross times 400 over denominator 2 cross times 4 end fraction space equals space 1246.5 space J

    Question 1585
    CBSEENPH11019857

    Two tumblers have stuck together into each other. How will you loosen them using hot water and cold water? 

    Solution

    Two tumblers which are stuck together can be separated using hot and cold water. Firstly, fill the glass on the to with cold water. The cold temperature will make the glasses contract. With cold water in the top glass, pour hot water in the glass which is at the bottom. The heat will make the bottom glass expand. Spontaneous contraction and expansion of the glasses will pull them apart. 

    Question 1586
    CBSEENPH11019858

    Show that kinetic energy of molecule of gas is independent of the nature of gas but depends only on the temperature of gas molecule.

    Solution

    Consider 1 mole of an ideal gas at absolute temperature T, of volume V and molecular weight M. 
    Let N be the Avogadro's number and m be the mass of each molecule of the gas. 
    Then, 
    M = m X N 
    Let, C be the r.m.s velocity of the gas molecules, then pressure P exerted by ideal gas is, 
    P = 1 third M over V C squared   or PV = 1 third M C squared
    Now, using the gas equation
                         PV = RT 
    Therefore, 
    space space space space space space space space 1 third M C squared space equals space R T space

rightwards double arrow space space space 1 half space M C squared space equals space 3 over 2 space R T
 
    Therefore, average K.E. of translation of one mole of the gas = 1 half space M C squared space equals space 3 over 2 space R T
    i.e., 1 half mNC squared space equals space 3 over 2 space straight R space straight T space space space space space left square bracket space because space straight M space equals space mN right square bracket space

rightwards double arrow space 1 half mC squared space equals space 3 over 2 space open parentheses straight R over straight N close parentheses space straight T space equals space 3 over 2 kT space space space

where comma space straight k space is space called space Boltzman space constant. space

Therefore comma space

Average space straight K. straight E space per space molecule space of space straight a space
gas space equals space 1 half mC squared space equals space 3 over 2 straight K space straight T space
    That is, the average K.E of the gas is independent of the nature of the gas and is dependent only on the temperature of the gas. 

    Question 1587
    CBSEENPH11019859

    Why light gases like hydrogen and helium are absent in the earth’s atmosphere?

    Solution

    The gas molecules with more mass will stay close to the surface of the planet. And the light gases like hydrogen and helium will escape from the surface of the earth. Therefore, light gases are absent in the atmosphere of the earth. 

    Question 1588
    CBSEENPH11019860

    Why the chimney of a burning lamp cracks if a drop of cold water suddenly falls on it?

    Solution

    A glass is a brittle material. As soon as droplet comes in contact with hot chimney then the water droplet absorbs heat from glass surface to evaporate and thus creates a cooling effect on the glass. This rapid cooling of the region causes it to develop a crack and the glass shatters. 

    Question 1589
    CBSEENPH11019861

    What is the difference between the internal energy of real gas and an ideal gas?

    Solution

    The internal energy of an ideal gas is a function of temperature only and is independent of pressure and volume of the gas.  
    The internal energy of real gas depends on the pressure and volume of the gas. 

    Question 1590
    CBSEENPH11019862

    What are the relative advantages of mercury over other liquids for ordinary thermometric use?

    Solution

    Advantages of mercury over other liquids for ordinary thermometric use is given by: 
    1.The mercury is a good conductor of heat and expand rapidly and thus the temperature changes quickly.
    2.Mercury has a low heat capacity. 
    3.The mercury is opaque to light and can be seen clearly.
    4.The mercury does not stick with the walls of container. 
    5.Mercury has a freezing point of -39o C and boiling point of 357. 

     

    Question 1591
    CBSEENPH11019863

    Why Cp is greater than Cv ?

    Solution

    The heat capacity at constant pressure CP is greater than the heat capacity at constant volume CV , because when heat is added at constant pressure, the substance expands and work.

    Question 1592
    CBSEENPH11019967

    A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

    Solution
    Length of the steel wire, L1 = 4.7 m
    Area of cross-section of the steel wire, A1 = 3.0 × 10–5 m2

    Length of the copper wire, L2 = 3.5 m
    Area of cross-section of the copper wire, A2 = 4.0 × 10–5 m2

    Change in length = ΔL1 = ΔL2 = Δ

    Force applied in both the cases = F

    Young ’ straight s space modulus space of space the space steel space wire space is comma space

straight Y subscript 1 space end subscript equals space open parentheses fraction numerator straight F subscript 1 space end subscript over denominator space straight A subscript 1 end fraction close parentheses open parentheses straight L subscript 1 over ΔL subscript 1 close parentheses

space space space space space space equals open parentheses fraction numerator straight F over denominator 3 space straight X space 10 to the power of negative 5 end exponent end fraction close parentheses space open parentheses fraction numerator 4.7 over denominator ΔL end fraction close parentheses space space space space space space space space space... space left parenthesis straight i right parenthesis

Young apostrophe straight s space modulus space of space the space copper space wire comma space

straight Y subscript 2 space end subscript equals space open parentheses fraction numerator straight F subscript 2 space over denominator straight A subscript 2 end fraction close parentheses open parentheses fraction numerator straight L subscript 2 over denominator space ΔL subscript 2 end fraction close parentheses space

space space space space equals space open parentheses fraction numerator straight F over denominator 4 space cross times space 10 to the power of negative 5 end exponent end fraction close parentheses open parentheses fraction numerator 3.5 space over denominator space ΔL end fraction close parentheses space space space space space space space... space left parenthesis ii right parenthesis

    Now, dividing equation (1) by (2), we get
    straight Y subscript 1 over straight Y subscript 2 space equals space open parentheses fraction numerator 4.7 space cross times space 4 space cross times space 10 to the power of negative 5 end exponent over denominator 3 space cross times space 10 to the power of negative 5 end exponent space cross times space 3.5 end fraction close parentheses equals space 1.79 colon 1

The space ratio space of space young apostrophe straight s space modulus space of space steel space to space that
space of space copper space is space 1.79 colon 1.
    Question 1593
    CBSEENPH11019968

    Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?


    Solution
    (a)
    It is clear from the given graph that for stress 150 × 106N/m2, strain is 0.002.
    ∴ Young’s modulus, Y = Stress / Strain
                                = 150 × 106 / 0.002 
                                =  7.5 × 1010 Nm-2 

    Hence, Young’s modulus for the given material is 7.5 ×1010N/m2
    (b)
    The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.
    It is clear from the given graph,
    The approximate yield strength of this material = 300 × 106 Nm/2 
                                                                     =3 × 108 N/m2
    Question 1594
    CBSEENPH11019969

    The stress-strain graphs for materials A and B are shown in Fig. 9.12.


    The graphs are drawn to the same scale. 
    (a) Which of the materials has the greater Young’s modulus? 
    (b) Which of the two is the stronger material?

    Solution
    (a) From the two graph's we note that for a given strain, stress for A is more than that of B. Hence, Young's modulus which is equal to stress/strain, is greater for A than that of B.
    (b) A is stronger than B.
    The strength of a material is measured by the amount of stress required to cause a fracture, corresponding to the point of fracture.
    Question 1595
    CBSEENPH11019970

    Read the following statement below carefully and state, with reasons, if it is true or false.


    A.

    (a) The Young’s modulus of rubber is greater than that of steel. 

    Solution
    A. FALSE

    Tips: -

    Because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.
    Question 1596
    CBSEENPH11019971

    Read the following statement below carefully and state, with reasons, if it is true or false

    A.

    The stretching of a coil is determined by its shear modulus.

    Solution
    A. TRUE

    Tips: -

    The stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.
    Question 1597
    CBSEENPH11019972

    Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.


    Solution

    Given, 
    Diameter of the wires, d = 0.25 m
    Hence, the radius of the wires, r = d/2  =  0.125 cm 
    Length of the steel wire, L1 = 1.5 m 
    Length of the brass wire, L2 = 1.0 m 
    Total force exerted on the steel wire is, 
    F1 = (4 + 6) g
        = 10 × 9.8
        = 98 N 
    Young’s modulus for steel is given by, 
    straight Y subscript 1 space equals space open parentheses straight F subscript 1 over straight A subscript 1 close parentheses open parentheses fraction numerator increment straight L subscript 1 over denominator straight L subscript 1 end fraction close parentheses

where comma space
increment straight L subscript 1 space equals space change space in space the space length space of space the space steel space wire comma
straight A subscript 1 space equals space Area space of space cross minus section space of space the space steel space wire space equals space πr subscript 1 squared space

Young ’ straight s space modulus space of space steel comma space straight Y 1 space equals space 2.0 space cross times space 10 to the power of 11 space end exponent Pa

therefore space increment straight L subscript 1 space equals space fraction numerator space straight F subscript 1 space end subscript space cross times space straight L subscript 1 over denominator left parenthesis straight A subscript 1 space end subscript space cross times space straight Y subscript 1 right parenthesis end subscript end fraction

space space space space space space space space space space space space space equals fraction numerator 98 space space cross times space 1.5 over denominator open square brackets straight pi left parenthesis 0.125 space space cross times space 10 to the power of negative 2 end exponent right parenthesis to the power of 2 space space end exponent cross times space 2 space space cross times space 10 to the power of 11 close square brackets end fraction space

space space space space space space space space space space space space space equals 1.49 space space cross times space 10 to the power of negative 4 space end exponent straight m

Total space Force space on space the space brass space wire space is comma

straight F subscript 2 space equals space 6 space cross times space 9.8 space equals space 58.8 space straight N

Young apostrophe straight s space modulus space for space brass space is comma space

straight Y subscript 2 space equals space open parentheses straight F subscript 2 over straight A subscript 2 close parentheses open parentheses fraction numerator increment straight L subscript 2 over denominator straight L subscript 2 end fraction close parentheses

where comma space

increment straight L subscript 2 space equals space space change space in space the space length space of space the space brass space wire comma

straight A subscript 2 space equals space Area space of space cross minus section space of space the space brass space wire space equals space πr subscript 2 squared

therefore space increment straight L subscript 2 space equals space fraction numerator space straight F subscript 2 space cross times space straight L subscript 2 over denominator left parenthesis straight A subscript 2 space cross times space straight Y subscript 2 right parenthesis end subscript end fraction

space space space space space space space space space space space space space space equals fraction numerator 58.8 space straight X space 1 over denominator left square bracket space left parenthesis straight pi space space cross times space left parenthesis 0.125 space space cross times space 10 to the power of negative 2 end exponent right parenthesis squared space space cross times space left parenthesis 0.91 space space cross times space 10 to the power of 11 right parenthesis space right square bracket space end fraction

space space space space space space space space space space space space space equals space 1.3 space space cross times space 10 to the power of negative 4 end exponent space straight m

So comma space

Elongation space of space the space steel space wire space equals space 1.49 space cross times space 10 to the power of – 4 end exponent space straight m
Elongation space of space the space brass space wire space equals space 1.3 space cross times space 10 to the power of – 4 space end exponent straight m.

    Question 1598
    CBSEENPH11019973

    The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

    Solution
    Given, 
    Edge of the aluminium cube, L = 10 cm = 0.1 m
    The mass attached to the cube, m = 100 kg
    Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
    Shear space modulus comma space straight eta space equals space fraction numerator Shear space stress space over denominator Shear space strain end fraction equals fraction numerator open parentheses begin display style bevelled straight F over straight A end style close parentheses over denominator left parenthesis begin display style bevelled fraction numerator straight L over denominator increment straight L end fraction end style right parenthesis end fraction

where comma space

straight F space equals space Applied space force space equals space mg space equals space 100 cross times 9.8 space equals space 980 space straight N

straight A space equals space Area space of space one space space of space the space faces space of space the space cube
space
space space space space equals space 0.1 space cross times space 0.1 space equals space 0.01 space straight m squared

increment straight L space equals space vertical space deflection space of space the space cube. space

therefore space increment straight L space equals space FL over Aη equals fraction numerator 980 space cross times space 0.1 over denominator open square brackets 10 to the power of negative 2 end exponent space cross times space left parenthesis 25 space cross times space 10 to the power of 9 right parenthesis close square brackets end fraction

space space space space space space space space space space space space equals space 3.92 space cross times space 10 to the power of negative 7 space end exponent straight m space

The space vertical space deflection space of space this space face space of space the space cube space
is space 3.92 space cross times 10 to the power of – 7 end exponent space straight m.

    Question 1599
    CBSEENPH11019974

    Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

    Solution
    Given,
    Mass of the big structure, = 50,000 kg
    Inner radius of the column, r = 30 cm = 0.3 m
    Outer radius of the column, R = 60 cm = 0.6 m
    Young’s modulus of steel, Y = 2 × 1011 Pa
    Total force exerted, F = Mg = 50000 × 9.8 N
    Stress = Force exerted on a single column
            = 50000 × fraction numerator 9.8 over denominator 4 end fraction 
            =  122500 N 
    Young's modulus, Y = fraction numerator Stress space space over denominator Strain end fraction
    Strain space equals space fraction numerator open parentheses begin display style bevelled straight F over straight A end style close parentheses over denominator straight Y end fraction
where comma space

Area comma space straight A space equals space straight pi space left parenthesis straight R squared space – space straight r squared right parenthesis space equals space straight pi space left square bracket left parenthesis 0.6 right parenthesis to the power of 2 space end exponent – space left parenthesis 0.3 right parenthesis squared right square bracket space

Strain space equals space fraction numerator 122500 over denominator open curly brackets space straight pi space open square brackets left parenthesis 0.6 right parenthesis to the power of 2 space end exponent – space left parenthesis 0.3 right parenthesis squared space cross times space 2 cross times 10 to the power of 11 close square brackets close curly brackets end fraction space

space space space space space space space space space space space equals 7.22 space cross times space 10 to the power of negative 7 space end exponent

Hence comma space the space compressional space strain space of space each space
column space is space 7.22 space cross times space 10 to the power of negative 7 space end exponent.

    Question 1600
    CBSEENPH11019975

    A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

    Solution
    Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
    Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
    Area of the copper piece, A = l × b
                 
                         = 19.1 × 10–3 × 15.2 × 10–3

                                       = 2.9 × 10–4 m2

    Tension force applied on the piece of copper, F = 44500 N
    Modulus of elasticity of copper, η = 42 × 109 N/m2

    Modulus space of space elasticity comma space straight eta space equals space Stress over strain

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals bevelled fraction numerator open parentheses begin display style straight F over straight A end style close parentheses over denominator Strain end fraction space
rightwards double arrow space Strain space space equals space straight F over Aη
space space space space space space space space space space space space space space space space space space equals fraction numerator space 44500 space over denominator left parenthesis 2.9 space cross times space 10 to the power of negative 4 end exponent space cross times space 42 space cross times space 10 to the power of 9 right parenthesis end fraction space

space space space space space space space space space space space space space space space space space equals space 3.65 space cross times space 10 to the power of negative 3 end exponent


    Question 1601
    CBSEENPH11019976

    A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10N m–2, what is the maximum load the cable can support?

    Solution
    Radius of the steel cable, r = 1.5 cm = 0.015 m
    Maximum allowable stress = 108 N m–2

    Maximum stress = fraction numerator Maximum space force over denominator Area space of space cross minus section end fraction = 10 to the power of 8 space cross times space straight pi space left parenthesis 0.015 right parenthesis squared
                                                               = 7.065 × 104 N
    Hence, the cable can support the maximum load of 7.065 × 104 N.
    Question 1602
    CBSEENPH11019977

    A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

    Solution

    Tension force acting on each wire is the same. Hence, the extension is the same for each wire. 
    Since, the wires are of the same length, the strain will also be same.
    Young's modulus is given by,
     straight Y space equals space Stress over Strain space equals space fraction numerator begin display style bevelled straight F over straight A end style over denominator Strain end fraction space equals space fraction numerator begin display style bevelled fraction numerator 4 straight F over denominator πd squared end fraction end style over denominator Strain end fraction space space space space space space space space space space... space left parenthesis 1 right parenthesis

where comma space
straight F space equals space Tension space force
straight A space equals space Area space of space cross minus section
space straight d space equals space Diameter space of space the space wire

From space equation space left parenthesis straight i right parenthesis comma space we space have

straight Y proportional to open parentheses 1 over straight d squared close parentheses

Young apostrophe straight s space modulus space for space iron comma space straight Y subscript 1 space equals space 190 space cross times space 10 to the power of 9 space Pa
Diameter space of space the space iron space wire space equals space straight d subscript 1 space end subscript
Young ’ straight s space modulus space for space copper comma space straight Y subscript 2 space equals space 120 space cross times space 10 to the power of 9 space Pa space
Diameter space of space the space copper space wire space equals space straight d subscript 2

Therefore comma space

Ratio space of space their space diameters space is space given space by comma space

straight d subscript 1 over straight d subscript 2 space equals space square root of straight Y subscript 1 over straight Y subscript 2 end root space equals space square root of fraction numerator 190 cross times 10 to the power of 9 over denominator 120 cross times 10 to the power of 9 end fraction end root space equals square root of 19 over 12 end root 

    Question 1603
    CBSEENPH11019978

    A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

    Solution
    Mass, m = 14.5 kg
    Length of the steel wire, l = 1.0 m
    Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
    Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
    Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
    When the mass is placed at the position of the vertical circle, the total force on the mass is given by, 
    F = mg + mlω2
     
      = 14.5 × 9.8 + 14.5 × 1 × (12.56)2

      = 2429.53 N
    Young apostrophe straight s space modulus comma space straight Y space equals space Stress over Strain

space space space space space space space space space space space straight Y space equals space fraction numerator open parentheses begin display style bevelled straight f over straight A end style close parentheses over denominator open parentheses begin display style bevelled fraction numerator increment straight l over denominator straight L end fraction end style close parentheses end fraction

therefore space space increment straight l space equals space fraction numerator straight F space straight l space over denominator straight A space straight Y end fraction space
Young ’ straight s space modulus space for space steel space equals space 2 space cross times space 10 to the power of 11 space Pa

increment straight l space equals space 2429.53 space cross times fraction numerator 1 over denominator left parenthesis 0.065 space cross times space 10 to the power of negative 4 end exponent space cross times space 2 space cross times space 10 to the power of 11 right parenthesis space end fraction space

space space space space space space equals space 1.87 space cross times space 10 to the power of negative 3 end exponent space straight m space

Hence comma space the space elongation space of space the space wire space equals space 1.87 space cross times space 10 to the power of – 3 space end exponent straight m.
    Question 1604
    CBSEENPH11019979

    Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

    Solution
    Initial volume, V= 100.0 l = 100.0 × 10 –3 m3
    Final volume, V= 100.5 l = 100.5 ×10 –3 m3 
    Increase in volume, ΔV = V2 – V= 0.5 × 10–3 m

    Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10Pa
    Bulk space modulus space equals space fraction numerator increment straight p over denominator left parenthesis begin display style bevelled fraction numerator increment straight V over denominator straight V subscript 1 end fraction end style right parenthesis end fraction space equals increment straight p cross times fraction numerator straight V subscript 1 over denominator increment straight V end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 100 space cross times space 1.013 space cross times space 10 to the power of 5 space end exponent cross times space 100 space cross times space 10 to the power of negative 3 end exponent over denominator left parenthesis 0.5 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 9 space Pa

Bulk space modulus space of space air space equals space 1 space cross times space 10 to the power of 5 space end exponent Pa

Therefore comma space

fraction numerator Bulk space modulus space of space water over denominator space Bulk space modulus space of space air end fraction space equals space fraction numerator 2.026 space cross times space 10 to the power of 9 over denominator left parenthesis 1 space cross times space 10 to the power of 5 right parenthesis end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.026 space cross times space 10 to the power of 4 space

This space ratio space is space very space high space because space air space is space
more space compressible space than space water.
    Question 1605
    CBSEENPH11019980

    What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103kg m–3?

    Solution
    Let the given depth be h.
    Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105Pa
    Density of water at the surface, ρ= 1.03 × 103 kg m–3

    Let ρ2 be the density of water at the depth h.
    Let V1 be the volume of water of mass m at the surface.
    Let V2 be the volume of water of mass m at the depth h.
    Let ΔV be the change in volume.
                       ΔV = V1 - V2
    straight m open square brackets open parentheses 1 over straight rho subscript 1 close parentheses minus open parentheses 1 over straight rho subscript 2 close parentheses close square brackets

Therefore comma space Volumetric space strain space equals space ΔV over straight V subscript 1 space
space space space space space space space space space space space equals straight m open square brackets open parentheses 1 over straight rho subscript 1 close parentheses minus open parentheses 1 over straight rho subscript 2 close parentheses close square brackets cross times open parentheses straight rho subscript 1 over straight m close parentheses space

rightwards double arrow space ΔV over straight V subscript 1 space equals space 1 space minus space open parentheses straight rho subscript 1 over straight rho subscript 2 close parentheses space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

Bulk space modulus comma space straight B space equals space fraction numerator ρV subscript 1 over denominator increment straight V end fraction
fraction numerator ΔV space over denominator space straight V subscript 1 end fraction space equals space straight p over straight B

Compressibilty space of space water space equals open parentheses 1 over straight B close parentheses space equals space 45.8 cross times 10 to the power of negative 11 end exponent space Pa to the power of negative 1 end exponent
Therefore comma space

fraction numerator ΔV space over denominator space straight V subscript 1 end fraction equals space 80 space cross times space 1.013 space cross times space 10 to the power of 5 space cross times space 45.8 space cross times space 10 to the power of negative 11 space end exponent

space space space space space space space space space equals space 3.71 space cross times space 10 to the power of negative 3 end exponent space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

From space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get

space 1 space minus space open parentheses straight rho subscript 1 over straight rho subscript 2 close parentheses space equals space 3.71 space cross times space 10 to the power of negative 3 end exponent
rightwards double arrow space straight rho subscript 2 space equals space fraction numerator 1.03 space cross times space 10 cubed over denominator space 1 space minus space left parenthesis 3.71 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction

space space space space space space space space space equals space 1.034 space cross times space 10 cubed space kg space straight m to the power of negative 3 end exponent space

Therefore comma space the space density space of space water space at space the space given space depth space left parenthesis straight h right parenthesis space
is space 1.034 space cross times space 10 to the power of 3 space end exponent kg space straight m to the power of – 3 end exponent.

    Question 1606
    CBSEENPH11019981

    Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

    Solution
    Hydraulic pressure exerted on the glass slab, = 10 atm = 10 × 1.013 × 105 Pa
    Bulk modulus of glass, B = 37 × 109 Nm–2

    Bulk modulus, B =fraction numerator straight p over denominator begin display style bevelled fraction numerator increment straight V over denominator straight V end fraction end style end fraction
    where, 
    V/V = Fractional change in volume
    ∴ ∆V/V = p / B

              =fraction numerator space 10 space cross times space 1.013 space cross times space 10 to the power of 5 over denominator left parenthesis 37 space cross times space 10 to the power of 9 right parenthesis end fraction
              = 2.73 × 10-5 
    Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.
    Question 1607
    CBSEENPH11019982

    Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×106 Pa.

    Solution
    Length of an edge of the solid copper cube, l = 10 cm = 0.1 m
    Hydraulic pressure, p = 7.0 × 106 Pa
    Bulk modulus of copper, B = 140 × 109 Pa
    Formula is given by, 
    Bulk space modulus comma space straight B space equals space fraction numerator straight p over denominator open parentheses begin display style bevelled fraction numerator increment straight V over denominator straight V end fraction end style close parentheses end fraction

where comma space
fraction numerator increment straight V over denominator straight V end fraction equals space Volumetric space strain
increment straight V space equals space change space in space volume space equals space pV over straight B
straight V space equals space original space volume

Original space volume space of space the space cube comma space straight V space equals space straight l cubed

therefore space increment straight V space equals space pl cubed over straight B
space space space space space space space space space space space space equals fraction numerator space 7 space cross times space 10 to the power of 6 space cross times space left parenthesis 0.1 right parenthesis cubed space over denominator left parenthesis 140 cross times 10 to the power of 9 right parenthesis end fraction
space space space space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 8 end exponent space straight m cubed

space space space space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 2 end exponent space cm to the power of negative 3 end exponent space

Therefore comma space space the space volume space contraction space of space the space
solid space copper space cube space is space 5 space cross times space 10 to the power of – 2 end exponent space cm to the power of – 3 end exponent.
    Question 1608
    CBSEENPH11019983

    How much should the pressure on a litre of water be changed to compress it by 0.10%?

    Solution
    Volume of water, = 1 L
    It is given that water is to be compressed by 0.10%.
    Therefore,
    Fractional change, ∆V / V = 0.1 / (100 × 1)  =  10-3

    Bulk modulus, B = ρ / (∆V/V
    rightwards double arrow              ρ = B × (∆V/V)
    Bulk modulus of water, B = 2.2 × 109 Nm-2

                                  ρ = 2.2 × 109 × 10-3  =  2.2 × 106 Nm-2 

    Therefore, the pressure on water should be 2.2 ×106 Nm–2
    Question 1609
    CBSEENPH11019984

    Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?


    Solution
    Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10–3 m
    Radius, r = d/2  =  0.25 × 10-3 m 
    Compressional force, F = 50000 N\
    Pressure at the tip of the anvil is given by, 

    straight P space space equals space straight F over straight A equals space fraction numerator 50000 over denominator straight pi space left parenthesis 0.25 cross times 10 to the power of negative 3 end exponent right parenthesis squared end fraction space

space space space space equals space 2.55 cross times 10 to the power of 11 space Pa space
    Therefore, the pressure at the tip of the anvil is 2.55 × 1011Pa. 
    Question 1610
    CBSEENPH11019985

    A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass be suspended in order to produce (a) equal stresses and b) equal strains in both steel and aluminium wires.


    Solution

    Given, 
    Cross-sectional area of wire A = a1 = 1.0 mm= 1.0 × 10–6m2
    Cross-sectional area of wire Ba2 = 2.0 mm2 = 2.0 × 10–6m2
    Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
    Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2

    a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
    Stress in the wire = Force / Area  =  F / a

    If the two wires have equal stresses, then 

    F1 / a1  =  F2 / a
    where,

    F1 = Force exerted on the steel wire
    F2 = Force exerted on the aluminum wire
    rightwards double arrow space straight F subscript 1 over straight F subscript 2 space equals space straight a subscript 1 over straight a subscript 2 space equals space 1 half space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis

Now comma space taking space torque space about space the space point space of space suspension comma space

fraction numerator left parenthesis 1.05 space minus space straight y right parenthesis over denominator straight y end fraction space equals space 1 half space
2 left parenthesis left parenthesis 1.05 space minus space straight y right parenthesis space equals space straight y

straight y space equals space 0.7 space straight m
    In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.
    b) Young apostrophe straight s space modulus space equals space Stress over strain
Strain space equals space fraction numerator Stress over denominator Young apostrophe straight s space Modulus end fraction equals fraction numerator begin display style bevelled fraction numerator left parenthesis straight F over denominator straight a end fraction end style right parenthesis over denominator straight Y end fraction

If space the space strain space in space the space two space wires space is space equal comma space then

fraction numerator bevelled fraction numerator left parenthesis straight F subscript 1 over denominator straight a subscript 1 end fraction right parenthesis over denominator straight Y subscript 1 end fraction space equals space fraction numerator open parentheses begin display style bevelled straight F subscript 2 over straight a subscript 2 end style close parentheses over denominator straight Y subscript 2 end fraction space

rightwards double arrow space straight F subscript 1 over straight F subscript 2 space equals space fraction numerator straight a subscript 1 straight Y subscript 1 over denominator straight a subscript 2 straight Y subscript 2 end fraction space

rightwards double arrow space straight a subscript 1 over straight a subscript 2 space equals space 1 half space
rightwards double arrow straight F subscript 1 over straight F subscript 2 space equals space open parentheses 1 half close parentheses cross times open parentheses fraction numerator 2 cross times 10 to the power of 11 over denominator 7 space cross times space 10 to the power of 10 end fraction close parentheses space equals space 10 over 7 space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis space

Taking space torque space about space the space point space where space mass space straight m subscript 1 space
is space suspended space at space straight a space distance space straight y subscript 1 space from space the space side space where space
wire space straight A space attached comma space we space get

straight F subscript 1 straight y subscript 1 space equals space straight F subscript 2 space end subscript left parenthesis 1.05 space – space straight y subscript 1 right parenthesis space

straight F subscript 1 over straight F subscript 2 space equals space fraction numerator left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis over denominator space straight y subscript 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis

Using space equations space left parenthesis ii right parenthesis thin space and space left parenthesis iii right parenthesis comma space we space get
fraction numerator left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis over denominator space straight y subscript 1 end fraction space equals space 10 over 7
rightwards double arrow space 7 left parenthesis 1.05 space minus space straight y subscript 1 right parenthesis space space equals space space 10 straight y subscript 1 space

rightwards double arrow space straight y subscript 1 space equals space 0.432 space straight m
    In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached. 
    Question 1612
    CBSEENPH11019987

    Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.

    Solution
    Diameter of the metal strip, d = 6.0 mm = 6.0 × 10–3 m
    Radius, r = d/2 = 3 × 10-3 m 
    Maximum shearing stress = 6.9 × 107 Pa 
    Maximum stress = fraction numerator Minimum space load space or space force over denominator Area end fraction
    Maximum force = Maximum stress × Area 
                         = 6.9 × 107 × π × (r) 2 

                         = 6.9 × 107 × π × (3 ×10–3)

                         = 1949.94 N 
    Each rivet carries one quarter of the load. 
    ∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N.
    Question 1613
    CBSEENPH11019988

    The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

    Solution
    Water pressure at the bottom, = 1.1 × 108 Pa 
    Initial volume of the steel ball, V = 0.32 m

    Bulk modulus of steel, B = 1.6 × 1011 Nm–2 

    The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface. 
    Let the change in the volume of the ball on reaching the bottom of the trench be ΔV
    Bulk modulus, B = p / (∆V/V
                    ∆V  =  B / pV 

                         = 1.1 × 108 × 0.32 / (1.6 × 1011 ) 
                         =  2.2 × 10-4 m

    Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3
    Question 1614
    CBSEENPH11019989

    Explain why:

    (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.

    Solution
    Near the sea level, density of air is the maximum. With increase in height from the surface, density of air decreases. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.
    Question 1615
    CBSEENPH11019990

    Explain why

    (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.

    Solution
    The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (θ).
    At the line of contact, the surface forces between the three media must be in equilibrium.
    i.e., cos θ = (Ssa - Sla) / Sla 
    where,
    sla is the liquid-air interface, 
    Ssa is the solid-air interface
    The angle of contact θ , is obtuse if Ssa < Sla (as in the case of mercury on glass). This angle is acute if Ssl < Sla (as in the case of water on glass).

    Question 1616
    CBSEENPH11019991

    Explain why?

    (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)

    Solution

    Mercury molecules, which make an obtuse angle with glass have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.
    On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.

    Question 1617
    CBSEENPH11019992

    Expalin why?

    (c) Surface tension of a liquid is independent of the area of the surface

    Solution

    Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

    Question 1618
    CBSEENPH11019993

    Explain why:

    (d) Water with detergent dissolved in it should have small angles of contact.

    Solution
    Water with detergent dissolved in it has small angles of contact (θ). This is because for a small θ, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (θ). If θ is small, then cosθ will be large and the rise of the detergent water in the cloth will be fast.
    Question 1619
    CBSEENPH11019994

    Explain why?

    (e) A drop of liquid under no external forces is always spherical in shape

    Solution
    A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take a spherical shape.
    Question 1620
    CBSEENPH11019995

    Explain why

    (a) To keep a piece of paper horizontal, you should blow over, not under it. 

    Solution

    When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal. 

    Question 1621
    CBSEENPH11019996

    Explain why:

    (b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers

    Solution

    According to equation of continuity, 
    Area cross times space velocity space equals space constant

    For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other. Therefore velocity of water increases. 

    Question 1622
    CBSEENPH11019997

    Explain why:

    (c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection

    Solution

    The small opening of a syringe needle controls the velocity of the blood oozing out. This is as a result of the equation of continuity.
    At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.

    Question 1623
    CBSEENPH11019998

    Explain why:

    (d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

    Solution

    When a fluid flows out from a small hole in a vessel receives a backward thrust.
    A fluid flowing out from a small hole has a large velocity according to the equation of continuity.
    Area × Velocity = Constant
    According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system

    Question 1624
    CBSEENPH11019999

    Explain why:

    (e) A spinning cricket ball in air does not follow a parabolic trajectory.

    Solution

    A spinning cricket ball has two simultaneous motions types of motion- rotatory and linear.
    These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. Therefore, the ball does not follow a parabolic path. 

    Question 1625
    CBSEENPH11020000

    A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

    Solution
    The maximum allowable stress for the structure, P = 109 Pa
    Depth of the ocean, d = 3 km = 3 × 103 m
    Density of water, ρ = 103 kg/m

    Acceleration due to gravity, g = 9.8 m/s

    The pressure exerted because of the sea water at depth, ρdg 

                                                                               = 3 × 103 × 103 × 9.8 
                                                                               = 2.94 × 107 Pa 
    The maximum allowable stress for the structure (109 Pa) is greater than the pressure of the sea water (2.94 × 107 Pa).
    The pressure exerted by the ocean is less than the pressure that the structure can withstand.
    Hence, the structure is suitable for putting up on top of an oil well in the ocean.
    Question 1626
    CBSEENPH11020001

    A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000
    kg. The area of cross-section of the piston carrying the load is 425 cm2. What
    maximum pressure would the smaller piston have to bear ?

    Solution
    The maximum mass of a car that can be lifted, m = 3000 kg
    Area of cross-section of the load-carrying piston, A = 425 cm= 425 × 10–4 m2

    The maximum force exerted by the load, F = m
                                                            = 3000 × 9.8
                                                            = 29400 N 
    The maximum pressure exerted on the load-carrying piston, P = F/A
                                                                                    =fraction numerator 29400 over denominator 425 cross times 10 to the power of negative 4 end exponent end fraction
                                                                                    = 6.917 × 105 Pa 
    Pressure is transmitted equally in all directions in a liquid.
    Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 105 Pa.
    Question 1627
    CBSEENPH11020002

    A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit? 

    Solution
    Height of the spirit column, h1 = 12.5 cm = 0.125 m
    Height of the water column, h2 = 10 cm = 0.1 m
    P0 = Atmospheric pressure,
    ρ1 = Density of spirit,
    ρ2 = Density of water,
    Pressure at point B = P0 + ρ1h1g
    Pressure at point D = P0 + ρ2h2g
    Pressure at points B and D is the same. 
                                 P0 + ρ1h1g  =  P0 + ρ2h2g
                                      ρ1 / ρ2  =  h2 / h

                                   10 / 12.5  =  0.8 
    Therefore, the specific gravity of spirit is 0.8.
    Question 1628
    CBSEENPH11020003

    Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

    Solution
    Length of the horizontal tube, l = 1.5 m
    Radius of the tube, r = 1 cm = 0.01 m
    Diameter of the tube, d = 2r = 0.02 m

    Glycerine is flowing at a rate of 4.0 × 10–3 kg s–1.
    Mass of the substance, M = 4.0 × 10–3 kg s–1
    Density of glycerine, ρ = 1.3 × 103 kg m–3
    Viscosity of glycerine, η = 0.83 Pa s
    Volume of glycerine flowing per sec,
    V = straight M over straight rho space equals space space fraction numerator 4 space cross times space 10 to the power of negative 3 space end exponent over denominator left parenthesis 1.3 space cross times space 10 cubed right parenthesis end fraction equals 3.08 space cross times space 10 to the power of negative 6 end exponent space m cubed s to the power of negative 1 end exponent
    According to Poiseville’s formula, we have the relation for the rate of flow,
    Error converting from MathML to accessible text.


    Question 1629
    CBSEENPH11020004

    In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s–1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.

    Solution
    Speed of wind on the upper surface of the wing, V1 = 70 m/s
    Speed of wind on the lower surface of the wing, V2 = 63 m/s
    Area of the wing, A = 2.5 m2

    Density of air, ρ = 1.3 kg m–3
    According to Bernoulli's Theorem, 

    straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space space plus space 1 half ρV subscript 2 squared space
space space space space space space space space space space straight P subscript 2 space space minus straight P subscript 1 space equals space 1 half straight rho space left parenthesis straight V subscript 1 squared space minus space straight V subscript 2 squared right parenthesis

where comma space

straight P subscript 1 space equals space Pressure space on space the space upper space surface space of space the space wing

straight P subscript 2 space equals space Pressure space on space the space lower space surface space of space the space wing 
    Pressure space difference space between space the space upper space and
lower space of space the space wing space provides space lift space to space the space aeroplane. space

Lift space on space the space wing space equals space left parenthesis straight P subscript 2 space minus straight P subscript 1 right parenthesis space straight A space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space bevelled 1 half space straight rho space space left parenthesis straight V subscript 1 squared space minus space straight V subscript 2 squared right parenthesis space straight A space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space bevelled 1 half space cross times 1.3 cross times left square bracket space 70 squared space minus space 63 squared space right square bracket space cross times space 2.5 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1512.87 space straight N space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.51 space cross times space 10 to the power of 3 space end exponent straight N

Therefore comma space the space lift space on space the space wing space of space the space aeroplane space
is space 1.51 space cross times space 10 cubed space straight N.
    Question 1630
    CBSEENPH11020005

    Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?




    Solution

    From the law of continuity, we have
       
                     straight A subscript 1 straight V subscript 1 space equals space straight A subscript 2 straight V subscript 2
    A= Area of pipe 1,
    A2 = Area of pipe 2, 
    V1 = Speed of fluid in pipe 1, and
    V2 = Speed of fluid in pipe 2.
    When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more.
    According to Bernoulli's principle, if velocity is more, pressure is less. 
    Pressure is directly proportional to height.
    Hence, the level of water in pipe 2 is less. Therefore, figure (a) is not possible.


    Question 1631
    CBSEENPH11020006

    The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
     

    Solution
    Area of cross-section of the spray pump, A1 = 8 cm= 8 × 10–4 m2

    Number of holes, n = 40
    Diameter of each hole, d = 1 mm = 1 × 10–3 m
    Radius of each hole, r = d/2 = 0.5 × 10–3 m
    Area of cross-section of each hole, = πr= π (0.5 × 10–3)2 m2

    Total area of 40 holes, An × a

                                     = 40 × π (0.5 × 10–3)2 m2

                                     = 31.41 × 10–6 m2

    Speed of flow of liquid inside the tube, V= 1.5 m/min = 0.025 m/s
    Speed of ejection of liquid through the holes = V2

    According to the law of continuity, we have 
                              A1V1 = A2V

                                 V2 = A1V1 / A

                                     = 8 × 10-4 × 0.025 / (31.61 × 10-6
                                     = 0.633 m/s 
    Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.
    Question 1632
    CBSEENPH11020007

    A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

    Solution
    The weight that the soap film supports, W = 1.5 × 10–2 N
    Length of the slider, l = 30 cm = 0.3 m
    A soap film has two free surfaces.
    ∴ Total length = 2l = 2 × 0.3 = 0.6 m 
    Surface tension, S = Force or Weight / 2l
                            = 1.5 × 10-2 / 0.6
                            = 2.5 × 10-2 N/m 
    Therefore, the surface tension of the film is 2.5 × 10–2 N m–1
    Question 1633
    CBSEENPH11020008

    Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.



    Solution
    Take case (a):
    The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm
    The weight supported by the film, W = 4.5 × 10–2 N
    A liquid film has two free surfaces.
    ∴ Surface tension = W / 2l
                           = 4.5 × 10-2 / (2 × 0.4)
                           = 5.625 × 10-2 N/m
    In all the three figures, the liquid is the same.
    Temperature is also the same for each case.
    Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a).
    i.e., 5.625 × 10–2 N m–1.
    Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10–2 N. 
    Question 1634
    CBSEENPH11020009

    What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

    Solution
    Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
    Surface tension of mercury, S = 4.65 × 10–1 N m–1

    Atmospheric pressure, P0 = 1.01 × 105 Pa
    Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
    equals space fraction numerator 2 straight S over denominator straight r space plus space straight P subscript straight o end fraction

equals space fraction numerator 2 space cross times space 4.65 space cross times space 10 to the power of negative 1 space end exponent over denominator left parenthesis 3 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction plus space 1.01 space cross times 10 to the power of 5 space

equals space 1.0131 space cross times space 10 to the power of 5

equals 1.01 space cross times space 10 to the power of 5 space end exponent Pa

Excess space pressure space equals space fraction numerator 2 straight S over denominator straight r end fraction

equals fraction numerator 2 space cross times space 4.65 space cross times space 10 to the power of negative 1 end exponent over denominator left parenthesis 3 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction space equals space 310 space Pa

    Tips: -

     
    Question 1636
    CBSEENPH11020011

    A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door close.

    Solution
    Base area of the given tank, A = 1.0 m2
    Given,
    Area of the hinged door, a = 20 cm= 20 × 10–4 m2

    Density of water, ρ1 = 103 kg/m3

    Density of acid, ρ2 = 1.7 × 103 kg/m3

    Height of the water column, h1 = 4 m
    Height of the acid column, h2 = 4 m
    Acceleration due to gravity, g = 9.8
    Pressure due to water is given using the relation, 
              P1 = h1ρ1
                  = 4 × 103 × 9.8 
                  =  3.92 × 104 Pa

    Pressure due to acid is given by the relation, 
    P2 = h2ρ2

        = 4 × 1.7 × 103 × 9.8 
        =  6.664 × 104 Pa
    Pressure difference between the water and acid columns:
                        ΔP2 - P1
                           
    = 6.664 × 104 - 3.92 × 104


                            =2.744 × 104 Pa
    Therefore, Force exerted on the door, F =  ΔP × a

                                                       = 2.744 × 104 × 20 × 10–4
                                                       = 54.88 N
    Hence, 54.88 N force is required to keep the door closed. 

    Question 1637
    CBSEENPH11020012

    Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?

    Solution
    Two vessels having the same base area have equal force and equal pressure acting on their common base area.
    The force exerted on the sides of the vessels has non-zero vertical components because the shapes of the two vessels are different. On adding these vertical components, the total force on one vessel comes out to be greater than that on the other vessel.
    Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.
    Question 1638
    CBSEENPH11020013

    During blood transfusion, the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].

    Solution
    Given,
    Gauge pressure, P = 2000 Pa 
    Density of whole blood, ρ = 1.06 × 10kg m–3 

    Acceleration due to gravity, g = 9.8 m/s

    Height of the blood container = 

    Pressure of the blood container, P = hρ
    Therefore comma space

straight h space equals space straight P over ρg space equals space fraction numerator 2000 over denominator 1.06 cross times 10 cubed cross times 9.8 end fraction space equals space 0.1925 space straight m 

    The blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.
    Question 1639
    CBSEENPH11020014

    In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

    Solution
    (a) If dissipative forces are present, then due to pressure difference spent against the dissipative forces, some forces in liquid flow. Therefore, the pressure drop becomes large.   
    (b) The dissipative forces become more important with increasing flow velocity, because of turbulence.
    Question 1640
    CBSEENPH11020015

    (a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).

    Solution
    Given,
    Radius of the artery, r = 2 × 10–3 
    Diameter of the artery, d = 2 × 2 × 10–3 m = 4 × 10– 3 
    Viscosity of blood, η = 2.084 X 10-3 Pa s 
    Density of blood, ρ = 1.06 × 103 kg/m

    Reynolds’ number for laminar flow, NR = 2000
    Average space velocity space of space blood space is space given space by space the space relation comma space

straight v subscript avg space equals space fraction numerator straight N subscript straight R straight eta over denominator ρd end fraction

space space space space space space space equals space fraction numerator 2000 space cross times space 2.084 space cross times space 10 to the power of negative 3 end exponent over denominator left parenthesis 1.06 space cross times space 10 to the power of 3 space end exponent cross times space 4 space cross times space 10 to the power of negative 3 end exponent right parenthesis end fraction
space
space space space space space space space equals space 0.983 space straight m divided by straight s
    b) Flow rate is given by the relation, 

    straight R space equals space πr squared straight v subscript avg

space space space equals space 3.14 space cross times space left parenthesis 2 space cross times space 10 to the power of negative 3 end exponent right parenthesis squared space cross times space 0.983

space space space equals space 1.235 space cross times space 10 to the power of negative 5 end exponent space straight m cubed straight s to the power of negative 1 end exponent space comma space is space the space corresponding
space rate space of space flow.
    Question 1641
    CBSEENPH11020016

    A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).

    Solution
    Given,
    Area of the wings of the plane, A = 2 × 25 = 50 m2

    Speed of air over the lower wing, V1 = 180 km/h = 50 m/s 
    Speed of air over the upper wing, V2 = 234 km/h = 65 m/s
    Density of air, ρ = 1 kg m–3

    Pressure of air over the lower wing = P1

    Pressure of air over the upper wing= P2

    The upward force on the plane can be obtained using Bernoulli’s equation,

    space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
fraction numerator straight P subscript 1 space minus space space straight P subscript 2 over denominator straight A end fraction space equals space 1 half space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
fraction numerator straight P subscript 1 space minus space space straight P subscript 2 over denominator straight A end fraction space equals space 1 half space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
space space left parenthesis straight P subscript 1 space minus space space straight P subscript 2 right parenthesis thin space straight A space equals space 1 half space straight rho space open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space straight A

space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 1 cross times open parentheses 65 space squared minus 50 squared space close parentheses space cross times 50 space

space space space space space space space space space space space space space space space space space space space space space space space space equals space 43125 space straight N space

Using space Newton apostrophe straight s space first space law comma space mass space of space the space plane space can space be space obtained. space

straight F space equals space mg

therefore space straight m space equals space fraction numerator 43125 over denominator 9.8 end fraction space equals space 4400.51 space kg

Therefore comma space the space mass space of space the space plane space is space
approximately space 4400 space kg.
    Question 1642
    CBSEENPH11020017

     In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

    Solution
    Given,
    Terminal speed = 5.8 cm/s
    Viscous force = 3.9 × 10–10 N
    Radius of the given uncharged drop, r = 2.0 × 10–5 m
    Density of the uncharged drop, ρ = 1.2 × 103 kg m–3

    Viscosity of air, η = 1.8 × 10-5 Pa s
    Density of air (ρ0) can be taken as zero in order to neglect buoyancy of air.
    Acceleration due to gravity, g = 9.8 m/s2

    Terminal velocity (v) is given by,
    straight v space equals space fraction numerator 2 straight r squared space cross times space left parenthesis straight rho space minus space straight rho subscript straight o right parenthesis straight g over denominator 9 straight eta end fraction

space space equals space fraction numerator 2 space cross times space left parenthesis 2 space cross times space 10 to the power of negative 5 end exponent right parenthesis squared space left parenthesis 1.2 space cross times space 10 cubed space minus space 0 space right parenthesis space cross times space 9.8 over denominator left parenthesis 9 space cross times space 1.8 space cross times space 10 to the power of negative 5 end exponent right parenthesis end fraction

space space equals space 5.8 space cross times space 10 to the power of negative 2 end exponent space straight m divided by straight s

space space equals 5.8 space cm space straight s to the power of negative 1 end exponent space
    Therefore, the terminal speed of the drop is 5.8 cm s–1.
    The viscous force on the drop is given by,
                      F = 6πηrv 

    ∴                F = 6 × 3.14 × 1.8 × 10-5 × 2 × 10-5 × 5.8 × 10-2

                        = 3.9 × 10-10 N
    Hence, the viscous force on the drop is 3.9 × 10–10 N
    Question 1643
    CBSEENPH11020018

    Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3.

    Solution

    Given, 

    Angle of contact between mercury and soda lime glass, θ = 140°
    Radius of the narrow tube, r = 1 mm = 1 × 10–3 m
    Surface tension of mercury at the given temperature, s = 0.465 N m–1

    Density of mercury, ρ =13.6 × 103 kg/m3

    Dip in the height of mercury = h

    Acceleration due to gravity, g = 9.8 m/s2

    Relation between surface tension, angle of contact and dip in height is given by, 

    straight s space equals space fraction numerator hρgr over denominator 2 space cos space straight theta end fraction

therefore space straight h space equals space fraction numerator 2 straight s space cos space straight theta over denominator ρgr end fraction

space space space space space space space equals fraction numerator 2 space cross times space 0.465 space cross times space Cos 140 to the power of 0 over denominator left parenthesis 1 space cross times space 10 to the power of negative 3 end exponent space cross times space 13.6 space cross times space 10 to the power of 3 space end exponent cross times space 9.8 right parenthesis end fraction

space space space space space space space equals negative 0.00534 space

space space space space space space equals negative 5.34 space straight m

Negative space sign space implies space decreasing space level space of space mercury. space

    Question 1644
    CBSEENPH11020019

    Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).

    Solution
    Diameter of the first bore, d1 = 3.0 mm = 3 × 10–3 m 
    So, the radius of the first bore, r1 = d1/ 2  =  1.5 × 10-3 m
    Diameter of the second bore, d2 = 6.0 mm = 6 × 10–3 mm
    Hence, the radius of the first bore, r2 = d2/ 2  =  3 × 10-3 m
    Surface tension of water, s = 7.3 × 10–2 N m–1

    Angle of contact between the bore surface and water, θ= 0
    Density of water, ρ =1.0 × 103 kg/m–3

    Acceleration due to gravity, g = 9.8 m/s2

    Let h1 and hbe the heights to which water rises in the first and second tubes respectively.

    The space heights space are space given space by space the space relation comma space

straight h subscript 1 space equals space fraction numerator 2 straight S space cosθ over denominator straight r subscript 1 ρg end fraction space space space space space space space space space... left parenthesis straight i right parenthesis
straight h subscript 2 space equals space fraction numerator 2 straight S space cosθ over denominator straight r subscript 2 ρg end fraction space space space space space space space space space... space left parenthesis ii right parenthesis

So comma space difference space between space the space levels space of space water space in space the space two space
limbs space of space the space tube space can space be space calculated space as colon

space space space space space space space equals space fraction numerator 2 straight S space cosθ over denominator straight r subscript 1 ρg end fraction minus fraction numerator 2 straight S space cosθ over denominator straight r subscript 2 ρg end fraction

space space space space space space equals fraction numerator 2 space cosθ over denominator ρg end fraction space open square brackets 1 over straight r subscript 1 minus 1 over straight r subscript 2 close square brackets space

space space space space space space equals space fraction numerator 2 cross times 7.3 cross times 10 to the power of negative 2 end exponent cross times 1 over denominator 1 cross times 10 cubed cross times 9.8 end fraction open square brackets fraction numerator 1 over denominator 1.5 cross times 10 to the power of negative 3 end exponent end fraction minus fraction numerator 1 over denominator 3 cross times 10 to the power of negative 3 end exponent end fraction close square brackets space

space space space space space space space equals space 4.966 space cross times space 10 to the power of negative 3 end exponent space straight m space

space space space space space space equals space 4.97 space mm space

Therefore comma space the space difference space between space levels space of
water space in space the space two space bores space is space 4.97 space mm.
    Question 1645
    CBSEENPH11020020

    (a) It is known that density ρ of air decreases with height y as ρ0e-y/y0
    Where ρ0 = 1.25 kg m-3
    is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of gremains constant.

    Solution
    Volume of the balloon, V = 1425 m

    Mass of the payload, m = 400 kg 
    Acceleration due to gravity, g = 9.8 m/s

    y0 = 8000 m 
    ρHe = 0.18 kg m-3 

    ρ0 = 1.25 kg m-3 

    Density of the balloon = ρ 

    Height to which the balloon rises = 

    Density (ρ) of air decreases with height (y) as,
                       ρ = ρ0e-y/y

                       ρ / ρ= e-y/y0                 ...(i)
    This density variation is called the law of atmospherics. 
    From equation (i), we can say that the rate of decrease of density with height is directly proportional to straight rho.
    That is, 

    space space space space space minus open parentheses dρ over dy close parentheses space proportional to space straight rho

rightwards double arrow space open parentheses dρ over dy close parentheses space equals space minus straight k space straight rho space

rightwards double arrow space space open parentheses fraction numerator dρ over denominator straight rho space end fraction close parentheses space equals space minus space straight k space dy space

straight k space is space the space constant space of space proportionality. space

Height space changes space from space 0 space to space straight y comma

Density space changes space from space space straight rho subscript straight o space to space straight rho space

On space integrating space sides space between space the space limits comma space

space space space space space space space space space space space integral subscript straight rho space end subscript superscript space straight rho subscript straight o end superscript fraction numerator dρ over denominator straight rho space end fraction space equals space minus integral subscript 0 superscript straight y straight k space dy

right enclose space space space space space space space space space space space space space space log subscript straight e space straight rho end enclose subscript straight rho subscript straight o end subscript superscript straight rho space end superscript space equals space minus space ky

space space space space space space space space open square brackets log subscript straight e space straight rho over straight rho subscript straight o close square brackets space equals space minus ky space

rightwards double arrow straight rho over straight rho subscript straight o space equals space straight e to the power of negative ky end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

On space comparing space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get

space space space space straight y subscript straight o space equals space 1 over straight k space
rightwards double arrow space straight k space equals space 1 over straight y subscript straight o space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis space

From space equations space left parenthesis straight i right parenthesis space and space left parenthesis iii right parenthesis comma space we space get

space space space space space space space space space space space space space space space space space space space space space box enclose space straight rho space equals space straight rho subscript straight o space straight e to the power of negative begin inline style bevelled straight y over straight y subscript straight o space end subscript end style end exponent end enclose
    Question 1647
    CBSEENPH11020022

    The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

    Solution
    Kelvin and Celsius scales are related as: 
    TC = TK – 273.15                  … (i) 

    Celsius and Fahrenheit scales are related as: 
    TF = (9/5)TC + 32                ...(ii)

    For neon: 
    TK = 24.57 K 
     T= 24.57 – 273.15 = –248.58°C 
    TF = (9/5)TC + 32 
        = (9/5) × (-248.58) +32 
        = 415.440 F
    For carbon dioxide:
    TK = 216.55 K 
     TC= 216.55 – 273.15
         = –56.60°C 
    TF = (9/5)TC + 32 
        = (9/5) × (-56.60) +32 
        = -69.880 C
    Question 1648
    CBSEENPH11020023

    Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation betweenTA and TB?

    Solution
    Triple point of water on absolute scale A, T1 = 200 A
    Triple point of water on absolute scale B, T2 = 350 B
    Triple point of water on Kelvin scale, TK = 273.15 K
    The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.
               T1 = T

          200 A = 273.15 K 
    ∴          A = 273.15/200 
    The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B. 
              T2 = T

         350 B = 273.15
    ∴        B = 273.15/350 
    TA is triple point of water on scale A. 
    TB is triple point of water on scale B. 
    ∴         273.15/200 × T = 273.15/350 × T

    Therefore, the ratio TA : Tis given as 4 : 7
    Question 1650
    CBSEENPH11020025

    The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

    Solution
    The triple point of water has a unique value of 273.16 K. At varying values of volume and pressure, the triple point of water is always 273.16 K.
    The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.
    Question 1652
    CBSEENPH11020027

    The absolute temperature (Kelvin scale) is related to the temperature tc on the Celsius scale by,
                    tc = – 273.15

    Solution
    The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice.
    Its corresponding value on Kelvin scale is 273.15 K.
    Hence, absolute temperature (Kelvin scale) T, is related to temperature tc, on Celsius scale as,

    tc = T – 273.15 
    Question 1653
    CBSEENPH11020028

    What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

    Solution
    Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale.
    Both the temperatures can be related as, 
    (TF - 32) / 180  =  (TK - 273.15) / 100                    ...(i)

    Let TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on absolute scale.
    Both the temperatures can be related as, 
    (TF1 - 32) / 180  =  (TK1 - 273.15) / 100                  ...(ii) 
     
    Given that, 
    TK1 – TK = 1 K 
    Subtracting equation (i) from equation (ii), we get 
                  (TF1 - TF) / 180  =  (TK1 - TK) / 100  =  1 / 100 
                             TF1 - TF = (1 ×180) / 100  =  9/5 
    Triple point of water = 273.16 K 
    ∴ Triple point of water on absolute scale = 273.16 × (9/5)  = 491.69
    Question 1654
    CBSEENPH11020029

    Two ideal gas thermometers Aand Buse oxygen and hydrogen respectively. The following observations are made:
    Temperature Pressure thermometer A Pressure thermometer B
    Triple-point of water 1.250 × 105 Pa 0.200 × 105 Pa
    Normal melting point of sulphur 1.797 × 105 Pa 0.287 × 105 Pa

    (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers Aand B?

    (b) What do you think is the reason behind the slight difference in answers of thermometers Aand B(The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
     

    Solution
    (a)
    Triple point of water, T = 273.16 K
     
    At this temperature, pressure in thermometer A, PA = 1.250 × 105 Pa 
    Let T1 be the normal melting point of sulphur.  
    At this temperature, pressure in thermometer A, P1 = 1.797 × 105 Pa
    According to Charles’ law, we have the relation,

    fraction numerator straight P subscript straight A over denominator space straight T space end fraction space equals space begin inline style fraction numerator space straight P subscript 1 over denominator space straight T subscript 1 end fraction end style space

therefore space straight T subscript 1 space equals space fraction numerator space straight P subscript 1 straight T over denominator straight P subscript straight A end fraction
space space space space space space space space space equals space fraction numerator 1.797 space cross times space 10 space cross times space 273.16 over denominator left parenthesis 1.250 space cross times space 10 to the power of 5 right parenthesis end fraction
space space space space space space space space space equals space 392.69 space straight K space

Therefore comma space the space absolute space temperature space of space the space normal
melting space point space of space sulphur space as space read space by space thermometer space straight A space
is space 392.69 space straight K.

At space triple space point space 273.16 space straight K comma space the space pressure space in space thermometer space straight B comma space
straight P subscript straight B space equals space 0.200 space cross times space 10 to the power of 5 space Pa

At space temperature space straight T subscript 1 comma space pressure space in space
thermometer space straight B comma space straight P subscript 2 space equals space 0.287 space cross times space 10 to the power of 5 space Pa space

According space to space Charle apostrophe straight s space Law comma space we space can space write space
the space relation comma space

straight P subscript straight B over straight T space equals space straight P subscript 1 over straight T subscript 1 space
fraction numerator left parenthesis 0.200 space cross times space 10 to the power of 5 right parenthesis over denominator 273.16 end fraction space equals space fraction numerator space left parenthesis 0.287 space cross times space 10 to the power of 5 right parenthesis over denominator straight T subscript 1 end fraction

therefore space straight T subscript 1 space equals space open square brackets fraction numerator left parenthesis 0.287 space cross times 10 to the power of 5 right parenthesis over denominator 0.200 cross times 10 to the power of 5 end fraction close square brackets cross times 273.16 space
space space space space space space space space space space equals space 391. space 98 space straight K space

Therefore comma space the space absolute space temperature space of space the space
normal space melting space point space of space sulphur space as space read space by space thermometer space
straight B space is space 391.98 space straight K.
    b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B.
    To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.
    Question 1655
    CBSEENPH11020030

    A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1.

    Solution

    Length of the steel tape at temperature T equal to 27°C, l = 1 m = 100 cm
    At temperature T1 equal to 45°C, the length of the steel rod, l1 = 63 cm 
    Coefficient of linear expansion of steel, α = 1.20 × 10–5 K–1 

    Let l2 be the actual length of the steel rod and l' be the length of the steel tape at 45°C.
                         l' = l + αl(T1 - T) 
    ∴                   l' = 100 + 1.20 × 10-5 × 100(45 - 27) 

                           = 100.0216 cm 
    Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as, 
    l2 = (100.0216 / 100) × 63 
       =  63.0136 cm 
    Therefore, the actual length of the rod at 45.0°C is 63.0136 cm.
    Its length at 27.0°C is 63.0 cm.

    Question 1656
    CBSEENPH11020031

    A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.

    Solution
    The given temperature, T = 27°C can be written in Kelvin as,
                         27 + 273 = 300 K
    Outer diameter of the steel shaft at Td1 = 8.70 cm 
    Diameter of the central hole in the wheel at Td2 = 8.69 cm 
    Coefficient of linear expansion of steel, αsteel = 1.20 × 10–5K–1 

    After the shaft is cooled using ‘dry ice’, its temperature becomes T1
    The wheel will slip on the shaft, if the change in diameter, Δd= 8.69 – 8.70
     
                                                                                   = – 0.01 cm 
    Temperature T1, can be calculated from the relation: 
                 Δd = d1αsteel (T1 – T
              0.01 = 8.70 × 1.20 × 10–5 (T1 – 300) 
      (T1 – 300) = 95.78 
    ∴           T1= 204.21 K  
                   = 204.21 – 273.16 
                   = –68.95°C 
    Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.
    Question 1657
    CBSEENPH11020032

    A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.

    Solution
    Given,
    Initial temperature, T1 = 27.0°C
    Diameter of the hole at T1d1 = 4.24 cm
    Final temperature, T2 = 227°C 
    Diameter of the hole at Td

    Co-efficient of linear expansion of copper, αCu= 1.70 × 10–5K–1
    Relation between coefficient of superficial expansion and change in temperature is given by, 
    space space space space space space space fraction numerator Change space in space Area space left parenthesis increment straight A right parenthesis over denominator Original space Area space left parenthesis straight A right parenthesis end fraction space equals space beta space increment T

rightwards double arrow space fraction numerator open parentheses begin display style fraction numerator pi d subscript 2 squared over denominator 4 end fraction end style space minus begin display style fraction numerator pi d subscript 1 squared over denominator 4 end fraction end style space close parentheses over denominator begin display style fraction numerator pi d subscript 1 squared over denominator 4 end fraction end style end fraction space equals space fraction numerator increment straight A over denominator A end fraction space

therefore space fraction numerator increment straight A over denominator A end fraction space equals fraction numerator d subscript 2 squared space minus space d subscript 1 squared over denominator d subscript 1 squared end fraction

But comma space given space that space beta space equals space 2 alpha space

Therefore comma space

space space space space space space space space fraction numerator straight d subscript 2 squared space minus space straight d subscript 1 squared over denominator straight d subscript 1 squared end fraction space equals space space 2 straight alpha space increment straight T
rightwards double arrow space space space space space fraction numerator straight d subscript 2 squared space over denominator straight d subscript 1 squared end fraction minus space 1 space equals space 2 straight alpha space left parenthesis straight T subscript 2 space minus space straight T subscript 1 right parenthesis space

rightwards double arrow space fraction numerator straight d subscript 2 squared space over denominator 4.24 space squared end fraction space equals space 2 cross times 1.7 cross times 10 to the power of negative 5 end exponent space left parenthesis 227 minus 27 right parenthesis space plus space 1

rightwards double arrow space space space space space straight d subscript 2 squared space equals space 17.98 space cross times space 1.0068 space equals space 18.1 space

therefore space space space space space space straight d subscript 2 space equals space 4.2544 space cm space

So comma space change space in space diameter space equals space straight d subscript 2 space minus space straight d subscript 1 space equals space 4.2544 space minus space 4.24 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.0144 space cm

Therefore comma space

Increase space in space diameter space equals space 1.44 space cross times space 10 to the power of negative 2 end exponent space cm
    Question 1658
    CBSEENPH11020033

    A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.

    Solution
    Given,
    Initial temperature, T1 = 27°C
    Length of the brass wire at T1l = 1.8 m
    Final temperature, T2 = –39°C
    Diameter of the wire, d = 2.0 mm = 2 × 10–3 m
    Tension developed in the wire = F

    Coefficient of linear expansion of brass, α = 2.0 × 10–5 K–1

    Young’s modulus of brass, = 0.91 × 1011 Pa 
    Young apostrophe straight s space modulus space is space given space by comma space

straight Y space equals space Stress over strain space equals space fraction numerator begin display style bevelled straight F over straight A end style over denominator begin display style bevelled fraction numerator increment straight L over denominator straight L end fraction end style end fraction

rightwards double arrow space increment straight L space equals space fraction numerator straight F space cross times straight L over denominator straight A cross times straight Y end fraction space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis thin space

where comma space

straight F space equals space Tension space in space the space wire comma
straight A space equals space Area space of space cross minus section space of space the space wire comma
increment straight L space equals space Change space in space length space equals space αL space left parenthesis straight T subscript 2 space minus space straight T subscript 1 space right parenthesis space space space space space space space space space... space left parenthesis ii right parenthesis space

Equating space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get space

αL space left parenthesis straight T subscript 2 space minus space straight T subscript 1 space right parenthesis space equals space fraction numerator FL over denominator left square bracket straight pi left parenthesis begin display style bevelled straight d over 2 end style right parenthesis squared cross times straight Y right square bracket end fraction space

rightwards double arrow space straight F space equals space straight alpha space left parenthesis straight T subscript 2 space minus space straight T subscript 1 space right parenthesis space πY left parenthesis bevelled straight d over 2 right parenthesis squared space

rightwards double arrow space straight F space equals space 2 cross times 10 to the power of negative 5 end exponent cross times left parenthesis negative 39 minus 27 right parenthesis cross times 3.14 cross times 0.91 cross times 10 to the power of 11 cross times open parentheses fraction numerator 2 space cross times space 10 to the power of negative 3 end exponent space over denominator 2 end fraction close parentheses squared

space space space space space space space space equals space minus 3.8 space cross times space 10 squared space straight N space

Negative space sign space implies space that space tension space is space directing
in space the space inward space direction. space
    Question 1659
    CBSEENPH11020034

    A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).

    Solution
    Here, we have
    Initial temperature, T1 = 40°C 
    Final temperature, T2 = 250°C 
    Change in temperature, ΔT = T2 – T= 210°C 
    Length of the brass rod at T1l1 = 50 cm
    Diameter of the brass rod at T1d1 = 3.0 mm
    Length of the steel rod at T2l2 = 50 cm
    Diameter of the steel rod at T2d2 = 3.0 mm 
    Coefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1 

    Coefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1
    Expansion in brass rod is given by, 
    fraction numerator increment straight L subscript 1 over denominator straight L subscript 1 end fraction space equals space straight alpha subscript 1 increment straight T space

therefore space increment straight l subscript 1 space equals 50 space cross times space left parenthesis 2.1 space cross times space 10 to the power of negative 5 end exponent right parenthesis space cross times space 210 space

space space space space space space space space space space space space space equals space 0.2205 space cm

Expansion space in space steel space rod comma space

fraction numerator increment straight L subscript 2 over denominator straight L subscript 2 end fraction space equals space space straight alpha subscript 2 increment straight T space

therefore space increment straight L subscript 2 space equals space space 50 space cross times space left parenthesis 1.2 space cross times space 10 to the power of negative 5 end exponent right parenthesis space cross times space 210

space space space space space space space space space space space space space equals 0.126 space cm

Therefore comma space

Total space change space in space the space length space of space brass space and space steel space is comma space

increment straight L space equals space increment straight l subscript 1 space plus space increment straight L subscript 2 space

space space space space space space space equals space 0.2205 space plus space 0.0126

space space space space space space space equals space 0.346 space cm
    No, thermal stress is developed at the junction because the rod expands freely from both the ends.
    Question 1660
    CBSEENPH11020035

    The coefficient of volume expansion of glycerin is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature?

    Solution
    Coefficient of volume expansion of glycerin, αV = 49 × 10–5K–1

    Rise in temperature, Δ= 30°C
    Fractional change in its volume = ΔV/V
    Relation between fractional change in volume and change in temperature is given by, 
    fraction numerator increment straight V over denominator straight V end fraction space equals space straight alpha subscript straight v increment straight T

straight V subscript straight T subscript 2 end subscript space minus space straight V subscript straight T subscript 1 end subscript space equals space space straight V subscript straight T subscript 1 end subscript straight alpha subscript straight V increment straight T space

open parentheses fraction numerator straight m space over denominator straight rho subscript straight T subscript 2 end subscript end fraction close parentheses minus space open parentheses straight m over straight rho subscript straight T subscript 1 end subscript close parentheses equals space open parentheses straight m over straight rho subscript straight T subscript 1 end subscript close parentheses space straight alpha subscript straight V space ΔT space

where comma space
straight m space equals space Mass space of space glycerine

straight rho subscript straight T subscript 1 space end subscript end subscript equals space Initial space density space at space straight T subscript 1
straight rho subscript straight T subscript 2 space end subscript end subscript equals space Initial space density space at space straight T subscript 2

Therefore comma space

fraction numerator straight rho subscript straight T subscript 1 space end subscript end subscript space minus straight rho subscript straight T subscript 2 space end subscript end subscript over denominator straight rho subscript straight T subscript 2 space end subscript end subscript end fraction space equals space Fractional space change space in space density
So comma space

Fractional space change space in space density space of space
glycerine space equals space 49 cross times 10 to the power of negative 5 end exponent cross times 30 space equals space 1.47 space cross times space 10 to the power of – 2 end exponent.

    Question 1661
    CBSEENPH11020036

    A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1.

    Solution
    Given that,
    Power of the drilling machine, P = 10 kW = 10 × 10W
    Mass of the aluminum block, m = 8.0 kg = 8 × 103 g
    Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s
    Specific heat of aluminium, c = 0.91 J g–1 K–1

    Rise in the temperature of the block after drilling = δT

    Total energy of the drilling machine = Pt 

                                                  = 10 × 10× 150
                                                  = 1.5 × 106 J 
    It is given that only 50% of the power is useful.

    Useful space energy comma space increment straight Q space equals space open parentheses 50 over 100 close parentheses cross times 1.5 cross times 10 to the power of 6
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 7.5 space cross times space 10 to the power of 5 space straight J

But comma space we space know space increment straight Q space equals space mc increment straight T

rightwards double arrow space increment straight T space equals space fraction numerator increment straight Q over denominator mc end fraction space equals space fraction numerator 7.5 space cross times space 10 to the power of 5 over denominator 8 space cross times space 10 cubed space cross times space 0.91 end fraction equals space 103 to the power of 0 space straight C
    Therefore, rise in temperature of the block is 103o C in 2.5 minutes of drilling. 
    Question 1662
    CBSEENPH11020037

     A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).

    Solution
    Given that,
    Mass of the copper block, = 2.5 kg = 2500 g
    Rise in the temperature of the copper block, Δθ = 500°C
    Specific heat of copper, C = 0.39 J g–1 °C–1

    Heat of fusion of water, L = 335 J g–1

    The maximum heat the copper block can lose, Q = mCΔθ 

                                                                   = 2500 × 0.39 × 500  
                                                                   = 487500 J 
    Let m1 g be the amount of ice that melts when the copper block is placed on the ice block. 
    The heat gained by the melted ice, Q = m1

    ∴                      m1 = Q / L 
                                =  487500 / 335 
                                =  1455.22 g 
    Hence, the maximum amount of ice that can melt is 1.45 kg.
    Question 1663
    CBSEENPH11020038

     In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

    Solution
    We have,
    Mass of the metal, m = 0.20 kg = 200 g
    Initial temperature of the metal, T1 = 150°C
    Final temperature of the metal, T2 = 40°C
    Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g
    Volume of water, V = 150 cm

    Mass (M) of water at temperature T = 27°C is,  
                       150 × 1 = 150 g 
    Fall in the temperature of the metal,
              ΔT1 – T2  
                  = 150 – 40
                  = 110°C 
    Specific heat of water, Cw = 4.186 J/g/°K 
    Specific heat of the metal = 

    Heat lost by the metal, θ = mCΔT                       … (i) 

    Rise in the temperature of the water and the calorimeter system is, 
                 ΔT = 40 – 27 = 13°C 
    Heat gained by the water and the calorimeter system is, 
                 Δθ′′ = m1 CwΔT’ 

                       = (M + m′) Cw ΔT                        … (ii)

    Heat lost by the metal = Heat gained by the water and colorimeter system.
    That is,
                     mC
    ΔT = (M + mCw ΔT’ 

         200 × C × 110 = (150 + 25) × 4.186 × 13
    therefore space straight C space equals space fraction numerator 175 cross times 4.186 cross times 13 over denominator 110 cross times 200 end fraction space equals space 0.43 space straight J space straight g to the power of negative 1 end exponent straight K to the power of negative 1 end exponent
    If some heat is lost to the surroundings, then the value of C will be less than the actual value. 
    Question 1664
    CBSEENPH11020039

    Given below are observations on molar specific heats at room temperature of some common gases.


    Gas Molar specific heat(CV)
    Hydrogen 4.87
    Nitrogen 4.97
    Oxygen 5.02
    Nitric Oxide 4.99
    Carbon Monoxide 5.01
    Chlorine 6.17


    The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

    Solution

    The above-listed gases are diatomic. Diatomic gases have other degree of freedom apart from the translational degree of freedom. 

    Inorder to increase the temperature of these gases, heat must be supplied to them. this heat increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.
    For only rotational motion, 
    Molar specific heat of the diatomic gas is = (5/2) R 
                                                         = open parentheses 5 over 2 close parentheses cross times 1.98 space equals space 4.95 space c a l space m o l to the power of negative 1 end exponent space K to the power of negative 1 end exponent
    Apart from Chlorine in the above list, for all gases molar specific heat is (5/2) R.
    At room temperature, Chlorine has vibrational modes apart from rotational and translational modes of motion. 
                                                      

    Question 1665
    CBSEENPH11020040

    Answer the following questions based on the P-phase diagram of carbon dioxide:

    (a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?

    (b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?

    (c) What are the critical temperature and pressure for CO2? What is their significance?

    (d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60° C under 10 atm,

    (c) 15 °C under 56 atm?

    Solution

    The P-T phase diagram for Carbon dioxide is as shown below: 



    a) At the triple point, that is temperature = -56.6 o C and pressure = 5.11 atm, solid, liquid and vapour phase of carbon dioxide exist in equilibrium.
    b) Fusion and boiling point of carbon dioxide will decrease when pressure decreases. 
    c) For carbon dioxide, the critical temperature is 31.1° C and critical pressure is 73.0 atm. If the temprature of carbon dioxide is more than 31.1° C, it can not be liquified, however large pressure we may apply.
    d) Carbon dioxide will be
       (a) a vapour, at =70° C under 1atm.
       (b) a solid, at -6° C under 10 atm
       (c) a liquid, at 15° C under 56 atm.

    Question 1666
    CBSEENPH11020041

    Answer the following questions based on the P–Tphase diagram of CO2:

    (a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?

    (b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?

    (c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.

    (d) CO2 is heated to a temperature 70° C and compressed isothermally. What changes in its properties do you expect to observe?

    Solution

    The P-T phase diagram of CO2 is as shown below:  
                      

    a) Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solid phase, so carbon dioxide will condense directly into the solid without becoming liquid.
    b) Since the pressure 4 atm is less than 5.11 atm the carbon dioxide will condense directly into solid without becoming liquid. 
    c) When a solid COat 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temprature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the T-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of COat 10 atm.
    d) Since 70° C is higher than the critical temprature of CO2, so the COgas can not be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.

    Question 1667
    CBSEENPH11020042

    A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.

    Solution
    Given,
    Initial temperature of the body of the child, T1 = 101°F 
    Final temperature of the body of the child, T2 = 98°F 
    Change in temperature, Δ= [ (101 - 98) × (5/9) ] o C 
    Time taken to reduce the temperature, t = 20 min 
    Mass of the child, m = 30 kg = 30 × 103 g 
    Specific heat of the human body = Specific heat of water = 

                                              = 1000 cal/kg/ °C 
    Latent heat of evaporation of water, L = 580 cal g–1 

    The heat lost by the child is given as, 
            ∆θ = mc

                = 30 × 1000 × (101 - 98) × (5/9) 
                = 50000 cal 
    Let m1 be the mass of the water evaporated from the child’s body in 20 min.
     
    Loss of heat through water is given by,
    ∆θ = m1
    therefore space straight m subscript 1 space equals space fraction numerator increment straight theta over denominator straight L end fraction space
space space space space space space space space space space equals space 50000 over 580

space space space space space space space space space space space equals space 86.2 space straight g

Therefore comma space

Average space rate space of space extra space evaporation space equals space straight m subscript 1 over straight t

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 86.2 over denominator 200 end fraction space equals space 4.3 space straight g divided by min
    Question 1668
    CBSEENPH11020043

    A ‘thermocole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermocole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]

    Solution

    Given, 
    Side of the given cubical ice box, s = 30 cm = 0.3 m
    Thickness of the ice box, l = 5.0 cm = 0.05 m
    Mass of ice kept in the ice box, m = 4 kg
    Time gap, t = 6 h = 6 × 60 × 60 s
    Outside temperature, T = 45°C
    Coefficient of thermal conductivity of thermocole, K = 0.01 J s–1 m–1 K–1 
    Heat of fusion of water, L = 335 × 103 J kg–1

    Let m be the total amount of ice that melts in 6 h.
    The amount of heat lost by the food, 

    straight theta space equals space fraction numerator KA left parenthesis straight T minus 0 right parenthesis straight t over denominator straight l end fraction

where comma space

straight A space equals space surface space area space of space box space equals space 6 space straight s squared space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 6 cross times left parenthesis 0.3 right parenthesis squared space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.54 space straight m cubed space

straight theta space equals space fraction numerator 0.01 space cross times space 0.54 cross times space 45 space cross times space 6 space cross times space 60 space cross times space 60 over denominator 0.05 end fraction
space space space equals space 104976 space straight J

But comma space

space space space space space space space space straight theta space equals space straight m apostrophe straight L

therefore space straight m apostrophe space equals space straight theta over straight L
space space space space space space space space space space equals fraction numerator 104976 over denominator 335 cross times 10 cubed end fraction space equals space 0.313 space kg

Mass space of space ice space left space equals space 4 space minus space 0.313 space equals space 3.687 space kg

That space is comma space

Amount space of space ice space remaining space after space 6 space straight h space is space 3.687 space kg. space

    Question 1669
    CBSEENPH11020044

    A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.

    Solution
    Given,
    Base area of the boiler, = 0.15 m

    Thickness of the boiler, l = 1.0 cm = 0.01 m 
    Boiling rate of water, R = 6.0 kg/min 
     
    Mass, m = 6 kg 
    Time, t = 1 min = 60 s 
    Thermal conductivity of brass, K = 109 J s –1 m–1 K–1 

    Heat of vaporisation, L = 2256 × 103 J kg–1 

    The amount of heat flowing into water through the brass base of the boiler is given by, 
    θ = KA(T1 - T2t / l                   ...(i)
    where, 
    T1 = Temperature of the flame in contact with the boiler,
    T2 = Boiling point of water = 100°C, 
    Heat required for boiling the water is,  
                       θ = mL                  … (ii)

    Now, on equating equations (i) and (ii), we get

    mL space equals space fraction numerator KA space left parenthesis straight T subscript 1 space minus space straight T subscript 2 right parenthesis straight t over denominator straight l end fraction

rightwards double arrow space straight T subscript 1 space minus space straight T subscript 2 space equals space mLl over KAt

space space space space space space space space space space space space space space space space space space space space equals fraction numerator 6 space cross times space 2256 space cross times space 10 cubed space cross times space 0.01 over denominator 109 space cross times space 0.15 space cross times space 60 end fraction

space space space space space space space space space space space space space space space space space space space space equals space 137.98 to the power of straight o space straight C 
    Therefore, temperature for the part of the flame in contact with the burner is 137.98o C
    Question 1670
    CBSEENPH11020045

    Explain why: 

    (a) a body with large reflectivity is a poor emitter

    Solution
     A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter. 
    Question 1671
    CBSEENPH11020046

    Explain why:

    (b) a brass tumbler feels much colder than a wooden tray on a chilly day

    Solution
    Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.
    Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool. 
    Thus, a brass tumbler feels colder than a wooden tray on a chilly day.
    Question 1672
    CBSEENPH11020047

    Explain why:

    (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

    Solution
    An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open. 
    Black body radiation equation is given by, 
                       E = σ (T4 - T04
    where, 
    E = Energy radiation,
    T = Temperature of optical pyrometer, 
    To = Temperature of open space, and
    σ = Constant.
    Hence, an increase in the temperature of open space reduces the radiation energy. 
    When the same piece of iron is placed in a furnace, the radiation energy, E = σ T4
    Question 1673
    CBSEENPH11020048

    Explain why:

    (d) the earth without its atmosphere would be inhospitably cold

    Solution
    Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface. 
    Question 1674
    CBSEENPH11020049

    Explain why:

    (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

    Solution
    A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g). 
    Question 1675
    CBSEENPH11020050

    A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

    Solution
    According space to space Newton ’ straight s space law space of space cooling comma

fraction numerator negative space dT over denominator straight T end fraction space equals space straight K space left parenthesis straight T space minus space straight T subscript straight o right parenthesis space

fraction numerator dT over denominator straight K thin space left parenthesis straight T space minus space straight T subscript straight o right parenthesis end fraction space equals space minus space Kdt space space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis space
    where, 
    T is the temperature of the body, 
    Temperature of the surroundings, To = 200 C, and
    K is a constant. 
    Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

    Now comma space integrating space equation space left parenthesis straight i right parenthesis comma space we space get

space space space space space space space space integral subscript 50 superscript 80 fraction numerator dT over denominator straight K left parenthesis straight T space minus space straight T subscript straight o right parenthesis end fraction space equals space minus space integral subscript 0 superscript 300 straight K. space dt space

space space space space space space space space space left square bracket log subscript straight e left parenthesis straight T space minus space straight T subscript straight o right parenthesis right square bracket subscript 50 superscript 80 space equals space minus straight K space left square bracket straight t right square bracket space subscript 0 superscript 300 space

rightwards double arrow fraction numerator 2.3026 over denominator straight K end fraction space log subscript 10 space 2 space space equals space minus 300 space

rightwards double arrow fraction numerator negative space 2.3026 over denominator 300 end fraction space log subscript 10 space 2 space equals space straight K space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

Temperature space of space the space body space falls space from space
60 to the power of straight o space straight C space to space 30 to the power of straight o space straight C space in space time space straight t apostrophe.

Hence comma space we space have

space space space space space space space fraction numerator 2.3026 over denominator straight K space end fraction space log subscript 10 space fraction numerator 60 minus 20 over denominator 30 space minus space 20 end fraction space equals space minus straight t to the power of 1 space

rightwards double arrow space space minus fraction numerator 2.3026 over denominator straight t apostrophe end fraction space log subscript 10 space 4 space equals space straight K space space space space space space space space space space... space left parenthesis iii right parenthesis

On space equating space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space get

space minus fraction numerator 2.3026 over denominator straight t apostrophe end fraction space log subscript 10 space 4 space equals space fraction numerator negative 2.3026 over denominator 300 end fraction space log subscript 10 space 2 space

rightwards double arrow space straight t apostrophe space equals space 300 cross times 2 space equals space 600 space straight s space equals space 10 space min

Therefore comma space 10 space minutes space is space taken space to space cool space
the space body space from space 60 to the power of straight o space straight C space to space 30 to the power of straight o space straight C.

    Question 1676
    CBSEENPH11020051

    A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston has to bear?

    Solution

    Area of cross-section of the piston, A = 425 cm2 = 4.25 cross times space 10 to the power of negative 2 end exponent space straight m squared 
    Maximum load on piston, F = 3000 kg = 3000 X 9.8 N 
    Therefore, 
    Maximum pressure on the piston on which car is, 

    straight P space equals space straight F over straight A space equals space fraction numerator 3000 space cross times space 9.8 over denominator 4.25 space cross times space 10 to the power of negative 2 end exponent end fraction space

space space space space space space space space space space space space space space space equals space 6.92 space cross times 10 to the power of 5 space Pa

    Question 1677
    CBSEENPH11020052

    Three vessels shown in the figure have same base area. These vessels are filled with same quantity of same liquid. In which vessel will the force on the base be maximum? 


    Solution
    The vessel (iii) is narrowest and vessel (i) is widest at the top. Therefore when the vessels are filled by same quantity of liquid, the height column of liquid in vessel (iii) will be maximum and in vessel (i) is minimum. Thus the pressure due to height column of liquid at the base is maximum in vessel (iii) and minimum in vessel (i).
    As the base area of all the vessels is same, hence the force at the base of vessel (iii) will be maximum and that at the base of vessel (i) will be minimum.
    Question 1679
    CBSEENPH11020167

    A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g? 

    Solution
    Water is flowing at a rate of 3.0 litre/min
    The geyser heats the water, raising the temperature from 27°C to 77°C. 
    Initial temperature, T1 = 27°C 
    Final temperature, T2 = 77°C 
    Rise in temperature, ΔT = T2 – T

                                          = 77 – 27
                                          = 50°C 
    Heat of combustion = 4 × 104 J/g 
    Specific heat of water, c = 4.2 J g–1 °C–1 

    Mass of flowing water, m = 3.0 litre/min
                                         = 3000 g/min 
    Total heat used, ΔQ = mc Δ

                                 = 3000 × 4.2 × 50 
                                 = 6.3 × 10J/min 
    Therefore,
    Rate of consumption = open parentheses fraction numerator 6.3 space cross times space 10 to the power of 5 over denominator 4 space cross times space 10 to the power of 4 end fraction close parentheses space 
                                    =  15.75 g/min.
    Question 1680
    CBSEENPH11020168

    What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R= 8.3 J mol–1 K–1.)

    Solution
    Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g
    Rise in temperature, ΔT = 45°C
    Molecular mass of N2M = 28 
    Universal gas constant, R = 8.3 J mol–1 K–1 

    Number of moles, n = straight m over straight M
                                  = (fraction numerator 2 space cross times space 10 to the power of negative 2 end exponent space cross times space 10 cubed over denominator 28 end fraction)
                                  = 0.714 
    Molar specific heat at constant pressure for nitrogen,
    Cp = 7 over 2 R
        = 7 over 2 × 8.3 
        = 29.05 J mol-1 K-1 

    The total amount of heat to be supplied is given by the relation, 
    ΔQ = nCΔ

         = 0.714 × 29.05 × 45 
         = 933.38 J 
    Therefore, the amount of heat to be supplied is 933.38 J.
    Question 1681
    CBSEENPH11020169

    Explain why

    (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2)/2.

    Solution
    When two bodies are in thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures becomes equal.
    The final temperature can be the mean temperaturefraction numerator straight T subscript 1 space plus space straight T subscript 2 over denominator 2 end fraction, only when thermal capacities of the two bodies are equal.
    Question 1682
    CBSEENPH11020170

    Explain why

    (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

    Solution
    (b) High specific heat capacity is required because the heat absorbed by a substance is directly proportional to the specific heat of the substance. Thus the coolant will work more effectively. 
    Question 1683
    CBSEENPH11020171

    Explain why?

    (c) Air pressure in a car tyre increases during driving.

    Solution
    During driving, due to motion of the car the temperature of air inside the tyre increases.
    According to Charle's law, P ∝T. 
    Therefore, air pressure inside the tyre increases.
    Question 1684
    CBSEENPH11020172

    Explain why:

    The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 

    Solution
    The climate of a harbour town is more temperate than that of a town in a desert because, in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.
    Question 1685
    CBSEENPH11020173

    A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

    Solution
    The cylinder is completely insulated from its surroundings.
    As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
    Initial pressure inside the cylinder = P1

    Final pressure inside the cylinder = P2

    Initial volume inside the cylinder = V1

    Final volume inside the cylinder = V2

    Ratio of specific heats, γ = 1.4
    For an adiabatic process, we have:
    P1V1γ = P2V2γ 

    The final volume is compressed to half of its initial volume.
    ∴ V2 = V1/2 
    P1V1γ = P2open parentheses straight V subscript 1 over 2 close parenthesesγ 

    P2/P1 = V1γ / open parentheses straight V subscript 1 over 2 close parentheses to the power of nu

             = 2γ 
             = 21.4 
             = 2.639 
    Hence, the pressure increases by a factor of 2.639. 
    Question 1686
    CBSEENPH11020174

    Two cylinders and of equal capacity are connected to each other via a stopcock. contains a gas at standard temperature and pressure. is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: 

    (a) What is the final pressure of the gas in and B? 

    (b) What is the change in internal energy of the gas? 

    (c) What is the change in the temperature of the gas? 

    (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its PVT surface?

    Solution
    (a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become double. Therefore, pressure will decrease to half, i.e., 0.5 atmosphere. 
    (b) Since no work is done on or by the gas, there will be no change in the internal energy of the gas.  
    (c) The temperature of the gas will remain constant because no work is being done by the gas during the expansion of the gas. 
    (d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. Hence, the gas does return to an equilibrium state. Intermediate state of the system does not lie on its PVT surface. 
    Question 1687
    CBSEENPH11020175

    A steam engine delivers 5.4×10J of work per minute and services 3.6 × 10J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

    Solution
    Work done by the steam engine per minute, W = 5.4 × 108 J 
    Heat supplied from the boiler, H = 3.6 × 109 J 
    Efficiency of the engine = fraction numerator Output space Energy over denominator Input space Energy end fraction
    ∴ η = straight W over straight H  =  fraction numerator 5.4 space cross times space 10 to the power of 8 over denominator left parenthesis 3.6 space cross times space 10 to the power of 9 right parenthesis end fraction space equals space 0.15 
    Hence, the percentage efficiency of the engine is 15 %. 
    Amount of heat wasted = 3.6 × 109 – 5.4 × 108
                                        = 30.6 × 108 
                                         = 3.06 × 109 J 
    Therefore, the amount of heat wasted per minute is 3.06 × 109 J. 
    Question 1688
    CBSEENPH11020176

    An electric heater supplies heat to a system at a rate of 100W. If the system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

    Solution
    Heat is supplied to the system at a rate of 100 W. 
    Therefore,
    Heat supplied, Q = 100 J/s 
    The system performs at a rate of 75 J/s. 
    ∴Work done, W = 75 J/s 
    From the first law of thermodynamics, we have
    Q = U + 

    where, 
    U = Internal energy 
    ∴ U = Q – W 

         = 100 – 75 
         = 25 J/s 
         = 25 W 
    Therefore, the internal energy of the given electric heater increases at a rate of 25 W. 
    Question 1689
    CBSEENPH11020177

    A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13).



    Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.

     

    Solution
    Total work done by the gas from D to E to F = Area of ΔDEF 
    Area of ΔDEF = 1 half DE × EF 
    where,
    DF = Change in pressure 
         = 600 N/m2 – 300 N/m

         = 300 N/m

    FE = Change in volume 
         = 5.0 m3 – 2.0 m

         = 3.0 m

    Area of ΔDEF = (1/2) × 300 × 3
                         = 450 J 
    Therefore, the total work done by the gas from D to E to F is 450 J. 
    Question 1690
    CBSEENPH11020178

    A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36° C, calculate the coefficient of performance.

    Solution
    Temperature inside the refrigerator, T1 = 9°C = 282 K 
    Room temperature, T2 = 36°C = 309 K 
    Coefficient of performance = fraction numerator straight T subscript 1 over denominator left parenthesis straight T subscript 2 space minus space straight T subscript 1 right parenthesis end fraction 
                                            =fraction numerator 282 over denominator left parenthesis 309 space minus space 282 right parenthesis end fraction
     
                                             = 10.44 
    Therefore, the coefficient of performance of the given refrigerator is 10.44.  
    Question 1691
    CBSEENPH11020179

    Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

    Solution
    Given,
    Diameter of an oxygen molecule, d = 3Å 
    Radius, r = straight d over 2 space equals space 3 over 2 space equals space 1.5 space A with o on top space
space space space space space space equals space 1.5 space cross times space 10 to the power of negative 8 end exponent space c m
    Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm

    Molecular volume of oxygen gas, V = 4 over 3πr3
    where,

    N
     is Avogadro’s number = 6.023 x 1023  molecules/mole 
    ∴ V = 4 over 3× 3.14 × (1.5 × 108)3 × 6.023 × 1023 
         = 8.51 cm

    Ratio of the molecular volume to the actual volume of oxygen =fraction numerator 8.51 over denominator 22400 end fraction 
                                                                                                          = 3.8 × 10-4.
    Question 1692
    CBSEENPH11020180

    Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

    Solution
    The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as, 
    PV = nRT 

    where, 
    R is the universal gas constant = 8.314 J mol–1 K–1 

    n = Number of moles = 1 
    T = Standard temperature = 273 K 
    P = Standard pressure = 1 atm = 1.013 × 105 Nm–2 

    V = nRT over straight P 

          = 1 × 8.314 × 273 / (1.013 × 105
          = 0.0224 m

         = 22.4 litres
    Hence, the molar volume of a gas at STP is 22.4 litres.
    Question 1693
    CBSEENPH11020181

    Figure 13.8 shows plot of PV/versus P for 1.00×10–3kg of oxygen gas at two different temperatures.


     


    (a) What does the dotted plot signify? 

    (b) Which is true: TT2 or T1 < T2

    (c) What is the value of PV/where the curves meet on they-axis? 

    (d) If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get the same value of PV over straight Tat the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV over straight T (for low-pressure high-temperature region of the plot)?

    (Molecular mass of H2= 2.02μ, of O2 = 32.0μ= 8.31 J mo1–1 K–1.) 

    Solution
    (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV over straight T is equal. 
    μR 
    where,
    μ
     is the number of moles, and
    R is the universal gas constant is a constant quality.
    It is not dependent on the pressure of the gas.
    (b) The dotted plot in the given graph represents an ideal gas.
    The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2.
    A real gas approaches the behaviour of an ideal gas when its temperature increases.
    Therefore, T1 > T2 is true for the given plot.
    (c) The value of the ratio PV over straight T, where the two curves meet, is μR. This is because the ideal gas equation is given as, 
            PV = μRT 

        PV over straight T = μR 
    where, 
    P is the pressure, 
    T is the temperature,
    is the volume,
    μ is the number of moles,
    R is the universal constant.
    Molecular mass of oxygen = 32.0 g 
    Mass of oxygen = 1 × 10–3 kg = 1 g 
    R = 8.314 J mole–1 K–1 

    ∴ open parentheses PV over straight T close parentheses = open parentheses 1 over 32 close parentheses × 8.314 
                = 0.26 J K-1 

    Therefore, the value of the ratio open parentheses PV over straight T close parentheses, where the curves meet on the y-axis, is
    0.26 J K–1.
    (d) If we obtain similar plots for 1.00 × 10–3 kg of hydrogen, then we will not get the same value of open parentheses PV over straight T close parentheses at the point where the curves meet the y-axis.
    This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). 
    We have, 
    open parentheses PV over straight T close parentheses = 0.26 J K-1
           R = 8.314 J mole–1 K–1 

    Molecular mass (M) of H2 = 2.02 u 
    open parentheses PV over straight T close parentheses = μR, at constant temperature
    where, μ = m/

      m = Mass of H

    ∴ m =open parentheses PV over straight T close parentheses × open parentheses straight M over straight R close parentheses
          = 0.26 × 2.02 / 8.31 
          = 6.3 × 10–2 g
          = 6.3 × 10–5 kg 
    Hence, 6.3 × 10–5 kg of H2 will yield the same value of open parentheses PV over straight T close parentheses.
    Question 1694
    CBSEENPH11020182

    An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (= 8.31 J mol–1 K–1, molecular mass of O2 = 32μ).

    Solution
    Volume of oxygen, V1 = 30 litres = 30 × 10–3 m

    Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa 
    Temperature, T1 = 27°C = 300 K 
    Universal gas constant, R = 8.314 J mole–1 K–1 

    Let the initial number of moles of oxygen gas in the cylinder be n1
    The gas equation is given as, 
    P1V1 = n1RT

    ∴ n1 = fraction numerator straight P subscript 1 straight V subscript 1 over denominator RT subscript 1 end fraction
           =fraction numerator 15.195 space cross times space 10 to the power of 5 space cross times space 30 space cross times space 10 to the power of negative 3 end exponent over denominator 8.314 space cross times space 300 end fraction
            =  18.276 
    But n1 = m1 / 

    where, 
    m1 = Initial mass of oxygen
    M = Molecular mass of oxygen = 32 g 
    Therefore, 
    m1 = n1= 18.276 × 32 = 584.84 g 
    After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces. 
    Volume, V2 = 30 litres = 30 × 10–3 m

    Gauge pressure, P2 = 11 atm
                                  = 11 × 1.013 × 105 Pa 
    Temperature, T2 = 17°C = 290 K 
    Let n2 be the number of moles of oxygen left in the cylinder. 
    The gas equation is given as, 
    P2V2 = n2RT

    ∴ n2 = P2V2RT

            =fraction numerator 11.143 space straight x 10 to the power of 5 space straight x space 30 space straight x space 10 to the power of negative 3 end exponent over denominator 8.314 space straight x space 290 end fraction
             =  13.86 
    But, 
         n2 = m2 / 

    where, 
    m2 is the mass of oxygen remaining in the cylinder 
    ∴ m2 = n2= 13.86 × 32 = 453.1 g 
    The mass of oxygen taken out of the cylinder is given by the relation, 
    Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder 
    m1 – m

    = 584.84 g – 453.1 g 
    = 131.74 g 
    = 0.131 kg 
    Therefore, 0.131 kg of oxygen is taken out of the cylinder.
    Question 1695
    CBSEENPH11020183

    An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

    Solution
    Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m

    Bubble rises to height, d = 40 m 
    Temperature at a depth of 40 m, T1 = 12°C = 285 K 
    Temperature at the surface of the lake, T2 = 35°C = 308 K 
    The pressure on the surface of the lake, 
    P2 = 1 atm = 1 ×1.013 × 105 Pa
    The pressure at the depth of 40 m, 
    P1 = 1 atm + 
    where, 
    ρ is the density of water = 103 kg/m

    g is the acceleration due to gravity = 9.8 m/s

    Therefore, 
    P1 = 1.013 × 105 + 40 × 103 × 9.8
         = 493300 Pa 
    We have,
     P1V1 / T1 = P2V2 / T

    where,
     V2 is the volume of the air bubble when it reaches the surface. 
    V2 = P1V1T2 / T1P

          = 493300 × 1 × 10-6 × 308 / (285 × 1.013 × 105
          = 5.263 × 10–6 m3 
     
           = 5.263 cm

    Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.
    Question 1696
    CBSEENPH11020184

    Estimate the average thermal energy of a helium atom at:

    (i) room temperature (27 °C),

    (ii) the temperature on the surface of the Sun (6000 K),

    (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

    Solution
    (i) At room temperature, T = 27°C = 300 K
    Average thermal energy = (3/2)kT

    Where is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

    ∴ (3/2)kT = (3/2) × 1.38 × 10-38 × 300
                   = 6.21 × 10–21J
    Hence, the average thermal energy of a helium atom at room temperature (27°C) = 6.21 × 10–21 J.
    (ii) On the surface of the sun, T = 6000 K
    Average thermal energy = (3/2)kT

                                         = (3/2) × 1.38 × 10-38 × 6000
                                         = 1.241 × 10-19 J
    Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.
    (iii) At temperature, T = 107 K
    Average thermal energy = (3/2)kT

                                         = (3/2) × 1.38 × 10-23 × 107

                                         = 2.07 × 10-16 J
    Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.
    Question 1697
    CBSEENPH11020185

    Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?

    Solution
    All the three vessels have the same capacity, they have the same volume. Hence, each gas has the same pressure, volume, and temperature.
    According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.
    Number is equal to Avogadro’s number, N = 6.023 × 1023.
    The root mean square speed (vrms) of a gas of mass m, and temperature T, is given by the relation,
      
                         vrms = (3kT/m)1/2  

    where, 
    k is Boltzmann constant 
    For the given gases, k and T are constants. 
    Hence vrms depends only on the mass of the atoms.
    i.e.,                      vrms ∝ (1/m)1/2 

    Therefore, the root mean square speed of the molecules in the three cases is not the same.
    Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest.
    Hence, neon has the largest root mean square speed among the given gases.
    Question 1698
    CBSEENPH11020186

    At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 

    Solution
    Temperature of the helium atom, THe = –20°C= 253 K
    Atomic mass of argon, MAr = 39.9 u
    Atomic mass of helium, MHe = 4.0 u
    Let, (vrms)Ar be the rms speed of argon.
    Let (vrms)He be the rms speed of helium.
    The rms speed of argon is given by, 
    open parentheses straight v subscript rms space close parentheses subscript A r end subscript space equals space square root of fraction numerator 3 R T subscript A r end subscript over denominator M subscript A r end subscript end fraction end root space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis
w h e r e comma space

R space equals space u n i v e r s a l space g a s space c o n s t a n t comma
T subscript A R end subscript space equals space t e m p e r t a u r e space o f space a r g o n space g a s space

T h e space r m s space s p e e d space o f space t h e space h e l i u m space g a s space i s space g i v e n space b y comma space

open parentheses v subscript r m s end subscript close parentheses subscript H e end subscript space equals space square root of fraction numerator 3 R T subscript H e end subscript over denominator M subscript H e end subscript end fraction end root space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis space

G i v e n space t h a t comma space

open parentheses v subscript r m s end subscript close parentheses subscript A r end subscript space equals space open parentheses v subscript r m s end subscript close parentheses subscript H e end subscript space

square root of fraction numerator 3 R T subscript A r end subscript over denominator M subscript A r end subscript end fraction end root space equals space space square root of fraction numerator 3 R T subscript H e end subscript over denominator M subscript H e end subscript end fraction end root space

T h e r e f o r e comma space

T subscript A r end subscript over M subscript A r end subscript space equals space T subscript H e end subscript over M subscript H e end subscript

T subscript A r end subscript space equals space T subscript H e end subscript over M subscript H e end subscript x space M subscript A r end subscript space

space space space space space space space equals space 253 over 4 x 39.9 space

space space space space space space space equals space 2523.675

space space space space space space space equals space 2.52 space x space 10 cubed space K
     
          = 2523.675
          = 2.52 × 103 K 
    Therefore, the temperature of the argon atom is 2.52 × 10K.
    Question 1699
    CBSEENPH11020187

    Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

    Solution

    Mean free path = 1.11 × 10–7 m
    Collision frequency = 4.58 × 109 s–1 

    Successive collision time ≈ 500 × (Collision time) 
    Pressure inside the cylinder containing nitrogen,

     P = 2.0 atm
        = 2.026 × 105 Pa 
    Temperature inside the cylinder, T = 17°C =290 K 
    Radius of a nitrogen molecule, r = 1.0 Å
                                                    = 1 × 1010 m 
    Diameter, d = 2 × 1 × 1010 = 2 × 1010 m 
    Molecular mass of nitrogen, M = 28.0 g
                                                 = 28 × 10–3 kg 
    The root mean square speed of nitrogen is given by the relation,
    straight v subscript rms space equals space square root of fraction numerator 3 RT over denominator straight M end fraction end root

where comma

straight R space is space the space universal space gas space constant space equals space 8.314 space straight J space mol to the power of negative 1 end exponent space straight K to the power of negative 1 end exponent

Therefore comma space

straight v subscript rms space equals space fraction numerator 3 straight x 8.314 space straight x 290 over denominator 28 space straight x space 10 to the power of negative 3 end exponent end fraction space equals space 508.26 space straight m divided by straight s

l italic space equals space fraction numerator KT over denominator square root of 2 straight x space straight d squared xP end fraction space

where comma space

straight k space is space the space boltzman space constant

straight k space equals space 1.38 space straight x space 10 to the power of negative 23 end exponent space kg space straight m squared space straight s to the power of negative 2 end exponent space straight K to the power of negative 1 end exponent

Therefore comma space

l italic space italic equals italic space fraction numerator space 1.38 space straight x space 10 to the power of negative 23 end exponent space straight x space 290 over denominator square root of 2 space straight x space 3.14 space straight x space left parenthesis 2 space straight x 10 to the power of negative 10 end exponent right parenthesis squared space straight x space 2.026 space straight x space 10 to the power of 5 end fraction

italic space italic space italic equals italic space 1.11 space straight x space 10 to the power of negative 7 end exponent space
    Collision frequency = straight v subscript rms over l 
                                 = 508.26 / (1.11 × 10-7
                                 =  4.58 × 109 s-1 

    Collision time is given as, 
    T = straight d over straight v subscript rms

      = fraction numerator 2 space cross times space 10 to the power of negative 10 end exponent over denominator 508.26 end fraction 
      =  3.93 × 10-13 s
    Time taken between successive collisions, 
    T ' = l / vrms 
        = 1.11 × 10-7 / 508.26 
        =  2.18 × 10-10 s
    Therefore, 
    fraction numerator straight T apostrophe over denominator straight T end fraction space equals space fraction numerator 2.18 space cross times space 10 to the power of negative 10 end exponent over denominator 3.93 space cross times space 10 to the power of negative 13 end exponent end fraction space equals space 500 
    Hence, the time taken between successive collisions is 500 times the time taken for a collision.

    Question 1700
    CBSEENPH11020188

    A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

    Solution
    Length of the narrow bore, L = 1 m = 100 cm 
    Length of the mercury thread, l = 76 cm 
    Length of the air column between mercury and the closed end, la = 15 cm 
    Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is,
     100 – (76 + 15) = 9 cm 
    Hence, the total length of the air column = 15 + 9
                                                                  = 24 cm 
    Let h cm of mercury flow out as a result of atmospheric pressure. 
    ∴ Length of the air column in the bore = 24 + h cm 
    And, length of the mercury column = 76 – h cm 
    Initial pressure, P1 = 76 cm of mercury 
    Initial volume, V1 = 15 cm

    Final pressure, P2 = 76 – (76 – h) = h cm of mercury 
    Final volume, V2 = (24 + h) cm

    Temperature remains constant throughout the process. 
    ∴                  P1V1 = P2V

                  76 × 15 = h (24 + h
    h2 + 24h – 1140 = 0

    therefore space straight h space equals space fraction numerator negative 24 space plus-or-minus space left parenthesis 24 right parenthesis squared space plus space 4 space straight x space 1 space straight x space 1140 over denominator 2 space straight x 1 end fraction
      
           = 23.8 cm
           = –47.8 cm 
    Height cannot be negative.
    Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it.
    The length of the air column will be 24 + 23.8 = 47.8 cm.
    Question 1702
    CBSEENPH11020190

    A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
    n2 = n1 exp [-mg (h– h1)/ kBT]
    Where n2n1 refer to number density at heights h2 and h1respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
    n2 = n1 exp [-mg NA(ρ - P′) (h2 –h1)/ (ρRT)]
    Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [NA is Avogadro’s number, and the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

    Solution
    According to the law of atmospheres, we have
    n2 = n1 exp [-mg (h– h1) / kBT]                    … (i) 

    where,
    nis the number density at height h1, and 
    n2 is the number density at height h

    mg is the weight of the particle suspended in the gas column 
    Density of the medium = ρ
    Density of the suspended particle = ρ 

    Mass of one suspended particle = m
    Mass of the medium displaced = 

    Volume of a suspended particle = 

    According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as, 

    Weight of the medium displaced – Weight of the suspended particle 
    mg – m'

    = mg - V ρ' g 
    =  mg - (m/ρ)ρ'

    mg(1 - (ρ'/ρ) )                                         ...(ii) 

    Gas constant, R = kB

    kB = R / N                                                  ...(iii) 

    Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get: 
    n2 = n1 exp [-mg (h– h1) / kBT
        = n1 exp [-mg (1 - (ρ'/ρ) )(h– h1)(N/RT) ] 
        = n1 exp [-mg (ρ - ρ')(h– h1)(N/RTρ) ]
    Question 1703
    CBSEENPH11020191

    Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms: 


     


    [Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

    Solution
    If r is the radius of the atom, then volume of each atom = 4/3 π r
    Volume of all atoms in one mole of substance = 4/3 π r3 × N = M/ρ 
    Therefore,
     r = [ 3M / 4πρN]1/3
    For Carbon, 
    M = 12.01 × 10-3 Kg
    ρ = 2.22 × 103 Kg m-3

     straight r space equals space fraction numerator 3 space straight x 12.01 straight x 10 to the power of negative 3 end exponent over denominator 4 space straight x space begin display style bevelled 22 over 7 end style straight x space left parenthesis 2.22 space straight x 10 to the power of 23 right parenthesis space straight x space left parenthesis 6.023 space straight x space 10 to the power of 23 right parenthesis end fraction space

space space space equals space 1.29 space straight x space 10 to the power of negative 10 space end exponent straight m
space space space equals space 1.29 space straight A with straight o on top 
    Similarly, 
     
    for gold, r = 1.59 Å 
    for liquid nitrogen, = 1.77 Å 
    for lithium, r = 1.73 Å 
    for liquid fluorine, r = 1.88 Å 
     
    Question 1704
    CBSEENPH11020192

    Which of the following examples represent periodic motion? 

    a) A swimmer completing one (return) trip from one bank of a river to the other and back.




    Solution
    The swimmer's motion is not periodic. Though the motion of a swimmer is to and fro but will not have a definite period.
    Question 1705
    CBSEENPH11020193

     Which of the following examples represent periodic motion?

    (b) A freely suspended bar magnet displaced from its N-S direction and released.

    Solution
    The motion of a freely suspended magnet, if displaced from its North-South direction is released, then the motion is periodic. This is because the magnet oscillates about its position with a definite period of time.
    Question 1706
    CBSEENPH11020194

     Which of the following examples represent periodic motion?

    (c) A hydrogen molecule rotating about its center of mass.

    Solution
    (c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such a motion is called as periodic motion. 
    Question 1707
    CBSEENPH11020195

    Which of the following examples represent periodic motion?

    d) An arrow released from a bow.

    Solution
    An arrow released from a bow moves only in the forward direction. It does not undergo a backward motion. Hence, this motion is not a periodic. 
    Question 1708
    CBSEENPH11020196

    Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? 

    (a) the rotation of earth about its axis. 

    (b) motion of an oscillating mercury column in a U-tube. 

    (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. 

    (d) general vibrations of a polyatomic molecule about its equilibrium position.

    Solution
    (a) It is periodic but not simple harmonic motion because it is not to and fro about a fixed point.
    (b) It is a simple harmonic motion (S.H.M) because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.
    (c) It is simple harmonic motion because the ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again.
    (d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.
    Question 1709
    CBSEENPH11020197

    Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): 

    (a) sin ω– cos ω

    (b) sin3 ω

    (c) 3 cos (π/4 – 2ωt

    (d) cos ω+ cos 3ωt + cos 5ω

    (e) exp (–ω2t2

    Solution
    (a) SHM 
    The given function is, 
           sin ω– cos ω
    equals space square root of 2 open square brackets fraction numerator 1 over denominator square root of 2 end fraction space sin space ωt space minus fraction numerator 1 over denominator square root of 2 end fraction space cos space ωt close square brackets

equals space square root of 2 space open square brackets sin space ωt space straight x space cos space straight pi over 4 space minus space cos space ωt space straight x space sin straight pi over 4 space close square brackets
equals square root of 2 space space sin space open parentheses ωt space space minus space straight pi over 4 space close parentheses space

     
    This function represents SHM as it can be written in the form of a sin (ωt + Φ).
    Its period is: 2π/ω

    (b) Periodic but not SHM 
    The given function is, 
    sin3 ω= 1/4 [3 sin ωt - sin3ωt
    The terms sin ωand sin ωt individually represent simple harmonic motion (SHM).
    However, the superposition of two SHM is periodic and not simple harmonic. 
    It has a period of 2π/ω

    (c) SHM 
    The given function is: 
    3 space cos space open square brackets straight pi over 4 space minus space 2 ωt close square brackets space
space equals space 3 space cos space open square brackets space 2 ωt space minus straight pi over 4 close square brackets
    This function represents simple harmonic motion because it can be written as, 
                               a cos (ωt + Φ)
    Its period is: 2π/2ω = π/ω

    (d) Periodic, but not simple harmonic motion. 
    The given function  is cosωt + cos3ωt + cos5ωt.
    Each individual cosine function represents SHM.
    However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.
    (e) Non-periodic motion 
    The given function exp(-ω2t2) is an exponential function.
    Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.
    (f) The given function 1 + ωt + ω2t2 is non-periodic.
    Question 1710
    CBSEENPH11020198

    A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is, 

    (a) at the end A, 

    (b) at the end B, 

    (c) at the mid-point of AB going towards A, 

    (d) at 2 cm away from B going towards A,
     
    (e) at 3 cm away from A going towards B, and 

    (f) at 4 cm away from B going towards A.

    Solution
     

    From above figure, where A and B represent the two extreme positions of a Simple Harmonic Motion. For velocity, the direction from A to B is taken to b positive. The acceleration and the force, along AP are taken as positive and along B are taken as negative. 
     
    (a) At the end, A, the particle executing SHM is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero. Acceleration is positive because it is directed along AP, Force is also Positive since the force is directed along AP.
     
    (b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP.
     
    (c) At the mid-point of AB going towards A, the particle is at its mean position P, with a tendency to move along PA. Hence, velocity is positive. Both acceleration and force are zero.
     
    (d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is the negative direction. Therefore,  velocity, acceleration, and force all are positive.
     
    (e) At 3 cm away from A going towards B, the particle is at R, with a tendency to move along RP, which is a positive direction. Here, velocity, acceleration all are positive.
     
    (f) At 4 cm away from A going towards A, the particles are at S, with a tendency to move along SA, which is the negative direction. Therefore, velocity is negative but acceleration is directed towards the mean position, along SP. Hence it is positive and also force is positive similarly.
     
    Question 1711
    CBSEENPH11020199

    Which of the following relationships between the acceleration and the displacement of a particle involve simple harmonic motion? 

    (a) = 0.7

    (b) = –200x

    (c) = –10x

    (d) = 100x3

    Solution
    In Simple Harmonic Motion, acceleration a is related to displacement by the relation of the form,
                         a = -kx,
    which is for relation (c). 
    Question 1712
    CBSEENPH11020200

    The motion of a particle executing simple harmonic motion is described by the displacement function,
     
    x (t) = cos (ω+ φ).  

    If the initial (= 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

    Solution
    Intially, at t = 0; 
    Displacement, x = 1 cm 
    Initial velocity, v = ω cm/ sec. 
    Angular frequency, ω = π rad/s–1 

    It is given that, 
                 x(t) = A cos(ωt + Φ
    1 = A cos(ω × 0 + Φ) = A cos Φ 
    A cosΦ = 1                                                ...(i) 
    Velocity, vdx/dt 
    ω = -A ωsin(ωt + Φ
    1 = -A sin(ω × 0 + Φ) = -sin Φ 

    sin Φ = -1                                               ...(ii) 

    Squaring and adding equations (i) and (ii), we get
    A2 (sin2 Φ + cos2 Φ) = 1 + 1 
                               A2 = 2 
    ∴                           A = √2 cm 
    Dividing equation (ii) by equation (i), we get
         tanΦ = -1 
    ∴        Φ = 3π/4 , 7π/4,...
    SHM is given as, 
    x = Bsin (ωt + α) 
    Putting the given values in this equation, we get
    1 = Bsin[ω × 0 + α] = 1 + 1 
    Bsin α = 1                                             ...(iii) 

    Velocity, v = ωBcos (ωt + α) 
    Substituting the given values, we get
       π = πBsin α 
    Bsin α = 1                                              ...(iv) 

    Squaring and adding equations (iii) and (iv), we get
    B2 [sin2 α + cos2 α] = 1 + 1 
                             B2 = 2 
    ∴                         B = √2 cm 
    Dividing equation (iii) by equation (iv), we get, 
    Bsin α / Bcos α = 1/1 
                  tan α = 1 = tan π/4
    ∴                α = π/4, 5π/4,......
    Question 1713
    CBSEENPH11020201

    A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

    Solution
    Maximum mass that the scale can read, M = 50 kg 
    Maximum displacement of the spring = Length of the scale, = 20 cm = 0.2 m 
    Time period, T = 0.6 s 
    Maximum force exerted on the spring, F = Mg 

    where, 
    g = acceleration due to gravity = 9.8 m/s

    = 50 × 9.8 = 490 
    ∴  Spring constant, k = straight F over straight l space equals space fraction numerator 490 over denominator 0.2 end fraction space equals space 2450 space N divided by m
    Mass m, is suspended from the balance. 
    Time space period comma space straight T space equals space 2 straight pi space square root of straight m over straight K end root space

therefore space space space space space space space space space space space space space space space space space space straight m space equals space open parentheses fraction numerator straight T over denominator 2 straight pi end fraction close parentheses squared space straight x space straight k space space

space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator 0.6 over denominator 2 space straight x space 3.14 end fraction close parentheses squared space straight x space 2450 space

space space space space space space space space space space space space space space space space space space space space space space space space equals space 22.36 space kg
     
    ∴Weight of the body = mg 
                                   = 22.36 × 9.8
                                    = 219.167 N 
    Hence, the weight of the body is about 219 N.
    Question 1714
    CBSEENPH11020202

    A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.


     

    Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

    Solution
    Spring constant, k = 1200 N m–1 

    Mass, = 3 kg 
    Displacement, A = 2.0 cm = 0.02 cm
    (i) Frequency of oscillation v , is given by the relation: 
    straight nu space equals space 1 over straight T space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root

where comma space straight T space is space the space time space period. space

Therefore comma space
straight nu space equals space fraction numerator 1 over denominator 2 space straight x 3.14 end fraction space square root of 1200 over 3 end root

space space space space equals space 3.18 space straight m divided by straight s
     
    Hence, the frequency of oscillations is 3.18 cycles per second.
    (ii) Maximum acceleration (a) is given by the relation, 
         a = ω2 

    where,
    straight omega space equals space Angular space frequency space equals space square root of straight k over straight m end root
straight A space equals space maximum space displacement

therefore space straight a space equals space straight k over straight m space straight A space
space space space space space space space space space equals space fraction numerator 1200 space straight x space 0.02 over denominator 3 end fraction space
space space space space space space space space space equals space 8 space straight m divided by straight s squared 
     
     
    Hence, the maximum acceleration of the mass is 8.0 m/s2.
    (iii) Maximum velocity, vmax = Aω 
                                   straight A space square root of straight k over straight m end root space equals space 0.02 space straight x space square root of 1200 over 3 end root space

space space space space space space space space space space space space space space space space space equals space 0.4 space straight m divided by straight s
     
    Hence, the maximum velocity of the mass is 0.4 m/s.
    Question 1715
    CBSEENPH11020203

    In Exercise 14.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis.

    Give as a function of time t for the oscillating mass if at the moment we start the stopwatch (= 0), the mass is, 

    (a) at the mean position, 

    (b) at the maximum stretched position, and

    (c) at the maximum compressed position. 

    In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

    Solution
    Distance travelled by the mass sideways, a = 2.0 cm 
    Angular frequency of oscillation, 
    straight omega space equals space square root of straight k over straight m end root space

space space space space equals space square root of 1200 over 3 end root space

space space space space equals square root of 400 space

space space space space space equals space 20 space rad divided by sec 
     
    (a)
    As time is noted from the mean position, hence using 
               x = a sin ω t, we have x = 2 sin 20 t
     
    (b)
    At maximum stretched position, the body is at the extreme right position, with an intial phase of π/2 rad.
    Then, 
    straight x space equals space straight a space sin space open parentheses ωt space plus straight pi over 2 close parentheses space
space space space space equals space straight a space cos space ωt
space
space space space space equals space 2 space cos space 20 space straight t 

    (c) At maximum compressed position, the body is at left position, with an initial phase of 3 π/2 rad.
    Then, 
    straight x space equals space straight a space sin space open parentheses ωt space plus space fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space

space space space equals space minus space straight a space cos space ωt space

space space space equals space minus space 2 space cos space 20 space straight t
     
    The functions neither differ in amplitude nor in frequency.
     They differ in initial phase.
    Question 1716
    CBSEENPH11020204

    Figures 14.29 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. 
    Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.


    Solution
    Time period, t = 2 s 
    Amplitude, A = 3 cm
    At time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis. 
    i.e., phase angle Φ = +π/2 
    Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by the displacement equation

    straight x space equals space straight A space cos space open square brackets fraction numerator 2 πt over denominator straight T end fraction plus straight ϕ close square brackets space

space space equals space 3 space cos space open parentheses fraction numerator 2 πt over denominator 2 end fraction plus straight pi over 2 close parentheses

space space equals space minus space 3 space sin space open parentheses fraction numerator 2 πt over denominator 2 end fraction close parentheses space

space space equals space minus space 3 space sin space straight pi space straight t space c m 
    b) Time Period, t = 4 s 
    Amplitude, a = 2 m 
    At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction.
    Hence, phase angle Φ = +π 
    Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by, 
    straight x space equals space straight a space cos space open square brackets fraction numerator 2 πt over denominator straight T end fraction space plus space straight phi close square brackets space

space space space equals space 2 space cos space open parentheses fraction numerator 2 πt over denominator 4 end fraction space plus space straight pi close parentheses

therefore space straight x space equals space minus 2 space cos space open parentheses straight pi over 2 straight t close parentheses straight m


    Question 1717
    CBSEENPH11020205

    Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (= 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (is in cm and is in s). 

    (a) = –2 sin (3+ π/3) 
    (b) = cos (π/6 – t
    (c) = 3 sin (2π+ π/4) 
    (d) = 2 cos πt

    Solution
     
    (a) 
     straight x space equals space minus space 2 space sin space open parentheses 3 straight t space plus straight pi over 3 close parentheses space

space space equals space plus space 2 space cos space open parentheses 3 straight t space plus space straight pi over 3 plus straight pi over 2 close parentheses

space space equals space 2 space cos space open parentheses 3 straight t space plus space fraction numerator 5 straight pi over denominator 6 end fraction close parentheses 
    If this equation is compared with the standard SHM equation, 
    straight x space equals space straight A space cos space open parentheses fraction numerator 2 straight pi over denominator straight T end fraction straight t space plus space straight ϕ close parentheses space

    We get, 
    Amplitude, = 2 cm 
    Phase angle, Φ = 5π/6 = 150° 
    Angular velocity = ω = 2π/= 3rad/sec
     
    The motion of the particle can be plotted as shown in fig. 10(a). 
    b) 
    straight x space equals space cos space open parentheses straight pi over 6 minus space straight t close parentheses space
space space equals space cos space open parentheses straight t space minus straight pi over 6 close parentheses
    If this equation is compared with the standard SHM equation, 
    straight x space equals space straight A space cos space open parentheses fraction numerator 2 straight pi over denominator straight T end fraction straight t space plus space straight ϕ close parentheses comma space then space we space get 
    Amplitude, A = 1
    Phase angle, Φ = -π/6 = -30°. 
    Angular velocity, ω = 2π/T = 1 rad/s. 
    The motion of the particle can be plotted as shown in fig. 10(b).
    c) 
    straight x space equals space 3 space sin space open parentheses 2 πt space plus straight pi over 4 close parentheses space

space space space equals space minus space 3 space cos space open square brackets space open parentheses 2 πt space plus straight pi over 4 close parentheses space plus straight pi over 2 close square brackets space

space space space equals space minus space 3 space cos space open parentheses 2 πt space plus fraction numerator 3 straight pi 4 over denominator 4 end fraction close parentheses
    If this equation is compared with the standard Simple Harmonic Motion the equation is, 
    straight x space equals space straight A space cos space open parentheses fraction numerator 2 straight pi over denominator straight T end fraction straight t space plus space straight ϕ close parentheses comma space then space we space get 
    Amplitude, A = 3 cm
    Phase angle, Φ = 3π/4 = 135°
    Angular velocity, ω = 2π/T = 2 rad/s.
    The motion of the particle can be plotted as shown in fig. 10(c).
    d) 
    straight x space equals space 2 space cos space πt
    If this equation is compared with the SHM equation, we get
    straight x space equals space straight A space cos open parentheses fraction numerator 2 straight pi over denominator straight T end fraction straight t space plus space straight ϕ close parentheses, we get
    Amplitude, A = 2 cm 
    Phase angle, Φ = 0 
    Angular velocity, ω = π rad/s. 
    The motion of the particle can be plotted as shown in fig. 10(d). 

    Question 1718
    CBSEENPH11020206

    Figure 14.30 (a) shows a spring of force constant clamped rigidly at one end and a mass attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows the same spring with both ends free and attached to a mass at either end. Each end of the spring in Fig. 14.30(b) is stretched by the same force 

     
    (a) What is the maximum extension of the spring in the two cases? 

    (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?


    Solution
    (a)
    The maximum extension of the spring in both cases =straight F over straight k, where k is the spring constant of the springs used.
     
    (b) In Fig.14.30(a) if x is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass is,
                                   F = -kx,
    i.e.,                           F ∝ 
    As, this F is directed towards the mean position of the mass, hence the mass attached to the spring will execute SHM. 
    Spring factor = spring constant = 
    Inertia factor = mass of the given mass = m
    As time period,
    straight T space equals space 2 straight pi space square root of fraction numerator Inertia space factor over denominator Spring space factor end fraction end root space

therefore space straight T space equals space space 2 straight pi space square root of straight m over straight K end root space 
     
    In Fig.14.30(b), we have a two body system of spring constant k and reduced mass,
                       µ = m × m / m + m = m/2. 
    Inertia factor = m/2
     
    Spring factor = k
    therefore space Time space period comma space straight T space equals space 2 straight pi space square root of fraction numerator straight m divided by 2 over denominator straight K end fraction end root space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 straight pi space square root of fraction numerator straight m over denominator 2 straight k end fraction end root

    Question 1719
    CBSEENPH11020207

    The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
     

    Solution
    Angular frequency of the piston, ω = 200 rad/ min.
    Stroke = 1.0 m
    Amplitude, A =fraction numerator 1.0 over denominator 2 end fraction = 0.5 m
    The maximum speed (vmax) of piston is given by the relation, 
    vmax = Aω
           = 200 × 0.5
           = 100m/min. 
    Question 1720
    CBSEENPH11020208

    The acceleration due to gravity on the surface of the moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of Earth is 3.5 s? (on the surface of earth is 9.8 ms–2)

    Solution
    Acceleration due to gravity on the surface of moon, g' = 1.7 m s–2 
    Acceleration due to gravity on the surface of earth, g = 9.8 m s–2 

    Time period of a simple pendulum on earth, T = 3.5 s

    Error converting from MathML to accessible text. 
    Hence, the time period of the simple pendulum on the surface of the moon is 8.4 sec. 

    Question 1721
    CBSEENPH11020209

    Answer the following: 

    (a) Time period of a particle in SHM depends on the force constant and mass of the particle: T = 2 straight pi space square root of straight m over straight k end root
    A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

    Solution
    For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in the denominator as well as in numerator. That is why the time period of a simple pendulum is independent of the mass of the bob. 
    Question 1722
    CBSEENPH11020210

    Answer the following:

    (b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that is greater than 2π √l/g

     
    Think of a qualitative argument to appreciate this result.

    Solution
    In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as, 
                             F = –mg sinθ 

    where, 
    = Restoring force
    m = Mass of the bob 
    g = Acceleration due to gravity
    θ = Angle of displacement 
    For small θ, sinθ ≈ θ 

    For large θ, sinθ is greater than θ
    This decreases the effective value of g
    Hence, the time period increases as: 
                          T = 2 straight pi space square root of straight l over straight g end root
    Question 1723
    CBSEENPH11020211

    Answer the question:

    (c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

    Solution
    Yes, the watch will give the correct time because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
    Question 1724
    CBSEENPH11020212

    Answer the Question:

    (d) What is the frequency of oscillation of a simple pendulum mounted in a cabinthat is freely falling under gravity?

    Solution
    The frequency of oscillation is zero because gravity disappears for a man under free fall. 
    Question 1725
    CBSEENPH11020213

    A simple pendulum of length and having a bob of mass is suspended in a car. The car is moving on a circular track of radius with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

    Solution
    The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. 

    Acceleration due to gravity = g 
    Centripetal acceleration = straight v squared over straight R 

    where, 
    v is the uniform speed of the car 
    R is the radius of the track 
    Effective acceleration (g') is given as,
    Error converting from MathML to accessible text.

    Question 1726
    CBSEENPH11020214

    Cylindrical piece of cork of density of base area and height floats in a liquid of density ρ1. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period,

    straight T space equals space 2 straight pi space fraction numerator hρ over denominator straight rho subscript 1 straight g end fraction
     
    where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

    Solution
    Given,
    Base area of the cork = A
    Height of the cork = h
    Density of the liquid = ρ
    Density of the cork = ρ

    In equilibrium:
    Weight of the cork = Weight of the liquid displaced by the floating cork
    Let the cork be depressed slightly by x.
    As a result, some extra water of a certain volume is displaced.
    Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
    Up-thrust = Restoring force, F = Weight of the extra water displaced 
    F = ­–(Volume × Density × g
    Volume = Area × Distance through which the cork is depressed
    Volume = Ax 

    Therefore, 
    F = – A ρg                              ...(i) 

    According to the force law: 
    F = kx 

    k = F/

    where, k is constant 
    k = F/x = -Aρg                          ...(ii) 

    The time period of the oscillations of the cork: 
    T = 2π √m/                              ...(iii) 

    where,
    m = Mass of the cork 
       = Volume of the cork × Density  
       = Base area of the cork × Height of the cork × Density of the cork
       = Ahρ 

    Hence, the expression for the time period is given by, 
    straight T space equals space 2 straight pi space square root of fraction numerator Ahρ over denominator Aρ subscript 1 straight g end fraction end root space equals space 2 straight pi space square root of fraction numerator hρ over denominator straight rho subscript 1 straight g end fraction end root space
    Question 1727
    CBSEENPH11020215

    One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

    Solution
    Area of cross-section of the U-tube = A

    Density of the mercury column = ρ 

    Acceleration due to gravity = 

    Restoring force, F = Weight of the mercury column of a certain height,
    F = –(Volume × Density × g
    F = –(A × 2h × ρ × g)
       = –2Aρgh = –k × Displacement in one of the arms (h)
    where,
    2h is the height of the mercury column in the two arms
    k is a constant, given by k =  negative straight F over straight h = 2Aρg
    Time space period comma space straight T space equals space 2 straight pi space square root of straight m over straight k end root space equals space 2 straight pi space square root of fraction numerator straight m over denominator 2 Aρg end fraction end root
    where, 
    m is the mass of the mercury column, 
    Let l be the length of the total mercury in the U-tube.
    Mass of mercury, m = Volume of mercury × Density of mercury = Alρ
    Error converting from MathML to accessible text.
    Hence, the mercury column executes simple harmonic motion
    with time period 2 pi square root of fraction numerator straight l over denominator 2 straight g end fraction end root .
     
    Question 1728
    CBSEENPH11020216

    An air chamber of volume has a neck area of cross section into which a ball of mass just fits and can move up and down without any friction (Fig.14.33). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.33]. 






    Solution
    Volume of the air chamber = V

    Area of cross-section of the neck = 

    Mass of the ball = 

    The pressure inside the chamber is equal to the atmospheric pressure. 
    Let the ball be depressed by units.
    As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber. 
    Decrease in the volume of the air chamber, ΔV = ax 

    Volumetric strain = Change in volume/original volume 
    ⇒                ΔV/V = ax/

    Bulk modulus of air, B = Stress/Strain = fraction numerator straight rho over denominator begin display style bevelled ax over straight V end style end fraction 

    In this case, stress is the increase in pressure.
    The negative sign indicates that pressure increases with decrease in volume. 
    p = -Bax/

    The restoring force acting on the ball, F = p × 

                                                               equals space minus Bax over straight V space cross times space a space

equals space minus fraction numerator B a x squared over denominator V end fraction        ...(i) 

    In simple harmonic motion, the equation for restoring force is, 
    F = -kx                                                          ...(ii) 

    where, k is the spring constant.
    Comparing equations (i) and (ii), we get:
    k = Ba2/

    Time Period,

    therefore space straight T space equals space 2 straight pi square root of straight m over straight k end root space

space space space space space space space space space equals space space 2 straight pi space square root of Vm over Ba squared end root 
    Question 1729
    CBSEENPH11020217

    You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant and (b) the damping constant for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

    Solution
    (a)
    Mass of the automobile, m = 3000 kg
    Displacement in the suspension system, x = 15 cm = 0.15 m
    There are 4 springs in parallel to the support of the mass of the automobile.
    The equation for the restoring force for the system:
     
                           F = –4kx = mg 

    where,
    is the spring constant of the suspension system
    Time period, T = 2π √m/4k

    Spring constant, k = mg/4x 
                            = 3000 × 10/ 4 × 0.15
                            = 5000
                            = 5 × 104 Nm 
    Spring Constant, = 5 × 104 Nm
    (b) Each wheel supports a mass, M = 3000/4 = 750 kg 
    For damping factor b, the equation for displacement is written as
    x = x0e-bt/2M

    The amplitude of oscilliation decreases by 50 %. 
    ∴  x = x0/2 
    x0/2 = x0e-bt/2M

    loge2 = bt/2

    Therefore,
         b = 2loge2 / t
    where,

    Time space Period comma space straight T space equals space 2 straight pi space square root of fraction numerator straight m over denominator 4 straight k end fraction end root space

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 straight pi space space square root of fraction numerator 3000 over denominator 4 space straight x space 5 space straight x space 10 to the power of 4 end fraction end root

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.7691 space straight s

therefore space straight b space equals space fraction numerator 2 space straight x space 750 space straight x 0.693 over denominator 0.7691 end fraction space equals space 1351. space 58 space kg divided by straight s
    Question 1730
    CBSEENPH11020218

    Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
     

    Solution
    The equation of displacement of a particle executing SHM at an instant t is given as: 
    x = Asin ω
    where, 
    A = Amplitude of oscillation 
    ω = Angular frequency = √k/
    The velocity of the particle, v = dx/dt = Aωcosω

    The kinetic energy of the particle is, 
    Ek = 1/2 Mv2 = 1/2 MA2 ω2 cos2 ω

    The portential energy of the particle is, 
    Ep = 1/2 kx2 = 1/2 M2 ω2 A2 sin2 ωt

    For time period T, the average kinetic energy over a single cycle is given as:

    open parentheses straight E subscript straight k close parentheses subscript a v g end subscript space equals space 1 over T space integral subscript 0 superscript T E subscript k space d t space

equals space 1 over T space integral subscript 0 superscript T space 1 half M A squared omega squared space cos squared space omega t space d t space
equals space fraction numerator 1 over denominator 2 T end fraction space M A squared omega squared italic space integral subscript italic 0 superscript T italic space open parentheses fraction numerator italic 1 italic space italic plus italic space italic cos italic space italic 2 omega t over denominator italic 2 end fraction close parentheses d t italic space

italic equals italic space fraction numerator italic 1 over denominator italic 4 T end fraction M A to the power of italic 2 omega to the power of italic 2 italic space open parentheses t italic space italic plus fraction numerator italic sin italic space italic 2 omega t over denominator italic 2 omega end fraction close parentheses subscript italic 0 superscript T italic space

italic equals italic space fraction numerator italic 1 over denominator italic 4 T end fraction M A to the power of italic 2 omega to the power of italic 2 italic space italic left parenthesis T italic right parenthesis italic space

italic equals italic space italic 1 over italic 4 M A to the power of italic 2 omega to the power of italic 2 italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic space left parenthesis 1 right parenthesis 
    Average potential energy over one cycle is given as, 
    open parentheses straight E subscript straight P close parentheses subscript A v g end subscript space equals space 1 over T integral subscript 0 superscript T E subscript P space d t space

space space space space space space space space space space space space space equals space 1 over T space integral subscript 0 superscript T 1 half M omega squared A squared space s i n squared omega t space d t

space space space space space space space space space space space space equals fraction numerator 1 over denominator 2 T end fraction M omega squared space A squared integral subscript 0 superscript T fraction numerator left parenthesis 1 space minus space c o s space 2 omega t right parenthesis over denominator 2 end fraction d t

space space space space space space space space space space space space equals space fraction numerator 1 over denominator 4 T end fraction M omega squared space A squared italic space open square brackets t italic space italic minus fraction numerator s i n italic space italic 2 omega t over denominator italic 2 omega end fraction close square brackets subscript italic 0 superscript T italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic equals italic space fraction numerator italic 1 over denominator italic 4 T end fraction italic space M omega to the power of italic 2 space A to the power of italic 2 italic space italic left parenthesis T italic right parenthesis italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic equals fraction numerator M omega to the power of italic 2 italic space A to the power of italic 2 over denominator italic 4 end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic space italic left parenthesis i i italic right parenthesis thin space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space
    From equations (i) and (ii) we can say that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

    Question 1731
    CBSEENPH11020219

    A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation = –α θ, where is the restoring couple and θ the angle of twist).

    Solution
    Mass of the circular disc, m = 10 kg
    Radius of the disc, r = 15 cm = 0.15 m
    The torsional oscillations of the disc has a time period, T = 1.5 s 
    The moment of inertia of the disc is, 
    =1 halfmr

      = 1 half× (10) × (0.15)

      = 0.1125 kg/m2 
    Time Period, T = 2π √I/α 
    α is the torsional constant. 
    α = fraction numerator 4 straight pi squared straight I over denominator straight T squared end fraction
       = 4 × (π)2 × fraction numerator 0.1125 over denominator left parenthesis 1.5 right parenthesis squared end fraction 

       = 1.972 Nm/rad 
    Hence, the torsional spring constant of the wire is 1.972 Nm rad–1
    Question 1732
    CBSEENPH11020220

    A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.

    Solution
    Given, 
    r
     = 5 cm = 0.05 m
    T = 0.2 s
    ω = 2π/T = 2π/0.2 = 10 π rad/s
    When displacement is y, then acceleration, A = -ω2 y
    Velocity, 
     straight v space equals space straight omega space square root of straight r squared minus straight y squared end root

Case space left parenthesis straight a right parenthesis

When space straight y space equals space 5 space cm space equals space 0.05 space straight m space

straight A space equals space minus space left parenthesis 10 space straight pi right parenthesis squared space straight x space 0.05 space

space space space space equals space minus space 5 space straight pi squared space straight m divided by straight s squared space

straight V space equals space 10 straight pi space square root of left parenthesis 0.05 right parenthesis squared space minus space left parenthesis 0.05 right parenthesis squared end root

space space space space space equals space 0

Case space left parenthesis straight b right parenthesis colon

When space straight y space equals space 3 space cm space equals space 0.03 space straight m space

space space space space straight A space equals space left parenthesis negative 10 space straight pi right parenthesis squared space straight x space 0.03 space

space space space space space space space equals space minus space 3 space straight pi squared space straight m divided by straight s squared space

space straight V space equals space 10 straight pi space straight x space square root of left parenthesis 0.05 right parenthesis squared space minus space left parenthesis 0.03 right parenthesis squared end root space

space space space space space space equals space 10 space straight pi space straight x space 0.04

space space space space space space equals space 0.4 space straight pi space straight m divided by straight s

Case space left parenthesis straight c right parenthesis colon

When space straight y space equals space 0

straight A space equals space minus space left parenthesis 10 straight pi right parenthesis squared thin space straight x space 0 space equals space 0

straight V space equals space 10 straight pi space straight x space square root of left parenthesis 0.05 right parenthesis squared minus left parenthesis 0 right parenthesis squared end root space

space space space space equals space 10 space straight pi space straight x space 0.05 space

space space space space equals space 0.5 space straight pi space straight m divided by straight s
     
    Question 1733
    CBSEENPH11020221

    A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation acos (ωt) and note that the initial velocity is negative.]

    Solution
    The displacement equation for an oscillating mass is given by, 
    x = Acos(ωt + θ
    where, 
    A is the amplitude 
    x is the displacement 
    θ is the phase constant
    Velcoity, v = dx/dt = -Aωsin(ωt + θ
    At t = 0, x = x

    x0 = Acos θ = x0        ...(i) 

    and,
    dx over dt = -v0 = Aωsinθ 

    Asinθ v0/ω            ...(ii) 
    Squaring and adding equations (i) and (ii), we get

    straight A squared space open parentheses cos squared space straight theta space plus space sin squared straight theta close parentheses space equals space straight x subscript straight o squared space plus space open parentheses straight v subscript straight o close parentheses squared over straight omega squared space
therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight A space equals space square root of straight x subscript straight o squared space plus space open parentheses straight v subscript straight o over straight omega close parentheses squared end root
    So, A is the required amplitude of oscillation. 
    Question 1734
    CBSEENPH11020222

    Use the formula v = square root of γP over straight rho end root to explain why the speed of sound in air:

    (a) is independent of pressure,

    Solution
    Take the relation: 
    v = √γP/ρ                          ...(i) 

    where, 
    Density, ρ = Mass/Volume = M/

    M = Molecular weight of the gas 
    V = Volume of the gas
    Hence, equation (i) reduces to: 
    v = √γPV/                        ...(ii) 

    Now from the ideal gas equation for n = 1: 
    PV RT 

    For constant TPV = Constant 
    Since both M and γ are constants, v = Constant 
    Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.
    Question 1735
    CBSEENPH11020223

    Use the formula v = square root of γP over straight rho end root to explain why the speed of sound in air:
    (b) increases with temperature,

    Solution
    Take the relation, 
    v = √γP/ρ                               ...(i) 

    For one mole of any ideal gas, the equation can be written as: 
    PV = RT 

    P = RT/V                                   ...(ii) 

    Substituting equation (ii) in equation (i), we get: 
    v = √γRT/Vρ  = √γRT/M            ...(iii) 

    where, 
    mass, M = ρV is a constant 
    γ and R are also constants. 
    We conclude from equation (iii) that v ∝ √

    Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa. 
    Question 1736
    CBSEENPH11020224

    Use the formula v = √γP/ρ to explain why the speed of sound in air 

    (c) increases with humidity.

    Solution
    Let vm and vd be the speed of sound in moist air and dry air respectively.
    Let ρm and ρd be the densities of the moist air and dry air respectively.

    Using space the space relation comma space

straight nu space equals space square root of γP over straight rho end root space

Speed space of space sound space in space moist space air space is comma space

straight v subscript straight m space equals space square root of γρ over straight rho subscript straight m end root space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis thin space

Speed space of space sound space in space dry space air space is comma

straight v subscript straight d space equals space square root of γρ over straight rho subscript straight d end root space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

On space dividing space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have space

straight v subscript straight m over straight v subscript straight d space equals space square root of γρ over straight rho subscript straight m straight x straight rho subscript straight d over γρ end root space space equals space straight rho subscript straight d over straight rho subscript straight m

 
    However, the presence of water vapour reduces the density of air.
    i.e.,                ρd < ρ
    ∴                    vm > v
    Hence, the speed of sound in moist air is greater than it is in dry air.
    Thus, in gaseous medium, the speed of sound increases with humidity. 
    Question 1737
    CBSEENPH11020225

    You have learnt that a travelling wave in one dimension is represented by a function (x, t)where and must appear in the combination x – v t or x + v t, i.e.y = f (x ± v t).

    Is the converse true? Examine if the following functions for can possibly represent a travelling wave:

    (a) (x – vt)2

    (b) log [(x + vt/ x0]

    (c) 1 / (x + vt)

    Solution
    No, the converse is not true. The basic requirements for a wave function to represent a travelling wave is that for all values of x and t, wave function must have finite value. Out of the given functions for y, no one satisfies this condition. Therefore, none can represent a travelling wave.
    Question 1738
    CBSEENPH11020226

    A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s–1 and in water 1486 m s–1.

    Solution
    (a) Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz 
    Speed of sound in air, va = 340 m/s 
    The wavelength (λr) of the reflected sound is given by the relation, 
    λv/

        = 340/106 
        = 3.4 × 10-4 m.
    (b) Frequency of the ultrasonic sound, ν = 1000 kHz
                                                                 = 106 Hz 
    Speed of sound in water, vw = 1486 m/s 
    The wavelength of the transmitted sound is given as, 
    λ= 1486 × 106
        
    = 1.49 × 10-3 m.
    Question 1739
    CBSEENPH11020227

    A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz.

    Solution
    Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 103m/s
    Operating frequency of the scanner, ν = 4.2 MHz
                                                              = 4.2 × 106 Hz
    The wavelength of sound in the tissue is given as,
    λ = v/v

       = 1.7 × 103 / 4.2 × 106 
        = 4.1 × 10-4 m.
    Question 1740
    CBSEENPH11020228

    For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

    Solution
    All the waves have different phases.
    The given transverse harmonic wave is, 
     
    straight y space left parenthesis straight x comma straight t right parenthesis space equals space 3.0 space sin space open parentheses 36 space straight t space plus space 0.018 space straight x space plus space straight pi over 4 close parentheses space space space space space space space space space space... space left parenthesis straight i right parenthesis space

For space straight x space equals space 0 comma space the space equation space reduces space to colon

straight y space left parenthesis straight x comma straight t right parenthesis space space 3.0 space sin space 3.0 space sin space open parentheses 36 space straight t space plus space straight pi over 4 close parentheses

Also comma space

straight omega space equals space fraction numerator 2 straight pi over denominator straight t end fraction space equals space 36 space rad divided by sec
therefore space space space space space space space space space space space space space straight t space equals space straight pi over 18 space straight s

    Now, plotting y vs. t graphs using the different values of t, as listed in the given table

    t(s) 0 T/8 2T/8 3T/8 4T/8 5T/8 6T/8 7T/8
    y (cm) fraction numerator 3 over denominator square root of 2 end fraction 3 fraction numerator 3 over denominator square root of 2 end fraction 0 -fraction numerator 3 over denominator square root of 2 end fraction -3 -fraction numerator 3 over denominator square root of 2 end fraction 0
     
    For x = 0, x = 2, and x = 4, the phases of the three waves will get changed.
    This is because amplitude and frequency are invariant for any change in x.
    The position-time plots of the three waves are shown in the given figure. 

    Question 1741
    CBSEENPH11020229

    For the travelling harmonic wave,

    (x, t) = 2.0 cos 2π (10– 0.0080+ 0.35)

    Where and are in cm and in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of 

    (a) 4 m, 

    (b) 0.5 m, 

    (c) λ/2 

    (d) 3λ/4

    Solution
    Equation for a travelling harmonic wave is given as:
    y (xt) = 2.0 cos 2π (10t – 0.0080x + 0.35) 
               = 2.0 cos (20πt – 0.016πx + 0.70 π) 
    where, 
    Propagation constant, k = 0.0160 π 
    Amplitude, = 2 cm 
    Angular frequency, ω= 20 π rad/s 
    Phase difference is given by the relation, 
    Φ kx = fraction numerator 2 straight pi over denominator straight lambda end fraction
    (a) For x = 4 m = 400 cm 
               Φ = 0.016 π × 400
                   = 6.4 π rad
    (b) For 0.5 m = 50 cm,
                      Φ = 0.016 π × 50
                          = 0.8 π rad
    (c) For x = λ/2 
              Φ = 2π/λ × λ/2 = π rad
    (d) For x = 3λ/4 
               Φ = 2π/λ × 3λ/4 = 1.5π rad.
    Question 1742
    CBSEENPH11020230

    The transverse displacement of a string (clamped at its both ends) is given by

    straight y space left parenthesis straight x comma straight t right parenthesis space equals space 0.06 space sin space 2 over 3 space straight x space cos space left parenthesis 120 space straight pi space straight t right parenthesis
    Where and are in m and in s.

    The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg. 

    Answer the following:

    (a) Does the function represent a travelling wave or a stationary wave?

    (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

    (c) Determine the tension in the string.

    Solution
    (a) The general equation representing a stationary wave is given by the displacement function: 
    y (xt) = 2a sin kx cos ω

    This equation is similar to the given equation, 
             straight y space left parenthesis straight x comma straight t right parenthesis space equals space 0.06 space sin space 2 over 3 space straight x space cos space left parenthesis 120 space straight pi space straight t right parenthesis
    Hence, the given equation represents a stationary wave.
     
    (b)
    A wave travelling along the positive x-direction is given as: 
    y1 = a sin (ωt kx
    The wave travelling along the positive x-direction is given as: 
    y2 = a sin (ωt + kx
    The supposition of these two waves yields: 
    y = y1 + y2
        = 
    a sin (ωt kx) - a sin (ωt + kx
       = a sin (ωt) cos (kx) - a sin (kx) cos (ωt) -  a sin (ωt) cos (kx) - a sin (kx) cos (ωt
       = 2 a sin (kx) cos (ωt
    negative space 2 space straight a space sin space open parentheses fraction numerator 2 straight pi over denominator straight lambda end fraction close parentheses space cos space left parenthesis space 2 πνt right parenthesis space space space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis
The space transverse space displacement space of space the space string space is space given space as colon

straight y space left parenthesis straight x comma space straight t right parenthesis thin space equals space 0.06 space sin space open parentheses fraction numerator 2 straight pi over denominator 3 end fraction space straight x close parentheses space space cos space left parenthesis 120 space straight pi space straight t right parenthesis space space space space space space space space space space... space left parenthesis ii right parenthesis space

Comparing space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have space

fraction numerator 2 straight pi over denominator straight lambda end fraction space equals space fraction numerator 2 straight pi over denominator 3 end fraction

Therefore comma space

Wavelength comma space straight lambda space equals space 3 space straight m space
It space is space given space that comma space

120 space straight pi space equals space 2 πν space

Frequency comma space straight nu space equals space 60 space Hz space

Wave space Speed comma space straight v space equals space νλ space equals space 60 space straight x space 3 space equals space 180 space straight m divided by straight s

    Therefore,
    Wavelength, λ = 3 m 
    It is given that, 
    120π = 2πν 
    Frequency, ν = 60 Hz 
    Wave speed, νλ 

                       = 60 × 3
                       = 180 m/s
    (c) The velocity of a transverse wave travelling in a string is given by the relation, 
    v = square root of straight T over straight mu end root                               ...(i) 

    where, 
    Velocity of the transverse wave, v = 180 m/s 
    Mass of the string, m = 3.0 × 10–2 kg 
    Length of the string, = 1.5 m 
    Mass per unit length of the string, µ = m/l 

                                                   =fraction numerator 3.0 over denominator 1.5 space cross times space 10 to the power of negative 2 end exponent end fraction

                                                   = 2 × 10-2 kg m-1
    Tension in the string = 

    From equation (i), tension can be obtained as
    T = v2μ 

       = (180)2 × 2 × 10–2 

       = 648 N 
    Question 1743
    CBSEENPH11020231

    i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

    Solution
    All the points on the string, 
    (i) have the same frequency except at the nodes (where frequency is zero) 
    (ii) have thse same phase everywhere in one loop except at the nodes. 
    However, the amplitude of vibration at different points is different.
    Error converting from MathML to accessible text.     
    Question 1744
    CBSEENPH11020232

    A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

    Solution
    (a) Mass of the wire, m = 3.5 × 10–2 kg 
    Linear mass density, μ =straight m over straight l = 4.0 × 10-2 kg m-1 

    Frequency of vibration, v = 45 Hz
    ∴ length of the wire, l =straight m over straight mu 
                                     = fraction numerator 3.5 space cross times space 10 to the power of negative 2 end exponent over denominator 4.0 space cross times space 10 to the power of negative 2 end exponent end fraction space equals space 0.875
    The wavelength of the stationary wave (λ) is related to the length of the wire by the relation, 
                                  straight lambda space equals space fraction numerator 2 space straight l over denominator straight m end fraction

    where, 
    n = Number of nodes in the wire. 
    For fundamental node, n = 1:
    λ = 2

    λ = 2 × 0.875
       = 1.75 m 
    The speed of the transverse wave in the string is given as, 
    = νλ
       = 45 × 1.75
       = 78.75 m/s
    (b) The tension produced in the string is given by the relation, 
    v2µ 

       = (78.75)2 × 4.0 × 10–2
       = 248.06 N
    Question 1745
    CBSEENPH11020233

    A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

    Solution
    Frequency of the turning fork, ν = 340 Hz 
    Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure. 
                              

    Such a system produces odd harmonics.
    The fundamental note in a closed pipe is given by the relation, 
    l1 = straight pi over 4 
    where, 
    length of pipe, l1 = 25.5 cm = 0.255 m 
    ∴ λ = 4l1 
          = 4 × 0.255
          = 1.02 m 
    The speed of the sound is given by the relation, 
    v = vλ
      = 340 × 1.02
      = 346.8 m/s.
    Question 1746
    CBSEENPH11020234

    A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

    Solution
    Length of the steel rod, l = 100 cm = 1 m 
    Fundamental frequency of vibration, ν = 2.53 kHz
                                                              = 2.53 × 10Hz 
    When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.



    The distance between two successive node is λ/2 
    ∴                          l = λ/2 
    λ = 2l = 2 × l = 2 m 
    The speed of sound in steel is given by the relation, 
    v = νλ 
    = 2.53 × 103 × 2
    = 5.06 × 103 m/s
    = 5.06 km/s
    Question 1747
    CBSEENPH11020235

    Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

    Solution
    Frequency of string A, fA = 324 Hz
    Frequency of string B = fB 
    Beat’s frequency, n = 6 Hz 
    Beat's Frequency is given as, 
    straight n space equals space open vertical bar straight f subscript straight A space plus space straight f subscript straight B close vertical bar space

6 space equals space 324 space plus-or-minus space space straight f subscript straight B space

straight f subscript straight B space equals space 330 space Hz space or space 318 space Hz
    Frequency decreases with a decrease in the tension in a string.
    This is because the frequency is directly proportional to the square root of tension.
    It is given as, 
    v ∝ √

    Hence, the beat frequency cannot be 330 Hz. 
    ∴                  fB = 318 Hz.
    Question 1748
    CBSEENPH11020236

    Explain why (or how):

    (a) In a sound wave, a displacement node is a pressure antinode and vice versa,

    Solution
    A node (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum. 
    An antinode (A) is a point where the amplitude of vibration is the maximum and pressure is the minimum.  
    Therefore, a displacement node is nothing but a pressure antinode, and vice versa. 
    Question 1749
    CBSEENPH11020237

    Explain why (or how):

    (b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

    Solution
    Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.
    Question 1750
    CBSEENPH11020238

     Explain why (or how):

    (c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,

    Solution
    The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Therefore, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration. 
    Question 1751
    CBSEENPH11020239

    Explain why (or how):

    (d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and

    Solution
    Solids have both, the elasticity of volume and elasticity of shape, whereas gases have only the volume elasticity. 
    Question 1752
    CBSEENPH11020240

     Explain why (or how):

    (e) The shape of a pulse gets distorted during propagation in a dispersive medium.

    Solution
    A sound pulse is a combination of waves of different wavelength.  
    As waves of different λ travel in a dispersive medium with different velocities, therefore, the shape of the pulse gets distorted. 
    Question 1753
    CBSEENPH11020241

    A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.

    (i) What is the frequency of the whistle for a platform observer when the train

    (a) approaches the platform with a speed of 10 m s
    –1,

    (b) recedes from the platform with a speed of 10 m s
    –1?

    (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s
    –1.

    Solution
    (i)
    (a)Frequency of the whistle, ν = 400 Hz 
    Speed of the train, vT= 10 m/s 
    Speed of sound, v = 340 m/s 
    The apparent frequency (v') of the whistle as the train approaches the platform is given by the  relation, 
    straight nu to the power of apostrophe space equals space open parentheses fraction numerator straight nu over denominator straight nu space minus space straight nu subscript straight r end fraction close parentheses space straight nu space

equals space open parentheses fraction numerator 340 space over denominator 340 space minus space 10 end fraction close parentheses space cross times 400 space

equals space 412.12 space Hz

straight b right parenthesis thin space The space apparent space frequency space left parenthesis straight nu apostrophe right parenthesis thin space of space the space whistle space as space the
train space recedes space from space the space platform space is space given space by space the space relation colon

straight nu to the power of apostrophe apostrophe end exponent space equals space open parentheses fraction numerator straight nu over denominator straight nu space minus space straight nu subscript straight r end fraction close parentheses space straight nu

space space space space space space equals space open parentheses fraction numerator 340 over denominator 340 space plus 10 end fraction close parentheses space cross times space 400 space

space space space space space space equals space 388. space 57 space Hz

  
    (ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e., 340 m/s.
    Question 1754
    CBSEENPH11020242

    A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s–1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s–1? The speed of sound in still air can be taken as 340 m s–1.

    Solution
    For the stationary observer: 
    Frequency of the sound produced by the whistle, ν = 400 Hz
    Speed of sound = 340 m/s
    Velocity of the wind, v = 10 m/s
    As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz. 
    The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units.
    i.e., Effective speed of the sound, ve = 340 + 10
                                                           = 350 m/s 
    The wavelength (λ) of the sound heard by the observer is given by the relation, 
    λ ve/v 
       = 350/400
       = 0.875 m 
    For the running observer: 
    Velocity of the observer, vo = 10 m/s 

    The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (v'). 
    This is given by the relation, 
    straight nu apostrophe space equals space open parentheses fraction numerator straight nu space plus space straight nu subscript straight o over denominator straight nu end fraction close parentheses straight nu

space space space space equals space open parentheses fraction numerator 340 space plus space 10 space over denominator 340 end fraction close parentheses space straight x space 400 space

space space space space equals space 411.76 space Hz 

    Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s 
    The source is at rest. Hence, the wavelength of the sound will not change.
     That is,  λ remains 0.875 m. 
    Hence, the given two situations are not exactly identical.
    Question 1755
    CBSEENPH11020243

    A travelling harmonic wave on a string is described by: 


    straight y space left parenthesis straight x comma straight t right parenthesis thin space equals space 7.5 space sin space left parenthesis 0.0050 space straight x space plus space 12 space straight t space plus space straight pi over 4 right parenthesis

    (a) What are the displacement and velocity of oscillation of a point at = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

    (b) Locate the points of the string which have the same transverse displacements and velocity as the x =1 cm point at t = 2 s, 5 s and 11 s.

    Solution
    (a) The given harmonic wave is:

    straight y space left parenthesis straight x comma straight t right parenthesis thin space equals space 7.5 space sin space left parenthesis 0.0050 space straight x space plus space 12 space straight t space plus space straight pi over 4 right parenthesis space
For space straight x space equals space 1 space cm space and space straight t space equals space 1 space straight s comma

straight y space left parenthesis 1 comma 1 right parenthesis space equals space 7.5 space sin space left parenthesis 0.0050 space straight x space plus space 12 space plus space straight pi over 4 right parenthesis space
space space space space space space space space space space space space equals space 7.5 space sin space straight theta

where comma space

straight theta space equals space 12.0050 space plus space straight pi over 4 space
space space space space equals space 12.0050 space plus space fraction numerator 3.14 over denominator 4 end fraction
space space space space equals space 12.79 space rad

space space space space equals fraction numerator 180 over denominator 3.14 end fraction space straight x space 12.79 space equals space 732.81 to the power of straight o space

therefore space straight y space equals space left parenthesis 1 comma 1 right parenthesis space equals space 7.5 space sin space left parenthesis 732.81 to the power of straight o space right parenthesis

space space space space space space space space equals space 7.5 space sin space left parenthesis 90 space cross times space 8 space plus space 12.81 to the power of straight o right parenthesis space

space space space space space space space space equals space 7.5 space sin space space 12.81 to the power of straight o space

space space space space space space space space equals space 7.5 space cross times space 0.2217 space

space space space space space space space space equals space 1.6629 space tilde space 1.663 space cm

Velocity space of space oscillation space at space any space point space is space given space by comma space

straight v space equals space fraction numerator d over denominator d straight t end fraction open square brackets 7.5 space sin space open parentheses space 0..50 space straight x space plus space 12 space straight t space plus space straight pi over 4 close parentheses close square brackets

space space space equals space 7.5 space cross times space 12 space cos space open parentheses 0.0050 space straight x space plus space 12 space straight t space plus space straight pi over 4 close parentheses
At space straight x space equals space 1 space cm space and space straight t equals 1 space sec comma space we space have

straight v space equals space straight y space left parenthesis 1 comma 1 right parenthesis space equals space 90 space cos space open parentheses 12.005 space plus space straight pi over 4 close parentheses

    = 90 coss (732.81°) = 90 cos (90 × 8 + 12.81°) 
    = 90 cos (12.81°) 
    = 90 × 0.975 =87.75 cm/s 
    Now, the equation of a propagating wave is given by, 
    y (xt) = asin (kx + wt + Φ
    where, 
       k = 2π/λ 

    ∴ λ = 2π/

    and,
       ω = 2π

    Therefore, 
       v = ω/2π 
    Speed, v = vλ = ω/

    where, 
    ω = 12rad/s 
    k = 0.0050 m-1 

    Therefore, 
    v = 12/0.0050 = 2400 cm/s 
    Hence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wave propagation. 
    (b) Propagation constant is related to wavelength as, 
       k = 2π/λ 

    ∴ λ = 2π/= 2 × 3.14 / 0.0050 
         = 1256 cm = 12.56 m \
    Therefore, all the points at distance nλ (n = ±1, ±2, .... and so on), i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.
    Question 1756
    CBSEENPH11020244

    A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz? 

    Solution
    (a) A short pipe be a whistle has neither a definite wavelength nor a definite frequency. However, its speed of propagation is fixed, being equal to the speed of sound in air.
    (b) No, frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pip of the whistle.  
    Question 1757
    CBSEENPH11020245

    One end of a long string of linear mass density 8.0 × 10–3 kg m–1is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At = 0, the left end (fork end) of the string = 0 has zero transverse displacement (= 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement as function of and that describes the wave on the string.

    Solution
    The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation: 
    y (xt) = a sin (wt – kx)                             … (i) 

    Linear mass density, μ = 8.0 × 10-3 kg m-1 
    Frequency of the tuning fork, ν = 256 Hz 
    Amplitude of the wave, = 5.0 cm
                                           = 0.05 m           … (ii) 

    Mass of the pan, = 90 kg 
    Tension in the string, T = mg
                                        = 90 × 9.8
                                         = 882 N 
    The velocity of the transverse wave v, is given by the relation, 

    straight v space equals space square root of straight t over straight mu end root space

space space equals space fraction numerator 882 over denominator 8.0 space straight x space 10 to the power of negative 3 end exponent end fraction space equals space 332 space straight m divided by straight s

Angular space frequency comma space straight omega space equals space 2 πν space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 space straight x space 3.14 space straight x space 256 space space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1608.5 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.6 space straight x space 10 cubed space rad divided by sec space space space space space space space space... space left parenthesis iii right parenthesis thin space

Wavelength comma space straight lambda space equals space straight v over bold nu space equals space 332 over 256 space straight m

therefore space Propagation space constant comma space straight k space equals space fraction numerator 2 straight pi over denominator straight lambda end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 space cross times space 3.14 over denominator begin display style bevelled 332 over 256 end style end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4.84 space straight m to the power of negative 1 end exponent space space space space space space space space space space space space space... space left parenthesis iv right parenthesis thin space

    Substituting the values from equations (ii)(iii), and (iv) in equation (i), we get the displacement equation, 
    (xt) = 0.05 sin (1.6 × 103t – 4.84 x) m.
    Question 1758
    CBSEENPH11020246

    A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h–1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s–1.

    Solution
    Operating frequency of the SONAR system, ν = 40 kHz 
    Speed of the enemy submarine, ve = 360 km/h
                                                          = 100 m/s 
    Speed of sound in water, = 1450 m/s 
    The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency (v') received and reflected by the submarine is given by the relation,

    Frequency space of space the space SONAR space system comma space
straight nu space equals space 40 space kHz space equals space 40 space cross times space 10 cubed space Hz

The space enrgy space submarine space first space acts space as space an space observer
which space is space approaching space the space source left parenthesis SONAR right parenthesis space with
straight a space speed space of colon

straight nu subscript straight o space space equals space 360 space km divided by hr space equals space 360 cross times 5 over 18 space straight m divided by straight s
equals space 100 space straight m divided by straight s

Speed space of space sound space in space water comma space straight nu space equals space 1450 space straight m divided by straight s

Apparent space frequency space of space sound space waves space is comma space

straight nu to the power of apostrophe space equals open parentheses space fraction numerator straight nu space plus space straight nu subscript straight o over denominator straight nu end fraction close parentheses straight nu space

space space space space equals space fraction numerator 1450 space plus space 100 space over denominator 1450 end fraction cross times space 4 space cross times space 10 to the power of 4 space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

This space frequency space is space the space frequency space of space sound space received
by space enemy space submarine. space

Now comma space the space enemy space submarine space will space reflect
the space waves space of space frequency space straight nu to the power of apostrophe space anf space thus space acts space as space straight a space
source space which space is space moving space towards space the space observer left parenthesis SONAR right parenthesis space
with space straight a space speed space given space by comma space

straight nu subscript straight s space equals space 100 space straight m divided by straight s

Therefore comma space

Apparent space frequency space is space given space by comma space

straight nu to the power of apostrophe subscript straight a space equals space open parentheses fraction numerator straight nu over denominator straight nu space minus space straight nu subscript straight s end fraction close parentheses space cross times space straight nu to the power of apostrophe space

space space space space space equals space fraction numerator 1450 over denominator 1450 space minus space 100 end fraction cross times fraction numerator 1450 space plus space 100 space over denominator 1450 end fraction space cross times space 4 space cross times space 10 to the power of 4 space

space space space space equals space 1550 over 1350 space cross times space 4 space cross times space 10 to the power of 4 space

space space space space equals space 45.92 space cross times space 10 cubed space Hz
space
space space space space equals space 45.92 space KHz
    Question 1759
    CBSEENPH11020247

    Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of Swave is about 4.0 km s–1, and that of wave is 8.0 km s–1. A seismograph records and waves from an earthquake. The first wave arrives 4 min before the first wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

    Solution

    Given here, 
    Speed of P waves, vP = 8.0 km/s
    Speed of S waves, vs = 4.0 km/s
    Let, t be the time taken by the first wave P to reach from point where earth quake occurs. 
    Time taken,
    t = 4 min = 4 x 60 s = 240 s
    And, let x be the distance upto  which the earthquake occurs is given by, 
    x = vpt
       = 8 km/s x 240 s
       = 1920 km

    Question 1760
    CBSEENPH11020248

    A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

    Solution
    Ultrasonic beep frequency emitted by the bat, ν = 40 kHz 
    Velocity of the bat, vb = 0.03 

    where,
     = velocity of sound in air 
    The apparent frequency of the sound striking the wall is given as: 
    straight nu apostrophe space equals open parentheses space fraction numerator straight nu over denominator straight nu space minus space straight nu subscript straight o end fraction close parentheses space straight nu space

space space space equals space open parentheses fraction numerator straight nu over denominator straight nu space minus space 0.03 space straight nu end fraction close parentheses space straight x space 40 space

space space space equals space open parentheses fraction numerator 40 over denominator 0.97 end fraction close parentheses space kHz 
    The frequency is reflected by the stationary wall (vs) toward the bat.
    straight nu to the power of apostrophe apostrophe end exponent space equals space open parentheses fraction numerator straight nu space plus space straight nu subscript straight o over denominator straight nu end fraction close parentheses straight nu to the power of apostrophe space

space space space space equals space open parentheses fraction numerator straight nu space plus space 0.03 space straight nu over denominator straight nu end fraction close parentheses space straight x space fraction numerator 40 over denominator 0.97 end fraction space

space space space space equals space open parentheses fraction numerator 1.03 space straight x space 40 over denominator 0.97 end fraction close parentheses

space space space space equals space 42.47 space kHz

    Question 1761
    CBSEENPH11020249

    A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

    • left parenthesis 92 space plus-or-minus 2 right parenthesis straight s
    • left parenthesis 92 space plus-or-minus 5 right parenthesis straight s
    • left parenthesis 92 space plus-or-minus space 1.8 right parenthesis straight s
    • left parenthesis 92 space plus-or-minus 3 right parenthesis straight s

    Solution

    A.

    left parenthesis 92 space plus-or-minus 2 right parenthesis straight s

    Arithmetic mean time of an oscillating simple pendulum
    fraction numerator straight capital sigma space straight x subscript straight i over denominator straight N end fraction space equals space fraction numerator 90 space plus 91 space plus 92 space plus 95 over denominator 4 end fraction space equals space 92 space straight s
    Mean deviation of a simple pendulum
    equals fraction numerator straight capital sigma space vertical line space begin display style straight x with minus on top end style space minus space straight x subscript straight i vertical line over denominator straight N end fraction space equals space fraction numerator 2 space plus space 1 space plus 3 space plus space 3 space plus space 0 over denominator 4 end fraction space equals space 1.5
    Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum = left parenthesis space 92 space plus-or-minus space 2 right parenthesis straight s

    Question 1764
    CBSEENPH11020257

    ‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

    • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction
    • fraction numerator 3 straight p subscript straight o straight V subscript straight o over denominator 2 nR end fraction
    • 9 over 2 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction
    • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator nR end fraction

    Solution

    A.

    fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction

    As, T will be maximum temperature where product of pV is maximum

    Equation of line AB, we have
    straight y minus straight y subscript 1 space equals space fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction space left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space straight p minus straight p subscript straight o space equals fraction numerator 2 straight p subscript straight o minus straight p subscript straight o over denominator straight V subscript straight o minus 2 straight V subscript straight o end fraction space left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p minus straight p subscript straight o space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction straight V squared space plus 3 straight p subscript straight o straight V
nRT space equals space fraction numerator negative straight p subscript straight o over denominator straight v subscript straight o end fraction straight V squared space plus space 3 straight p subscript straight o straight V
straight T space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o straight V squared over denominator straight v subscript straight o end fraction space plus 3 straight p subscript straight o straight V close parentheses
For space maximum space temperature
fraction numerator partial differential straight T over denominator partial differential straight V end fraction space equals space 0
fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis 2 straight V right parenthesis space equals space minus space 3 straight p subscript straight o
straight V space equals space 3 divided by 2 straight V subscript straight o
left parenthesis condition space for space maximum space temperature right parenthesis
Thus comma space the space maximum space temperature space of space the space gas space during space the space process space will space be
straight T subscript max space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction space straight x 9 over 4 straight V subscript 0 superscript 2 space plus space 3 straight p subscript straight o space straight x space 3 over 2 straight V subscript straight o close parentheses
equals space 1 over nR open parentheses negative 9 over 4 straight p subscript straight o straight V subscript straight o space plus space 9 over 2 straight p subscript straight o straight V subscript straight o close parentheses space equals space 9 over 4 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction

    Question 1769
    CBSEENPH11020267

    Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is: open parentheses straight gamma space equals space straight C subscript straight p over straight C subscript straight v close parentheses

    • fraction numerator 3 straight gamma plus 5 over denominator 6 end fraction
    • fraction numerator 3 straight gamma minus 5 over denominator 6 end fraction
    • fraction numerator straight gamma plus 1 over denominator 2 end fraction
    • fraction numerator straight gamma minus 1 over denominator 2 end fraction

    Solution

    C.

    fraction numerator straight gamma plus 1 over denominator 2 end fraction

    For an adiabatic process TVγ-1 = constant
    We know that average time of collision between molecules
    straight tau space equals space fraction numerator 1 over denominator nπ square root of 2 vrms end root straight d squared end fraction
    Where n= number of molecules per unit volume
    vrms = rms velocity of molecules
    straight n space proportional to 1 over straight V space and space space straight v subscript rms space proportional to space square root of straight T
straight tau space proportional to fraction numerator straight V over denominator square root of straight T end fraction
    Thus, we can write
    n =K1V-1 and Vrms  = K2T1/2
    Where K1 and K2 are constants.
    For adiabatic process TVγ-1 = constant. Thus we can write
    straight tau space proportional to space VT to the power of negative 1 divided by 2 end exponent space proportional to space straight V space left parenthesis straight V to the power of 1 minus straight gamma end exponent right parenthesis to the power of negative 1 divided by 2 end exponent
straight tau proportional to space straight V to the power of fraction numerator straight gamma plus 1 over denominator 2 end fraction end exponent

    Question 1770
    CBSEENPH11020268

    For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?

    Solution

    A.

    During oscillation, the motion of a simple pendulum KE is maximum of mean position where PE is minimum.

    Question 1771
    CBSEENPH11020274

    The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 )

    • 2.2 x 108 Pa

    • 2.2 x 109 Pa

    • 2.2 x 107 Pa

    • 2.2 x 106 Pa

    Solution

    A.

    2.2 x 108 Pa

    According to Hooke's law, i.e,
    Young apostrophe straight s space modulus space left parenthesis straight Y right parenthesis space equals space fraction numerator Tensile space stress over denominator Tensile space strain end fraction
So comma space straight Y space equals fraction numerator straight F divided by straight A over denominator increment straight L divided by straight L end fraction space equals space fraction numerator FL over denominator straight A increment straight L end fraction
    If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for the tensile case. 
    Give, length of a steel wire (L) = 10 cm,
    Temperature (θ) = 100oC
    As length is constant,
    ∴ Strain space equals space fraction numerator increment straight L over denominator straight L end fraction space equals space straight alpha increment straight theta
    Now, pressure = stress =y x strain
    [Given, Y = 2 x 1011N/M2 and α = 1.1 x 10-5 K-1]
    = 2 x 1011 x1.1 x 10-5 x 100
    =2.2 x 108

    Question 1776
    CBSEENPH11020279

    An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

    • 16 cm

    • 22 cm

    • 38 cm

    • 6 cm

    Solution

    A.

    16 cm

    Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.
    straight p subscript 1 space equals straight p subscript 0 space plus ρgh
    .
    For air trapped in tube, p1V1 = p2V2
    p1 = patm = pg76
    V1 = A.8 [ A = area of cross section]
    p2 = patm - ρg(54-x) =  ρg(22+x)
    V2 = A.x
    ρg76 x 8A = ρg (22+x) Ax
    x2+22x-78x 8
    by solving, x = 16

    Question 1778
    CBSEENPH11020281
    Question 1785
    CBSEENPH11020294

    Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, the density of liquid is ρ and L is its latent heat of vaporisation. 

    • ρL/T

    • square root of straight T divided by ρL end root
    • T/ρL

    • 2T/ρL

    Solution

    D.

    2T/ρL

    Decrease in surface energy = Heat required in vapourisation.... (i)
    Decrease in surface energy = T x ΔA
     = 2T x 4πrdr  [ For two surface]
    heat required in vapourisation = Latent heat
    = ML = V x ρL
    = (4πr2dr) ρL
    Now from eq (i)
    2T x 4πrdr = 4πr2ρL
    r = 2T/ρL

    Question 1788
    CBSEENPH11020305

    Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in the figure. Efficiency of this cycle is nearly:(Assume the gas to be close to ideal gas)

    • 15.4%

    • 9.1%

    • 10.5%

    • 12.5%

    Solution

    A.

    15.4%

    The efficiency of a process is defined as the ratio of work done to energy supplied.
    Here,straight eta space equals space fraction numerator increment straight W over denominator increment straight Q end fraction space equals space fraction numerator Area space under space straight p minus straight V space diagram over denominator increment straight Q subscript AB plus space increment straight Q subscript BC end fraction
    Where Cp and Cv are two heat capacities (molar)
    therefore space space fraction numerator straight p subscript straight o straight V subscript straight o over denominator nC subscript straight v increment straight T subscript 1 space plus nC subscript straight p space increment straight T subscript 2 end fraction
space equals fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 end style nR left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis plus begin display style 5 over 2 end style nR left parenthesis straight T subscript straight C minus straight T subscript straight D right parenthesis end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 left parenthesis 2 straight p subscript straight o straight V subscript straight o minus straight p subscript straight o straight V subscript straight o right parenthesis plus 5 over 4 left parenthesis 4 straight p subscript straight o straight V subscript straight o minus 2 straight p subscript straight o straight V subscript straight o right parenthesis end style end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 straight p subscript straight o straight V subscript straight o plus 5 over 4 2 straight p subscript straight o straight V subscript straight o end style end fraction
equals space fraction numerator 1 over denominator 6.5 end fraction space equals space 15.4 percent sign

    Question 1793
    CBSEENPH11020317

    100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)

    • 8.4 kJ

    • 84 kJ

    • 2.1 kJ

    • 4.2 kJ

    Solution

    A.

    8.4 kJ

    ΔQ = M,S,ΔT
    = 100 × 10-3 × 4.184 × 20 = 8.4 × 103
    ΔQ = 84 kJ, ΔW = 0
    ΔQ = ΔV + ΔW
    ΔV = 8.4 kJ

    Question 1799
    CBSEENPH11020327

    A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

    Solution

    C.

     ρoil < ρ < ρwater 
    Oil is the least dense of them, so it should settle at the top with water at the base. Now, the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So, it will stay at the oil -water interface.

    Question 1802
    CBSEENPH11020332
    Question 1803
    CBSEENPH11020337

    The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y= 0.02(m) sin open square brackets 2 straight pi open parentheses fraction numerator straight t over denominator 0.04 space left parenthesis straight s right parenthesis end fraction minus fraction numerator straight x over denominator 0.50 space left parenthesis straight m right parenthesis end fraction close parentheses close square brackets The tension in the string is 

    • 4.0 N

    • 12.5

    • 0.5 N

    • 6.25 N

    Solution

    D.

    6.25 N

    The tension in the string
    on comparing with the standard equation,
    y = a sin (ω t = kx)
    T = μv2 = μω2/k2 = 0.04
     = 0.04 (2π/0.004)2/(2π/0.50)2
     = 6.25

    Question 1805
    CBSEENPH11020339

    The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be

    • 2.5 x 1025

    • -2.5 x 1025

    • -1.61 x 1023

    • 1.38 x 1023

    Solution

    B.

    -2.5 x 1025

    Using ideal gas equation
    PV = nRT
    (N is number of moles)
    P0V0 = niR × 290 ...... (1)
    [Ti = 273 + 17 = 290 K]
    After heating
    P0V0 = NfR × 300 ...... (2)
    [Tf = 273 + 27 = 300 K]
    from equation (1) and (2)
    straight N subscript straight f minus straight N subscript straight i space equals space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript 0 over denominator straight R space straight x 290 end fraction
difference space in space number space of space moles space minus fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 290 end fraction
difference space in space number space of space moles space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets
Hence comma space straight n subscript straight f space minus straight n subscript straight i space is
equals space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets straight x space 6.023 space straight x space 10 to the power of 23

    putting P0 = 105 PA and V0 = 30 m3
    Number of molecules nf – ni = – 2.5 × 1025

    Question 1808
    CBSEENPH11020343
    Question 1811
    CBSEENPH11020348

    A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T.Time taken for 10 complete oscillations is :

    • 6.98 s 

    • 8.76 s

    • 6.65 s

    • 8.89 s

    Solution

    C.

    6.65 s

    straight T space equals space 2 straight pi space square root of straight I over MB end root
    I = 7.5 × 10–6 kg – m2
    M = 6.7 × 10–2 Am2
    By substituting value in the formula
    T = .665 sec
    for 10 oscillation, time taken will be
    Time = 10 T = 6.65 sec

    Question 1815
    CBSEENPH11020380
    Question 1816
    CBSEENPH11020383
    Question 1819
    CBSEENPH11020482

    Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity v. The retarding viscous force acting on the spherical ball is

    • directly proportional to R but inversely proportional to v.

    • directly proportional to both radius R and velocity v.

    • inversely proportional to both radius R and velocity v.

    • inversely proportional to R but directly proportional to velocity v.

    Solution

    B.

    directly proportional to both radius R and velocity v.

    Retarding force acting on a ball falling into a viscous fluid
    F = 6πηRv
    where R = radius of the ball
    v = velocity of ball and
    η = coefficient of viscosity
    ∴ F ∝ R and F ∝ v
    Or in words, retarding force is proportional to both R and v

    Question 1820
    CBSEENPH11020483

    If two soap bubbles of different radii are connected by a tube,

    • air flows from the bigger bubble to the smaller bubble till the sizes are interchanged.

    • air flows from bigger bubble to the smaller bubble till the sizes are interchanged

    • air flows from the smaller bubble to the bigger.

    • there is no flow of air.

    Solution

    C.

    air flows from the smaller bubble to the bigger.

    The excess pressure inside the soap bubble is inversely proportional to the radius of soap bubble i.e. P ∝1/r, r being the radius of the bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.

    Question 1822
    CBSEENPH11020510

    The charge following through a resistance R varies with time t as Q = at - bt2, where a and b are positive constants. The total heat produced in R is,

    • a3R/3b

    • a3R/2b

    • a3R/b

    • a3R/6b

    Solution

    D.

    a3R/6b

    Given,
    Charge, Q = at - bt2   ... (i)
    We know that,

    Current, I = dθ over dt
    So, equation (i) can be written as,
    space space space space space straight I space equals space straight d over dt left parenthesis at minus bt squared right parenthesis
rightwards double arrow space straight I space equals space straight a space minus space 2 bt   ... (ii)
    For maximum value of t, the current is given by,
    a-2bt = 0
    Therefore, straight t space equals space fraction numerator straight a over denominator 2 straight b end fraction   ...(iii)
    Total heat produced (H) can be given as,
    straight H space equals space integral subscript 0 superscript straight t straight I squared straight R space dt
space space space space equals space integral subscript 0 superscript straight a divided by 2 straight b end superscript left parenthesis straight a minus 2 bt right parenthesis squared space straight R. space dt open curly brackets because space straight t space equals space fraction numerator straight a over denominator 2 straight b end fraction close curly brackets
    space space space space equals space integral subscript 0 superscript straight a divided by 2 straight b end superscript left parenthesis straight a squared plus 4 straight b squared straight t squared minus 4 abt right parenthesis space Rdt
straight H space equals space open square brackets straight a squared straight t plus 4 straight b squared straight t cubed over 3 minus fraction numerator 4 abt squared over denominator 2 end fraction close square brackets subscript 0 superscript begin inline style bevelled fraction numerator straight a over denominator 2 straight b end fraction end style end superscript space straight R
    On solving the above equation, we get
    rightwards double arrow space space straight H space equals space fraction numerator straight a cubed straight R over denominator 6 straight b end fraction
    Question 1823
    CBSEENPH11020512

    Coefficient of linear expansion of brass and steel rods are straight alpha subscript 1 and straight alpha subscript 2. Lengths of brass and steel rods are l1 and l2 respectively. If (I2 - I1) is maintained same at all temperatures, which one of the following relations holds good?

    • straight alpha subscript 1 straight I subscript 2 squared space equals space straight alpha subscript 2 straight I subscript 1 squared
    • straight alpha subscript 1 squared straight I subscript 2 space equals space straight alpha subscript 2 squared straight I subscript 1 space
    • straight alpha subscript 1 straight I subscript 1 space equals space straight alpha subscript 2 straight I subscript 2
    • straight alpha subscript 1 straight I subscript 2 space equals space straight alpha subscript 2 straight I subscript 1

    Solution

    C.

    straight alpha subscript 1 straight I subscript 1 space equals space straight alpha subscript 2 straight I subscript 2

    Coefficient of linear expansion of brass = straight alpha subscript 1
    Coefficient of linear expansion = straight alpha subscript 2
    Length of brass and steel rods are l1 and l2 respectively.
    Given,
    Increase in length (l2'-l1' ) is same for all temperature.
    So,
    l2'-l1' = l2 - l1
    rightwards double arrow space straight l subscript 2 space left parenthesis 1 plus straight alpha subscript 2 increment straight t right parenthesis minus space straight l subscript 1 space left parenthesis 1 plus straight alpha subscript 1 increment straight t right parenthesis space equals space straight l subscript 2 minus straight l subscript 1
rightwards double arrow space straight l subscript 2 space straight alpha subscript 2 space equals space straight l subscript 1 straight alpha subscript 1

    Question 1824
    CBSEENPH11020516

    An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is,

    • 100 cm

    • 150 cm

    • 200 cm

    • 66.7 cm

    Solution

    B.

    150 cm

    b) The smallest length of the air column is associated with the fundamental mode of vibration of the air column.

    Lminstraight lambda over 4
    rightwards double arrow space 50 space cm space equals space straight lambda over 4
    rightwards double arrow space straight lambda space equals space 200 space cm
    The next higher length of the air column is,
    straight L space equals space straight lambda over 4 plus straight lambda over 2
space space space equals space fraction numerator straight lambda space plus space 2 straight lambda over denominator 4 end fraction
space space equals space fraction numerator 3 straight lambda over denominator 4 end fraction
space space equals space 3 over 4 straight x 200
space space equals space 150 space cm

    Question 1825
    CBSEENPH11020517
    Question 1826
    CBSEENPH11020519

    A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced its half. Then,

    • compressing the gas through adiabatic process will require more work to b done

    • compressing the gas through isothermally or adiabatically will require the same amount of work.

    • which of the case (whether compression through isohermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.

    • compressing the gas isothermally will require more work to be done

    Solution

    B.

    compressing the gas through isothermally or adiabatically will require the same amount of work.

    Plotting P-V graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume.

    An isothermal curve is less steeper than the adiabatic curve. So, area under the P-V curve for adiabatic process has more magnitude than the isothermal curve. Hence, work done in adiabatic process will be more than in isothermal process.
    Question 1827
    CBSEENPH11020524

    A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4 x 105 J/kg and g =10 N/kg]

    • 544 km

    • 136 km

    • 68 km

    • 34 km

    Solution

    B.

    136 km

    As per conservation of energy, energy gained by the ice during its fall from height h is given by,
    straight E space equals space mgh
    Given, only one-quarter of its energy is absorbed by the ice.
    So, 
    mgh over 4 space equals space m L subscript f
rightwards double arrow space h space equals space fraction numerator m L subscript f space x space 4 over denominator m g end fraction
space space space space space space space space space space equals space fraction numerator L subscript f x 4 over denominator g end fraction
space space space space space space space space space equals space fraction numerator 3.4 space x space 10 to the power of 5 x 4 over denominator 10 end fraction
space space space space space space space space equals space 13.6 space x space 10 to the power of 4
space space space space space space space space equals space 13600 space m space
space space space space space space space space equals space 136 space k m

    Question 1828
    CBSEENPH11020531

    The approximate depth of an ocean is 2700 m.The compressibility of water is 45.4 x 10-11 Pa-1 and density of water is 103 kg/m3. What obtained at  the bottom of the ocean?

    • 0.8 x10-2

    • 1.0 x 10-2

    • 1.2 x 10-2

    • 1.4 x 10-2

    Solution

    C.

    1.2 x 10-2

    Given d = 2700m
    straight rho space equals space 10 cubed space kg divided by straight m cubed
    Compressibility =454 x 10-11 per pascal
    The pressure at the bottom of ocean is given by
    straight p equals ρgd
    = 103 x 10 x 2700 = 27 x 106 Pa
     So,fractional compression =compressibilty x pressure
    = 45.4 x 10-11 x 27 x 106 = 1.2 x 10-2

    Question 1829
    CBSEENPH11020532

    The two ends of a metal rod are maintained at temperatures 100o C and 110o C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200o C and 210o C, the rate of heat flow will be

    • 44.0 J/s

    • 16.8 J/s

    • 8.0 J/s

    • 4.0 J/s

    Solution

    D.

    4.0 J/s

    Here, Δ T1 = 110-100 = 10o
    dQ subscript 1 over dt space equals space 4 straight J divided by straight s
increment straight T subscript 2 space equals space 210 minus 200 space equals space 10 to the power of straight o space straight C
dQ subscript 2 over dt space equals space ?
    As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. 10o C. So, the same rate of heat will flow in the second case. 
    Hence,
    dQ over dt space equals space 4 straight J divided by straight s

    Question 1830
    CBSEENPH11020533

    The ratio of the specific heats straight C subscript straight p over straight C subscript straight v space equals gamma in terms of degrees of freedom (n) is given by

    • open parentheses 1 plus 1 over straight n close parentheses
    • open parentheses 1 plus straight n over 3 close parentheses
    • open parentheses 1 plus 2 over straight n close parentheses
    • open parentheses 1 plus straight n over 2 close parentheses

    Solution

    C.

    open parentheses 1 plus 2 over straight n close parentheses

    The specific heat of gas at constant volume in terms of degree of freedom 'n' is
    straight C subscript straight v space equals straight n over 2 straight R
also space straight C subscript straight p space minus straight C subscript straight v space equals space straight R
So space straight C subscript straight p space equals straight n over 2 straight R space plus straight R space
equals space straight R open parentheses space 1 plus straight n over 2 close parentheses
Now comma space straight gamma space equals straight C subscript straight p over straight C subscript straight v space equals space fraction numerator space straight R open parentheses space 1 plus straight n over 2 close parentheses over denominator straight n over 2 straight R end fraction space equals space 2 over straight n space plus 1

    Question 1833
    CBSEENPH11020536

    The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

    • 80 cm

    • 100 cm

    • 120 cm

    • 140 cm

    Solution

    C.

    120 cm

    The fundamental frequencies of closed and open organ pipe are given as,
    straight v subscript straight o space equals fraction numerator straight v over denominator 4 space l end fraction
straight v subscript straight o space equals space fraction numerator straight v over denominator 2 space straight l apostrophe end fraction
    Given the second overtone (i.e. third harmonic) of open pipe is equal to the fundamental frequency of close pipe


    i.e 3vo =vo
    rightwards double arrow space 3 fraction numerator straight v over denominator 2 straight l apostrophe end fraction space equals space fraction numerator straight v over denominator 4 space straight l end fraction
rightwards double arrow space straight l apostrophe space equals space 6 straight l space equals space 6 space straight x space 20 space equals space 120 space cm
    Question 1835
    CBSEENPH11020546

    A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength straight lambda subscript 1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is straight lambda subscript 2. The ratio straight lambda subscript 2 over straight lambda subscript 1 is,

    • square root of fraction numerator straight m subscript 1 plus straight m subscript 2 over denominator straight m subscript 2 end fraction end root
    • square root of straight m subscript 2 over straight m subscript 1 end root
    • square root of fraction numerator straight m subscript 1 plus straight m subscript 2 over denominator straight m subscript 1 end fraction end root
    • square root of straight m subscript 1 over straight m subscript 2 end root

    Solution

    A.

    square root of fraction numerator straight m subscript 1 plus straight m subscript 2 over denominator straight m subscript 2 end fraction end root

    Wavelength of the transverse pulse is,
    straight lambda space equals space straight nu over straight f            ... (i);
    where v is the velocity of the wave and f is the frequency of the wave.
    We know that,

    straight nu space equals space square root of straight T over straight mu end root space space space space space space space space space space space space space... space left parenthesis ii right parenthesis
    T is the tension in the spring
    straight mu is the mass per unit length of the rope
    From eqns. (i) and (ii), we have

    space space space space space straight lambda space equals space 1 over straight f square root of straight T over straight mu end root
rightwards double arrow space straight lambda space proportional to space square root of straight T
    For two different cases, we have
    straight lambda subscript 2 over straight lambda subscript 1 space equals square root of T subscript 2 over T subscript 1 end root
space space space space space space equals space square root of fraction numerator m subscript 1 plus m subscript 2 over denominator m subscript 2 end fraction end root

    Question 1836
    CBSEENPH11020555

    A refrigerator works between 4o C and 30o C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 Joules)

    • 23.65 W

    • 236.5 W

    • 2365 W

    • 2.365 W

    Solution

    B.

    236.5 W

    Given,
    Temperature of the source, T = 30o C = 30 + 273  = 303 K
    Temperature of sink, T2 = 4o C = 4 + 273 = 277 K
    We know,
    space space space space space straight Q subscript 1 over straight Q subscript 2 space equals space T subscript 1 over T subscript 2

rightwards double arrow space fraction numerator Q subscript 2 space plus space W over denominator straight Q subscript 2 end fraction space equals space T subscript 1 over T subscript 2 space open curly brackets W space equals space Q subscript 1 space minus space Q subscript 2 close curly brackets
    where, Q2 is the amount of heat drawn from the sink at T2,
    W is the work done on the working substance,
    Q1 is the amount of heat rejected to source at room temperature T1.
    That is,
    rightwards double arrow space WT subscript 2 space plus space straight T subscript 2 straight Q subscript 2 space equals space straight T subscript 1 straight Q subscript 2

rightwards double arrow space WT subscript 2 space equals space straight T subscript 1 straight Q subscript 2 space minus straight T subscript 2 straight Q subscript 2

rightwards double arrow space WT subscript 2 space equals space straight Q subscript 2 space left parenthesis straight T subscript 1 space minus space straight T subscript 2 right parenthesis
rightwards double arrow space straight W space equals space straight Q subscript 2 space open parentheses straight T subscript 1 over straight T subscript 2 minus 1 close parentheses
rightwards double arrow space straight W space equals space 600 space straight x space 4.2 space straight x space open parentheses 303 over 277 minus 1 close parentheses
rightwards double arrow space straight W space equals space 600 space straight x space 4.2 space straight x space open parentheses 26 over 277 close parentheses
rightwards double arrow space straight W space equals space 236. space 5 space straight J

Power space equals space fraction numerator Work space done over denominator time end fraction
space space space space space space space space space space space space equals space straight W over straight t space equals space fraction numerator 236.5 over denominator 1 end fraction
space space space space space space space space space space space space equals 236.5 space straight W

    Question 1841
    CBSEENPH11020572

    The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos (straight omegat)
    where, X is the displacement at time t

    straight omega is the frequency of the oscillation
    Which one of the following graphs shows correctly the variation of a with t?

    Solution

    A.

    Putting values in the given expression,
    x = A cos (straight omegat)
    At t = 0, x =+ A
    At, t = straight T over 4, x = A cos open parentheses fraction numerator 2 straight pi over denominator straight T end fraction x T over 4 close parenthesesequals space straight A space cos space left parenthesis straight pi divided by 2 right parenthesis space equals 0
    Again, at t = T/2, x = A cos open parentheses fraction numerator 2 straight pi over denominator straight T end fraction x T over 2 close parentheses = A cos straight pi = -A
    As per the above results, only graph (i) will satisfy the above results.

    Question 1842
    CBSEENPH11020575

    If the radius of a star is  R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?

    • Q/4πR2σ

    • (Q/4πR2σ)-1/2

    • (4πR2Q/σ)1/4

    • (Q/4πR2σ)1/4

    Solution

    D.

    (Q/4πR2σ)1/4

    From Stefan' law
    E =σT4
    so, the rate energy production
    Q=E x A

    Q = σT4 x  4πR2
    Temperature of star
    straight T space equals open parentheses fraction numerator straight Q over denominator 4 πR squared straight sigma end fraction close parentheses to the power of 1 divided by 4 end exponent
    Question 1843
    CBSEENPH11020578

    Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is increment straight l. Which of the following graphs is a straight line?

    • increment straight l space versus space 1 over straight l
    • increment straight l space versus space straight l squared
    • increment straight l space versus space 1 over straight l squared
    • increment straight l space versus space straight l

    Solution

    B.

    increment straight l space versus space straight l squared

    According to Hooke's law,
    Young's Modulus is given by,
    straight y space equals space fraction numerator begin display style bevelled straight F over straight A end style over denominator begin display style bevelled fraction numerator increment straight l over denominator straight l end fraction end style end fraction space equals space fraction numerator straight F space xl over denominator increment straight l space straight x space straight A end fraction     ... (i)
    V = A xl = constant             ... (ii)
    From equations (i) and (ii), we have
    space space space straight y space equals space fraction numerator straight F space straight x space straight l space straight x space straight l over denominator increment straight l space straight x space straight V end fraction
increment straight l space equals space fraction numerator straight F over denominator straight V space straight x space straight y end fraction space straight l squared

space increment straight l space proportional to space straight l squared

    Question 1844
    CBSEENPH11020579

    A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

    • Energy = 4VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released.

    • Energy = 3VTopen parentheses 1 over straight r plus 1 over R close parentheses is absorbed

    • Energy = 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released

    • Energy is neither absorbed nor released.

    Solution

    C.

    Energy = 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses is released

    If the surface area changes, it will change the surface energy also.
    When the surface area decreases, it means energy is released ad vice-versa.
    Change in surface energy is increment straight A space straight x space straight T   ... (i)
    Let, there be n number of drops initially.
    So, increment straight A space equals space 4 πR squared space minus space straight n space left parenthesis 4 πr squared right parenthesis       ... (ii)
    Volume is constant.
    So, straight n space 4 over 3 πr cubed space equals space 4 over 3 πr cubed space equals space straight V         ... (iii)
    From equations (ii) and (iii), we have
    increment straight A space equals space 3 over straight R fraction numerator 4 straight pi over denominator 3 end fraction xR cubed space minus space 3 over straight r open parentheses straight n fraction numerator 4 straight pi over denominator 3 straight r cubed end fraction close parentheses
space space space space space space space space equals space space 3 over straight R xV space minus space 3 over straight r straight V
increment straight A space space equals space 3 straight V space open parentheses 1 over straight R minus 1 over straight r close parentheses
space space space space space space space space equals space left parenthesis negative right parenthesis ve space value
    Since, R > r, incrementA is negative.
    That is, the surface area is decreased.
    Hence, energy must be released.
    Energy released = increment straight A space straight x space straight T space equals space minus 3 VT open parentheses 1 over straight r minus 1 over straight R close parentheses
    In the above expression, (-)ve sign shows that amount of energy is released.

    Question 1845
    CBSEENPH11020581

    Steam at 100o C is passed into 20 g of water at 10o C. When the water acquires a temperature of 80o C, the mass of water present will be,

    • 24 g

    • 31.5 g

    • 42.5 g

    • 22.5 g

    Solution

    D.

    22.5 g

    Heat lost by the steam = heat gained by water
    Let, m' be the amount of heat converted into water.
    m' x L = ms ms space increment straight t
    m' x 540 = 20 x 1 x (80 - 10)
    m' = fraction numerator 20 space straight x space 70 over denominator 540 end fraction space equals space 2.5 space g
    Therefore, net amount of water = 20 + 2.5 = 22.5 g

    Question 1846
    CBSEENPH11020582

    Certain quantity of water cools from 70o C to 60o C in the first 5 min and to 54o  C in the next 5 min. The temperature of the surroundings is,

    • 45o C

    • 20o C

    • 42o C

    • 10o C

    Solution

    A.

    45o C

    Assume, the temperature of the surrounding be to C.
    Case 1:
    fraction numerator 70 space minus 60 over denominator 5 space min end fraction space equals space K space left parenthesis 65 to the power of o space C space minus space t to the power of o space C right parenthesis
    The average of 70 and 60 is 65.
    fraction numerator 10 over denominator 5 space min end fraction space equals space K space left parenthesis 65 to the power of o space C space minus space t to the power of o space C right parenthesis         ... (i)
    Case 2:
    fraction numerator 60 space minus space 54 over denominator 5 space min end fraction space equals space K space left parenthesis 57 space minus space t right parenthesis             ... (ii)
    Dividing equations (i) and (ii), we get
    10 over 6 equals fraction numerator left parenthesis 65 minus t right parenthesis over denominator left parenthesis 57 space minus t right parenthesis end fraction
    On solving, we get
    Temperature, t = 45o C

    Question 1847
    CBSEENPH11020584

    A monoatomic gas at a pressure p, having a volume V expands isothermally to a volume 16 V. The final pressure of the is (take straight gamma space equals space 5 over 3),

    • 64 p

    • 32 p

    • straight p over 64
    • 16 p

    Solution

    C.

    straight p over 64

    For isothermal expansion,
    pV = p' x 2 V [because space straight V apostrophe space equals space 2 straight V]
    p' = p/2
    For adiabatic expansion,
    pv to the power of straight nu space equals space constant

straight p apostrophe straight v apostrophe to the power of straight nu space equals space straight p apostrophe apostrophe straight V apostrophe apostrophe to the power of straight nu

straight p over 2 open square brackets 2 straight V close square brackets to the power of begin inline style bevelled 5 over 3 end style end exponent space equals space straight p apostrophe apostrophe space left square bracket 16 space straight V right square bracket to the power of begin inline style bevelled 5 over 3 end style end exponent

straight p apostrophe apostrophe space equals space straight p over 2 open square brackets fraction numerator 2 straight V over denominator 16 space straight V end fraction close square brackets to the power of begin inline style bevelled 5 over 3 end style end exponent space

space space space space space equals space straight p over 2 open parentheses 1 over 8 close parentheses to the power of begin inline style bevelled 5 over 3 end style end exponent space
space space space space space equals space straight p over 2 open square brackets 0.03125 close square brackets
space space space space space equals space 0.0156 space straight p
space space space space space equals space straight p over 64

    Question 1848
    CBSEENPH11020585

    A thermodynamic system undergoes cyclic process ABCDA as shown in the figure. The work done by the system in the cycle is:

    • po Vo

    • 2 po Vo

    • fraction numerator straight p subscript straight o space straight V subscript straight o over denominator 2 end fraction
    • zero

    Solution

    D.

    zero

    Work done in the cyclic process = Area bound by the closed configuration
    = Area of closed configuration
    equals space open curly brackets 2 space open square brackets 1 half left parenthesis straight V subscript straight o divided by 2 right parenthesis space straight x space straight p subscript straight o close square brackets space plus space minus 2 open square brackets 1 half left parenthesis straight V subscript straight o divided by 2 right parenthesis space straight p subscript straight o close square brackets close curly brackets

equals space zero

    Question 1849
    CBSEENPH11020586

    The mean free path of molecules of a gas, (radius r) is inversely proportional to,

    • r3

    • r2

    • r

    • square root of straight r

    Solution

    B.

    r2

    Mean free path, l = 1 over nπr squared
    That is, straight l space proportional to space 1 over straight r squared

    Question 1850
    CBSEENPH11020587

    If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by,

    • 1 over straight n space equals space 1 over n subscript 1 plus 1 over n subscript 2 plus 1 over n subscript 3
    • fraction numerator 1 over denominator square root of straight n end fraction space equals space fraction numerator 1 over denominator square root of n subscript italic 1 end root end fraction plus fraction numerator 1 over denominator square root of n subscript italic 2 end root end fraction plus fraction numerator 1 over denominator square root of n subscript 3 end root end fraction
    • square root of straight n space equals square root of n subscript italic 1 end root space plus square root of n subscript italic 2 end root space plus space space square root of n subscript 3 end root
    • n = n1 + n2 + n3

    Solution

    A.

    1 over straight n space equals space 1 over n subscript 1 plus 1 over n subscript 2 plus 1 over n subscript 3
    For first part, n1fraction numerator straight v over denominator 2 straight l subscript 1 end fraction space rightwards double arrow space l subscript 1 space equals space fraction numerator v over denominator 2 n subscript 1 end fraction
    For second part, n2fraction numerator straight v over denominator 2 straight l subscript 2 end fraction space rightwards double arrow space l subscript 2 space equals space fraction numerator v over denominator 2 n subscript 2 end fraction
    For third part, n3fraction numerator straight v over denominator 2 straight l subscript 3 end fraction space rightwards double arrow space l subscript 3 space equals space fraction numerator v over denominator 2 n subscript 3 end fraction
    For the complete wire,

    straight n space equals space fraction numerator straight v over denominator 2 straight l end fraction space rightwards double arrow space straight l space equals space fraction numerator straight v over denominator 2 straight n end fraction
    We have, l = l1 + l2 + l3
    That is,

    fraction numerator straight v over denominator 2 straight n end fraction space equals space fraction numerator v over denominator 2 n subscript 1 end fraction space plus space fraction numerator v over denominator 2 n subscript 2 end fraction plus fraction numerator v over denominator 2 n subscript 3 end fraction

space 1 over n space equals space 1 over n subscript 1 space plus space 1 over n subscript 2 plus 1 over n subscript 3
    Question 1853
    CBSEENPH11020596

    Liquid oxygen at 50  K is heated to 3000 K at a constant pressure of 1 atm. The rate of heating is constant. which one of the following graphs represents the variation of temperature with time?

    Solution

    A.

    Graph (a) shows the variation temperature with time. At first temperature will increase then there will be state change from liquid to gas.

    Question 1854
    CBSEENPH11020598

    One mole of an ideal gas goes from an initial state A to final state B via two processes. It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct p-V diagram representing the two processes is

    Solution

    D.

    According to the question, first gas goes from volume V to 3V and after this volume is reduced from 3V to V at constant pressure. In the isothermal expansion, the p-V curve is a rectangular hyperbola.

    Question 1856
    CBSEENPH11020600

    A thermodynamic system is taken through the cycle ABCD as shown in a figure. Heat rejected by the gas during the cycle is

    • 2 pV

    • 4 pV

    • 1/2 pV

    • pV

    Solution

    A.

    2 pV

    For given cyclic process, ΔU = 0
    therefore, Q =W
    Also, W = - area enclosed by the curve
     = AB x AD
     = (2p-p) (3V-V)
    =-p x 2V
    therefore,
    Heat rejected  = 2pV

    Question 1858
    CBSEENPH11020612

    The wettability of a surface by a liquid depends primarily on

    • viscosity

    • surface tension

    • density

    • angle of contact between the surface and the liquid

    Solution

    D.

    angle of contact between the surface and the liquid

    The value of angle of contact determines whether a liquid will spread on the surface or not.

    Question 1861
    CBSEENPH11020615

    A gas is taken through the cycle A → B → C → A, as shown, What is the net work done by the gas?


    • 2000 J

    • 1000 J

    • Zero

    • -2000 J 

    Solution

    B.

    1000 J

    Net work done = Area enclosed in pV curve i.e. Δ ABC
     = 1 half space straight x space 5 space straight x space 10 to the power of negative 3 end exponent space straight x space 4 space straight x space 10 to the power of 5 space straight J = 1000 J

    Question 1863
    CBSEENPH11020617

    In the given (V-T )diagram, what is the relation between pressures p1 and p2?

    • p2 =p1

    • p2>p1

    • p2<p1

    • Cannot be predicted

    Solution

    C.

    p2<p1

    The slope of the graph directly proportional to 1/p

    Question 1873
    CBSEENPH11020633

    During an isothermal expansion, a confined ideal gas does - 150 J of work against its surroundings. This implies that

    • 300 J of heat has been added to the gas

    • no heat is transferred because the process is isothermal

    • 150 J of heat has been added to the gas

    • 150 J of heat has been removed from the gas

    Solution

    C.

    150 J of heat has been added to the gas

    From the first law of thermodynamics
    ΔU = Q + W
    For isothermal process, ΔU = 0
    therefore, Q = - W
    Given W = - 150 
    Therefore, Q = + 150
    When Q is positive, the heat is added to the gas.

    Question 1874
    CBSEENPH11020643

    If dimensions of critical velocity vc of a liquid flowing through  a tube are expressed as left square bracket straight eta to the power of straight x space straight rho to the power of straight y space straight r to the power of straight z right square bracket comma space where space straight eta comma space straight rho space and space straight r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by

    • 1,-1,-1

    • -1,-1,1

    • -1,-1,-1

    • 1,1,1

    Solution

    A.

    1,-1,-1

    According to the principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct if the dimension of all the terms occurring on both sides of the equations is same.
    Given critical velocity of liquid flowing through a tube are expressed as
    straight v subscript straight c space equals space straight eta to the power of straight n straight rho to the power of straight n straight r to the power of straight z
Coefficient space of space viscosity space of space liquid comma space straight eta space equals space left square bracket ML to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket
Density space of space liquid comma space straight rho space equals space left square bracket ML to the power of negative 3 end exponent right square bracket
Radius space of space straight a space tube comma space straight r space equals space left square bracket straight L right square bracket
critical space velcity space of space liquid space straight v subscript straight c space equals space left square bracket straight M to the power of straight o LT to the power of negative 1 end exponent right square bracket

equals space left square bracket straight M to the power of straight o straight L to the power of 1 straight T to the power of negative 1 end exponent right square bracket space equals space left square bracket straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket to the power of straight x. left square bracket ML to the power of negative 3 end exponent right square bracket to the power of straight y. left square bracket straight L right square bracket to the power of straight z

left square bracket straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent right square bracket space equals space left square bracket straight M to the power of straight x plus straight y end exponent straight L to the power of negative straight x minus 3 straight y plus straight z end exponent straight T to the power of negative straight x end exponent right square bracket

Comparing space exponents space of space straight M comma space straight L space space and space straight L comma space we space space get comma
    x+ y = 0, - x-3y+z = 1, -x = -1
    z = -11 x = 1, y = -1

    Question 1876
    CBSEENPH11020646

    A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is

    • 155 Hz

    • 205 Hz

    • 10.5 Hz

    • 105 Hz

    Solution

    D.

    105 Hz

    Given l = 75 cm, f1 = 420 Hz and f2 = 315 Hz
    As two consecutive resonant frequencies for a string fixed at both ends will be +
    straight f subscript 1 space equals space fraction numerator nv over denominator 2 l end fraction space and space straight f subscript 2 space equals fraction numerator left parenthesis straight n plus 1 right parenthesis over denominator 2 straight R end fraction

rightwards double arrow space straight f subscript 2 minus straight f subscript 1 equals space 420 space minus space 315 space

rightwards double arrow space fraction numerator left parenthesis straight n plus 1 right parenthesis straight v over denominator 2 l end fraction space minus fraction numerator nv over denominator 2 space l end fraction space equals space 105 space Hz

rightwards double arrow space fraction numerator straight v over denominator 2 space l end fraction space equals space 105 space Hz
    Thus, lowest resonant frequency of a string is 105 Hz.

    Question 1877
    CBSEENPH11020647

    The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20o C, th temperature of the surroundings to which it rejects heat is

    • 31o C

    • 41o C

    • 11o C

    • 21o C

    Solution

    A.

    31o C

    COfficient of performance (β) of a refrigerator is defined as the ratio of the quantity of heat removed per cycle to the work done on the substance per cycle to remove this heat.
    Given, coefficient of performance of a refrigerator, β = 5
    Temperature of surface i.e., inside freezer,
    T2 = - 20o C = - 20 +273 = 253 K
    The temperature of surrounding i.e heat rejected outside T1 =?
    So, straight beta space equals space fraction numerator straight T subscript 2 over denominator straight T subscript 1 minus straight T subscript 2 end fraction

rightwards double arrow space 5 space equals space fraction numerator 253 over denominator straight T subscript 1 space minus space 253 end fraction

rightwards double arrow space 5 space straight T subscript 1 space equals space 1265 space equals space 253
rightwards double arrow space 5 space straight T subscript 1 space equals space 1518

straight T subscript 1 space equals space 1518 over 5 space equals space 303.6 space straight K
straight T subscript 1 space equals space 303.6 space minus 273 space equals space 31 to the power of straight o space straight C

    Question 1878
    CBSEENPH11020648

    The value of the coefficient of volume expansion of glycerin is 5 x 10-4 K-1. The fractional change in the density of glycerin for a rise of 400C in its temperature is 

    • 0.015

    • 0.020

    • 0.025

    • 0.010

    Solution

    B.

    0.020

    Given the value of coefficient of volume expansion of glycerin  is  5 x 10-4 K-1
    As, original density of glycerine 
    bold rho bold space bold equals bold space bold rho subscript bold o bold space left parenthesis 1 plus space straight Y increment straight T right parenthesis
bold rightwards double arrow bold space bold rho bold space bold minus bold space bold rho subscript bold o space equals space straight rho subscript straight o straight Y increment straight T
space
    Thus, fractional change in the density of glycerine for a rise of 40o C in its temperature,

    fraction numerator bold rho bold space bold minus bold rho subscript bold o over denominator bold p subscript bold o end fraction bold space equals space Y increment T space equals space 5 space x space 10 to the power of negative 4 end exponent space x space 40 space equals space 200 space x space 10 to the power of negative 4 end exponent space equals space 0.020

    Question 1879
    CBSEENPH11020650

    Water rises to a height 'h' in a capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then

    • water rises upto the tip of the capillary tube and then starts overflowing like a fountain

    • water rises  upto the top of capillary tube and stays there without overflowing 

    • water rises upto a point little below the top and stays there

    • Water does not rise at all

    Solution

    A.

    water rises upto the tip of the capillary tube and then starts overflowing like a fountain

    It is given that water rises to a height 'h' in a capillary tube. so, the length of the capillary tube above the surface water is made less than 'h' then the height of water > length of capillary tube ⇒ So, the liquid will be staying there.

    Question 1883
    CBSEENPH11020659

    An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?

    • Adiabatic

    • Isobaric

    • Isochoric 

    • Isothermal 

    Solution

    A.

    Adiabatic

    Given, ideal gas is compressed to half its initial volume i.e.,
    Vo = V/2

    The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero  ie.e Wisochoric  = 0
    As we know, work done on the gas  = Area under curve  i.e,
    Wadiabatic > Wisothrmal > Wisochoric

    Question 1884
    CBSEENPH11020660

    A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60o with the source observer is at rest. the apparent frequency observed by the observer (velocity of sound in air 330 ms-1) is 

    • 100 Hz

    • 103 Hz

    • 106 Hz

    • 97 Hz

    Solution

    B.

    103 Hz

    Given, as a source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance. Such that, source is moving with a speed of 19.4 m/s  at angle 60o with source- observer line as shown in figure

    The apparent frequency heard by observer
    straight f subscript straight o space equals space straight f subscript straight s open square brackets fraction numerator straight V over denominator straight v minus straight v subscript straight s space cos space 60 to the power of straight o end fraction close square brackets space equals space 100 space open square brackets fraction numerator 330 over denominator 330 minus 19.4 space straight x begin display style 1 half end style end fraction close square brackets

equals space 100 space open square brackets fraction numerator 330 over denominator 300 minus 9.7 end fraction close square brackets space equals space 100 open square brackets fraction numerator 330 over denominator 320.3 end fraction close square brackets

equals space 103.02 space Hz

    Question 1885
    CBSEENPH11020661

    If space vectors space straight A space equals space space cos space ωt space straight i with hat on top space plus space sin space straight omega straight j with hat on top space and space straight B space equals space cos space ωt over 2 straight i with hat on top space plus space sin ωt over 2 straight j with hat on top are functions of time, then the value of t at which they are orthogonal to each other
    • straight t space equals space fraction numerator straight pi over denominator 4 straight omega end fraction
    • straight t space equals space fraction numerator straight pi over denominator 2 straight omega end fraction
    • t equals space straight pi over straight omega
    • t = 0 

    Solution

    C.

    t equals space straight pi over straight omega

    For a perpendicular vector, we have A. B  = 0 
    left square bracket cos space ωt space straight i with hat on top space plus space sin space ωt space straight j with hat on top right square bracket space open square brackets cos space ωt over 2 straight i with hat on top space plus space fraction numerator sin space ωt over denominator 2 end fraction space straight j with hat on top close square brackets space equals space 0
rightwards double arrow space left square bracket therefore space cos space left parenthesis straight A minus straight B right parenthesis space equals space cos space straight A space Cos space straight B space plus space sin space straight A space Sin space straight B right square bracket

cos space open parentheses ωt space minus space ωt over 2 close parentheses space equals space 0

cos space ωt over 2 space equals space 0 space rightwards double arrow space ωt over 2 space equals space straight pi over 2 space rightwards double arrow space straight t space equals space straight pi over straight omega
Thus comma space time space taken space by space vectors space which space are space orthogonal space to space each space other space is space straight pi over straight omega

    Question 1888
    CBSEENPH11020681

    The displacement of a particle along the x-axis is given by x = a sin2 ωt. The motion of the particle corresponds to 

    • simple of harmonic motion of frequency ω/π

    • simple harmonic motion of frequency 3ω/2π

    • non simple harmonic motion

    • simple harmonic motion of frequency  ω/2π

    Solution

    D.

    simple harmonic motion of frequency  ω/2π

    For space straight a space particle space executing space SHM
acceleration space left parenthesis straight a right parenthesis space proportional to space minus straight omega squared space displacement space left parenthesis straight x right parenthesis space.... space left parenthesis straight i right parenthesis space
Given space straight x space equals space straight a space sin squared space ωt space space space..... space left parenthesis ii right parenthesis

Differentiating space the space above space equation space straight w. straight r. straight t space we space get
dx over dt space equals 2 aω left parenthesis space sin space ωt right parenthesis left parenthesis cost space ωt right parenthesis

Again space differentiating comma space we space get
fraction numerator straight d squared straight x over denominator dt squared end fraction space equals space straight a space equals space 2 aω squared space left square bracket space cos squared space ωt minus space sin space ωt right square bracket

equals space 2 aω squared space cos space 2 ωt
    The given equation does not satisfy the condition for SHM (Eq. (i) ]. Therefore, motion is not simple harmonic.
    Question 1894
    CBSEENPH11020700

    An alpha nucleus of energy mv2/2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to 

    • 1/Ze

    • v2

    • 1/m

    • 1/v4

    Solution

    C.

    1/m

    An α -  a particle of mass m possesses initial velocity when it is at large distance from the nucleus of an atom having atomic number Z. At the distance of closest approach, the kinetic energy of α - particle is completely converted into potential energy. Mathematically,

    12mv2 = 14πε0(2e)(Ze)ror0 = 14πε02Ze212mv2 =1/m

    Question 1896
    CBSEENPH11020702

    The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by

    • σr2T4/R2

    • σr2T4/4πR2

    • σr4T4/r4

    Solution

    A.

    σr2T4/R2

    If r is the radius of the star and T its temperature, then the energy emitted by the star per second through radiation in accordance with Steafan's law will be given by 
    AσT4 = 4πr2σ T4

    In reaching a distance R this energy will spread over a sphere of radius R; so the intensity of radiation will be given by 

    straight S space equals space fraction numerator straight P over denominator 4 πR squared end fraction space equals space fraction numerator 4 straight pi squared σT to the power of 4 over denominator 4 πR squared end fraction

    Question 1899
    CBSEENPH11020711
    Question 1901
    CBSEENPH11020719

    A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x -axis with a speed of 128 ms-1 and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is

    • y = (0.02) m sin (7.85 x +1005t)

    • y = (0.02) m sin (15.7 x -2010t)

    • y = (0.02) m sin (15.7 x + 2010t)

    • y = (0.02) m sin (7.85 x - 1005t)

    Solution

    D.

    y = (0.02) m sin (7.85 x - 1005t)

    Find the parameters and put in the general wave equation.
     here, A = 2 cm
    direction = + ve x direction
     v = 128 ms-1
    and 5 λ = 4


    k = 2π /  λ = 2π x 5/4 = 7.85
    and v = ω / k = 128 ms-1

    ω = v s k = 128 7.85 = 1005
    as  y = A sin (kx - ωt)
    y = 2 sin (7.85 x - 1005 t)
    = 0.02 m sin (7.85 x -1005 t)
    Question 1904
    CBSEENPH11020727

    The internal energy change in a system that has absorbed 2 Kcal of heat and done 500 J of work is 

    • 8900 J

    • 6400 J

    • 5400 J

    • 7900 J

    Solution

    D.

    7900 J

    Heat given to a system (ΔQ) is equal to the sum of the increase in the internal energy (Δu) and the work done (ΔW) by the system against the surrounding and 1 cal = 4.2. J
    According to first law of thermodynamics
    ΔU = Q-W
    = 2 x 4.2 x 1000-500
     = 8400-500
    =7900 J

    Question 1905
    CBSEENPH11020728
    Question 1907
    CBSEENPH11020730

    In a thermodynamic process which of the following statements is not true?

    • In an adiabatic process, the system is insulated from the surroundings

    • In an isochoric process pressure remains constant

    • In an isothermal process, the temperature remains constant

    • In an adiabatic process pVγ = constant

    Solution

    B.

    In an isochoric process pressure remains constant

    For an adiabatic process, there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes, gases obey Poisson's law, ie pVγ = constant. In an isochoric process volume remains constant and for isobaric process pressure remains constant.

    Question 1909
    CBSEENPH11020733

    Sodium has body centred packing.Distance between two nearest atoms is 3.7 A. The lattice parameter is

    • 6.8 A

    • 4.3 A

    • 3.0 A

    • 8.6 A

    Solution

    B.

    4.3 A

    The neighbour distance of a body centred cubic cell,straight d space equals space fraction numerator square root of 3 straight a end root over denominator 2 end fraction where a is the lattice parameter.

    3.7 space equals space square root of fraction numerator 3 straight a over denominator 2 end fraction end root
straight a space equals space fraction numerator 2 space straight x space 3.7 over denominator square root of 3 end fraction space equals space 4.3 space straight A

    Question 1911
    CBSEENPH11020738

    The driver of a car travelling with speed 30 ms-1 towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 ms-1, the frequency of reflected sound as heard by driver is

    • 550 Hz

    • 555.5 Hz

    • 720 Hz

    • 500 Hz

    Solution

    C.

    720 Hz

    Apply Doppler's effect.
    Whenever there is a relative motion between a source of the sound and the observer (listener), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by the source.
    For space case space space straight I space space
straight n apostrophe space equals space fraction numerator straight v over denominator straight v minus 30 end fraction straight n space space space... space left parenthesis straight i right parenthesis

For space case space II

straight n apostrophe apostrophe space equals space fraction numerator straight v plus 30 over denominator straight v end fraction straight n apostrophe space space space..... space left parenthesis ii right parenthesis

From space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
straight n apostrophe apostrophe space equals space fraction numerator straight v plus 30 over denominator straight v minus 30 end fraction straight n space equals space 360 over 300 space straight x space 600 space equals space 720 space Hz

    Question 1912
    CBSEENPH11020745

    On a new scale of temperature (which is linear), an called the W scale, the freezing and boiling point of water are 39o W and 239o W respectively. what will be the temperature of 39o C  celsius scale?

    • 78o W

    • 117o C

    • 200o C

    • 139o

    Solution

    B.

    117o C

    In general, whenever we are to go from any known scale to any unknown scale, then we follow the equation
    (temperature on known scale) - (LFP for known scale)
    ______________________________________________
    (UFP - LFP ) Known (temerature on unknown scale)
    = - (LEP for unknown scale)
    ________________________________________
    (UFP-LFP) unknown
    or fraction numerator 390 minus 0 over denominator 100 minus 0 end fraction space equals space fraction numerator t minus 39 over denominator 239 minus 39 end fraction 
    t= 117o W
    LFP = Lower fixed point
    UFP  = Upper fixed point

    Question 1913
    CBSEENPH11020746

    If Q, E andW denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then

    • W =0

    • Q = W = 0

    • E=0

    • Q =0

    Solution

    C.

    E=0

    In a cyclic process, the system is not isolated from the surrounding.
    In a cyclic process, a system starts from one point and ends at the same point. In this case, the change in the internal energy must again be zero and therefore the thermal energy added to the system must equal the work done during the cycle. That is, in a cyclic process.

    ΔU = 0 or E = 0

    Question 1915
    CBSEENPH11020749

    An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 200 C? The temperature boiling water is 100o C.

    • 6.3 min

    • 8.4 min

    • 12.6 min

    • 4.2 min

    Solution

    A.

    6.3 min

    Heat is evolved due to joule's effect is used up in boiling water.
    such as
    Vlt = msΔt
    t = msΔt/VI
    Putting under given values
    I= 4A, V= 220 volt, m =1 kg,
    Δt = (100-20)oC, s = 4200 J/kgo C
    t = 1 x 4200 x 80 / 220 x 4 = 6.3 min

    Question 1916
    CBSEENPH11020755

    A point performs simple harmonic oscillation of period T and the equation of motion is given x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity? 

    • T/8

    • T/6

    • T/3

    • T/12

    Solution

    D.

    T/12

    Velocity is the time derivative of displacement.
    Performing SHM

    straight x space equals space straight a space sin space open parentheses ωt plus straight pi over 6 close parentheses space space space.... space left parenthesis straight i right parenthesis
Differentiating space eq space left parenthesis straight i right parenthesis space straight w. straight r. straight t space time space comma space we space obtain

straight v space equals space dx over dt space equals space straight a space straight omega space cos space open parentheses ωt plus straight pi over 6 close parentheses
It space is space given space that space straight v space equals space aω over 2 comma space so space that
aω over 2 space equals space aω space cos space open parentheses ωt plus straight pi over 6 close parentheses
or
1 half space equals space cos space open parentheses ωt plus straight pi over 6 close parentheses
Or space
ωt plus straight pi over 6 space equals space straight pi over 3
rightwards double arrow space ωt space equals space straight pi over 6
or space
straight t space equals space fraction numerator straight pi over denominator 6 straight omega end fraction space equals space fraction numerator straight pi space straight x space straight T over denominator 6 space straight x space 2 straight pi end fraction space equals space straight T over 12
    Thus, at a T/12 velocity of the point will be equal to half of its maximum velocity.

    Question 1917
    CBSEENPH11020760

    The wave described by y = 0.25 sin (10 π x -2 πt), where x and y are in metre and t in second, is a wave travelling along the 

    • - ve x direction with frequency 1 Hz

    • +ve x direction with frequency π  Hz and wavelength λ = 0.2 m

    • +ve x direction with frequency 1 Hz and wavelength λ = 0.2 m

    • -ve x direction with amplitude 0.25 m and wavelength λ = 0.2 m

    Solution

    C.

    +ve x direction with frequency 1 Hz and wavelength λ = 0.2 m

    The sign between two terms in the argument of sine will define its direction. 
    Writing the given wave equation
    y= 0.25 sin (10 πx- 2πt) ... (i)
    The minus (-) between (10πx) and (2πt) implies that the waves is travelling along the positive x-direction.
    Now comparing eq (i) with standard wave equation
    y = a sin (kx -ωt) ..... (ii)
    we have 
    a= 0.25m, ω = 2π, k = 10 π m
    therefore,
    2π/T = 2π
    f= 1Hz
    λ = 2π/k = 2π/10π = 0.2 m
    Therefore, the wave is travelling along +ve x direction wth frequency 1 Hz and wavelength 0.2m

    Question 1919
    CBSEENPH11020763

    If the cold junction of a thermocouple is kept at 0o C and the hot junction is kept at To C, then the relation between neutral temperature (Tn) and temperature of inversion (Ti) is:

    • straight T subscript straight n space equals space straight T subscript straight i over 2
    • Tn = 2Ti

    • Tn = Ti -T

    • Tn = Ti + T

    Solution

    A.

    straight T subscript straight n space equals space straight T subscript straight i over 2

    It is found that temperature of inversion (Ti) is as much above the neutral temperature (Tn) as neutral temperature is above the temperature of the cold junction (T), ie, 
    Ti -Tn = Tn-T
    Ti = 2Tn-T
    But, here the cold junction is kept at 0o C, hence
     
    T = 0
    Thus, Ti = 2Tn
    Or Tn = Ti/2

    Question 1920
    CBSEENPH11020764

    Nickel shows the ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show:

    • paramagnetism

    • anti -ferromagnetism

    • no magnetic property

    • diamagnetism

    Solution

    A.

    paramagnetism

    Nickel exhibits ferromagnetism because of a quantum physical effect called exchange couplingin which the electron spins of one atom interact with those of neighbouring atoms. The result is aligment of the magentic dipole moments of the atoms, inspite of the randomizing tendency of atomic collisions. The presistent aligment is what gives ferromagnetic material their permanent magnetism.
    If the temerature of ferromagnetic material is raised above a certain critical value, called the Curie temperature, the exchange coupling ceases to be effective. Most such material then become simply tend to align with an external field but much more weakly,  and thermal agitation can now more easily distrupt the alignmet.

    Question 1921
    CBSEENPH11020771

    A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

    • T/4

    • T/8

    • T/12

    • T/2

    Solution

    C.

    T/12

    Let displacement equation of particle executing SHM is
    y = a sin ωt
    As particle travels half of the amplitude from the equilibrium position, so
    y= a/2
    therefore,
    straight a over 2 space equals space straight a space sin space ωt
or space sin space ωt space equals space 1 half space equals space sin space straight pi over 6
or space ωt space equals space straight pi over 6
space straight t equals space fraction numerator straight pi over denominator 6 straight omega end fraction
space straight t space equals fraction numerator straight pi over denominator 6 begin display style fraction numerator 2 straight pi over denominator straight T end fraction end style end fraction space space space space space space space space space space space space therefore open parentheses straight omega space equals fraction numerator 2 straight pi over denominator straight T end fraction space space close parentheses
straight t space equals space straight T over 12
    Hence, the particle travels half of the amplitude from the equilibrium in T/12 sec.

    Question 1922
    CBSEENPH11020774

    The particle executing simple harmonic motion has a kinetic energy Ko cos2 ωt. The maximum values of the potential energy and the toatal energy are respectively:

    • 0 and 2Ko

    • Ko/2 and K0

    • Ko and 2Ko

    • Ko and Ko

    Solution

    D.

    Ko and Ko

    In simple harmonic motion, the total energy of the particle is constant at all instants which are totally kinetic when the particle is passing through the mean position and is totally potential when the particle is passing through the extreme position.

    The variation of PE and KE with time is shown in the figure, by the dotted parabolic curve and solid parabolic curve respectively.
    Figure indicated that maximum values of total energy KE and PE of SHM are equal.
    Now, EK = Ko cos2 ωt
    therefore, (EK)max = Ko
    So, (EP)max = Ko
    and (E)Total  = Ko

    Question 1923
    CBSEENPH11020775

    A mass of 2.0 Kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan?

    Take g = `10 m/s2

    • 8.0 cm

    • 10.0 cm

    • any value less than 12.0 cm

    • 4.0

    Solution

    B.

    10.0 cm

    Let he minimum amplitue of SHM is a.
    Restoring force on spring
    F = ka
    Restoring  force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a.
    ka = mg
    a = mg/k
    Here,  m =2 kg, k = 200 N/m, g = 10 m/s2
    therefore, a = 2 x 10 / 200 = 10/100 m.
    10 cm

    Question 1924
    CBSEENPH11020777

    A black body is at 727o C. It emits energy at a rate which is proportional to:

    • (727)2

    • (1000)4

    • (1000)2

    • (727)2

    Solution

    B.

    (1000)4

    Amount of heat energy radiated per secnd by unit area of black body is directly proportional to fourth power of absoulte temperature.
    Accordign to steafan's law,
    straight E space proportional to straight T to the power of 4
space straight E thin space equals σT to the power of 4
    Where σ is constatn of proportionality and called Steafan's constant. Its value is
    5.67 x 10-8 Wm-2 K-4.
    hence, E = (727+273)4
    E = (1000)4

    Question 1926
    CBSEENPH11020779

    For a cubic structure which one of the following relations indicating the cell characteristic is correct?

    • straight a not equal to straight b not equal to straight c space and space space straight alpha space not equal to straight beta space and space straight gamma space space 90 to the power of straight o
    • straight a not equal to straight b not equal to straight c space and space space straight alpha space equals space straight beta space equals straight gamma space equals space 90 to the power of straight o
    • straight a equals straight b equals straight c space and space space straight alpha space not equal to straight beta space not equal to straight gamma space not equal to space 90 to the power of straight o
    • straight a equals straight b equals straight c space and space space straight alpha space equals straight beta space equals straight gamma space equals space 90 to the power of straight o

    Solution

    D.

    straight a equals straight b equals straight c space and space space straight alpha space equals straight beta space equals straight gamma space equals space 90 to the power of straight o

    In cubic crystal, the crystal axes are perpendicular to one another (α= β =γ=90) and the repetitive interval is the same along the three axes (a= b=c).

    Question 1927
    CBSEENPH11020780

    The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

    • 0.5π

    • π

    • 0.707 π

    • zero

    Solution

    A.

    0.5π

    The displacement equation of particle executing SHM is
    x = a cos ( ωt +Φ)  .. (i)
    velocity, v = dx/dt = - aω sin (ωt + Φ)  .. (ii)
    and acceleration,
    straight A space equals space dv over dt space equals space minus space aω squared space cos space left parenthesis ωt space plus straight ϕ right parenthesis space... space left parenthesis iii right parenthesis

    Fig (i) is a plot of eq (i) with Φ = 0. Fig (ii) shows Eq. (ii) also with Φ = 0 fig (iii) is a plot of Eq (iii) . it should be noted that in the figures the curve of v is shifted (to the left) from the curve of x by one - equate period (T/4).Similarly, the acceleration curve of A is shifted (to the left) by T/4 relative to the velocity curve of v. This implies that velocity is 90o C (0.5 π) out phase with the displacement and the acceleration is 90o C (0.5 π) out of phase with the velocity but 180o  π out phase with displacement.

    Question 1928
    CBSEENPH11020781

    A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically in the spring. so that the spring is compressed by a distance d. The net work done in the process is:

    • mg space left parenthesis straight h plus straight d right parenthesis space space plus space 1 half space kd squared
    • mg space left parenthesis straight h plus straight d right parenthesis space space minus space 1 half space kd squared
    • mg space left parenthesis straight h minus straight d right parenthesis space space minus space 1 half space kd squared
    • mg space left parenthesis straight h minus straight d right parenthesis space space plus space 1 half space kd squared

    Solution

    B.

    mg space left parenthesis straight h plus straight d right parenthesis space space minus space 1 half space kd squared

    Work done is equal to change in energy of the body.
    The situation is shown in the figure. When mass m falls vertically on spring, then spring si compressed by distance d.

    Hence, network done in the process is 
    W= Potential energy stored in the spring + Loss of potential energy of mass 

    mg space left parenthesis straight h plus straight d right parenthesis space space minus space 1 half space kd squared

    Question 1929
    CBSEENPH11020785

    In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per min is liberated? ECE of chlorine is 0.367 x 10-6 kg/C

    • 17.6 mg

    • 21.3 mg

    • 24.3 mg

    • 13.6 mg

    Solution

    A.

    17.6 mg

    The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity i.e., charge passed through the electrolyte.
    Power consumed in electrolysis, P = 100 W
    Voltage applied,  V = 125 V
    So, current in the solution,
                   straight i space equals space straight P over straight V equals space 100 over 125 space equals space 0.8 space straight A
    According to first law of Faraday, mass liberated at an electrode is directly proportional to charge passed through the electrolyte i.e.,
      straight m proportional to straight Q space rightwards double arrow space space straight m space equals space zQ
    where z is a constant called 'electro chemical equivalent' (ECE). 
    Also,           Q = it
    therefore space space space space straight m space equals space zit
    Given,
    straight z equals 0.367 space cross times space 10 to the power of negative 6 end exponent space kg divided by straight C comma space space straight I space equals space 0.8 space straight A comma space space straight t space equals space 60 straight s
space space space
    Hence comma space space straight m space equals space 0.367 space cross times space 10 to the power of negative 6 end exponent space cross times space 0.8 space cross times space 60
space space space space space space space space space space space space space space space space equals space 17.6 space cross times space 10 to the power of negative 6 end exponent kg
space space space space space space space space space space space space space space space space equals space 17.6 space mg

    Question 1930
    CBSEENPH11020786

    A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then:

    • straight T proportional to square root of straight rho
    • straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction
    • straight T proportional to 1 over straight rho
    • straight T proportional to fraction numerator 1 over denominator square root of straight m end fraction

    Solution

    B.

    straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction

    Force applied on the body will be equal to upthrust for vertical oscillations.
    Let block is displaced through x m, then weight of displaced water or upthrust (upwards)
        equals negative Axρg
    where A is area of cross-section of the block and straight rho is its density. This must be equal to force (=ma) applied, where m is mass of the block and straight alpha is acceleration.
    therefore space space space space ma space equals space minus Axρg
     or                straight alpha space equals space minus Aρg over straight m straight x space equals space minus straight omega squared straight x
    This is the equation of simple harmonic motion.
    Time period of oscillation
                  straight T equals fraction numerator 2 straight pi over denominator straight omega end fraction equals 2 straight pi square root of straight m over Aρg end root
rightwards double arrow space space space space space straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction

    Question 1931
    CBSEENPH11020787

    A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency?

    • 275 K

    • 175 K

    • 250 K

    • 225 K

    Solution

    C.

    250 K

    The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e., 
                              straight eta space equals space fraction numerator Work space done over denominator Heat space supplied end fraction space equals straight W over straight Q subscript 1 space equals space fraction numerator straight Q subscript 1 minus straight Q subscript 2 over denominator straight Q subscript 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 minus straight Q subscript 2 over straight Q subscript 1 space equals space 1 minus straight T subscript 2 over straight T subscript 1
    Here, straight T subscript 1 is the temperature of source and T2 is the temperature of sink
        As given,             straight eta space equals space 40 percent sign space equals space 40 over 100 space equals 0.4
    and                straight T subscript 2 space equals space 300 space straight K
    So,        
           0.4 space equals space 1 minus 300 over straight T subscript 1
rightwards double arrow space space space space straight T subscript 1 space equals space fraction numerator 300 over denominator 1 minus 0.4 end fraction equals fraction numerator 300 over denominator 0.6 end fraction space equals 500 space straight K
    Let temperature of the source be increased by x K, then efficiency becomes
    straight eta apostrophe space equals space 40 percent sign space plus space 50 percent sign space of space straight eta
space space space space space equals space 40 over 100 plus 50 over 100 cross times 0.4
space space space space space space equals 0.4 space plus space 0.5 space cross times space 0.4
space space space space space space space equals space 0.6
Hence comma space space space space space 0.6 space equals space 1 minus fraction numerator 300 over denominator 500 plus straight x end fraction
rightwards double arrow space space space space space space fraction numerator 300 over denominator 500 plus straight x end fraction space equals space 0.4
rightwards double arrow space space space space space space 500 plus straight x space equals space fraction numerator 300 over denominator 0.4 end fraction space equals space 750
    therefore space space space space straight x space equals space 750 minus 500 space equals space 250 space straight K

    Question 1932
    CBSEENPH11020788

    A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at

    • 4000 Å

    • 5000 Å

    • 6000 Å

    • 3000 Å

    Solution

    D.

    3000 Å

    The product of wavelength corresponding to maximum intensity of radiation and temperature of the body in Kelvin is constant.
    According to Wien's law
       straight lambda subscript straight m straight T space equals space constant space left parenthesis say space straight b right parenthesis
    where straight lambda subscript straight m is wavelength corresponding to maximum intensity of radiation and T is temperature of the body in Kelvin.
    therefore space space space space straight lambda subscript straight m apostrophe end subscript over straight lambda subscript straight m space equals space fraction numerator straight T over denominator straight T apostrophe end fraction
    Given,   T = 1227 + 273 = 1500 K,
                 T' = 1227 + 1000 + 273 = 2500 K
                  straight lambda subscript straight m space equals space 5000 space straight Å
Hence comma space space space straight lambda subscript straight m apostrophe end subscript space equals space 1500 over 2500 cross times 5000 space equals space 3000 space straight Å

    Question 1933
    CBSEENPH11020789

    A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

    • f/4

    • 8f

    • straight f divided by 2 square root of 2
    • f/2

    Solution

    C.

    straight f divided by 2 square root of 2

    In a series LC circuit, frequency of LC oscillations is given by
                            straight f space equals fraction numerator 1 over denominator 2 straight pi square root of LC end fraction
    or           straight f proportional to fraction numerator 1 over denominator square root of LC end fraction
    rightwards double arrow space space space space space straight f subscript 1 over straight f subscript 2 space equals space square root of fraction numerator straight L subscript 2 straight C subscript 2 over denominator straight L subscript 1 straight C subscript 1 end fraction end root
      Given comma space space space straight L subscript 1 space equals space straight L comma space space straight C subscript 1 equals straight C comma space space straight L subscript 2 space equals space 2 straight L comma space space straight C subscript 2 space equals space 4 straight C comma space space straight f subscript 1 space equals space straight f
therefore space space space space space straight f over straight f subscript 2 space equals space square root of fraction numerator 2 straight L space cross times space 4 straight C over denominator LC end fraction end root space equals space square root of 8
rightwards double arrow space space space space space space space straight f subscript 2 space equals space fraction numerator straight f over denominator 2 square root of 2 end fraction

    Question 1934
    CBSEENPH11020790

    The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8cm the potential energy stored in it is

    • 4U

    • 8U

    • 16U

    • U/4

    Solution

    C.

    16U

    Let extension produced in a spring be x initially in stretched condition spring will have potential energy
               straight U space equals 1 half kx squared
    where k is spring constant or force constant.
                 therefore space space space space space space straight U subscript 1 over straight U subscript 2 space equals space fraction numerator straight x subscript 1 superscript 2 over denominator straight x subscript 2 superscript 2 end fraction space space space space space space space space space space... left parenthesis straight i right parenthesis
Given comma space straight U subscript 1 space equals space straight U comma space space space space space space straight x subscript 1 space equals space 2 space cm comma space space space space straight x subscript 2 space equals space 8 space cm
putting space these space values space in space Eq. space left parenthesis straight i right parenthesis comma space we space have
space space space space space space space space space space space space space space space space space space space space space space space space straight U over straight U subscript 2 space equals space fraction numerator left parenthesis 2 right parenthesis squared over denominator left parenthesis 8 right parenthesis squared end fraction space equals space 4 over 64 equals 1 over 16
therefore space space space space space space space straight U subscript 2 space space equals space 16 straight U space space space space space space space

    Question 1935
    CBSEENPH11020798

    The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant pressure to that at constant volume is

    • 7/5

    • 8/7

    • 5/7

    • 9/7

    Solution

    A.

    7/5

    We have given molar specific heat at constant pressure
             straight C subscript straight P space equals space 7 over 2 straight R
    Mayor's relation can be written as:
    Molar specific heat at constant volume = Gas constant
    i.e.,     CP - CV = R
        rightwards double arrow space space space space straight C subscript straight V space equals space straight C subscript straight P space minus space straight R
space space space space space space space space space space space equals space 7 over 2 straight R space minus space straight R space equals space 5 over 2 straight R
    Hence, required ratio is
            straight gamma space equals space straight C subscript straight P over straight C subscript straight V space equals fraction numerator left parenthesis 7 divided by 2 right parenthesis straight R over denominator left parenthesis 5 divided by 2 right parenthesis straight R end fraction equals 7 over 5

    Question 1936
    CBSEENPH11020800

    Two sound waves with wavelengths 5.0 m and 5.5 m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second

    • 12

    • 0

    • 1

    • 6

    Solution

    D.

    6

    Let straight lambda subscript 1 space equals space 5.0 space straight m comma space space straight v space equals space 330 space straight m divided by straight s space and space straight lambda subscript 2 space equals space 5.5 space straight m
    The relation between frequency , wavelength and velocity is given by
                                   straight v equals nλ
        rightwards double arrow space space space space straight n space equals space straight v over straight lambda                          ...(i)
    The frequency corresponding to wavelength, straight lambda subscript 1
                            straight n subscript 1 space equals space straight v over straight lambda subscript 1 space equals space fraction numerator 330 over denominator 5.0 end fraction space equals space 66 space Hz
    The frequency corresponding to wavelength straight lambda subscript 2 comma
                                straight n subscript 2 space equals straight v over straight lambda subscript 2 space equals space fraction numerator 330 over denominator 5.5 end fraction space equals space 60 space Hz
    Hence, no of beats per second            
                                  equals straight n subscript 1 minus straight n subscript 2
equals 66 minus 60 equals space space 6

    Question 1937
    CBSEENPH11020801

    A transverse wave propagating along x-axis is represented by:

    straight y left parenthesis straight x comma space straight t right parenthesis space equals space 8.0 space sin space open parentheses 0.5 πx minus 4 πt minus straight pi over 4 close parentheses
    where x is in metres and t is in seconds. The speed of the wave is

    • 4 straight pi m/s
    • 0.5 straight pi space straight m divided by straight s
    • straight pi over 4 straight m divided by straight s
    • 8 m/s

    Solution

    D.

    8 m/s

    The standard transverse wave propagating along x-axis can be written as
                         straight y equals straight a space sin space left parenthesis kx space minus space ωt plus space straight ϕ right parenthesis
    The given equation is
     straight y left parenthesis straight x comma space straight t right parenthesis space equals space 8.0 space sin space open parentheses 0.5 space straight pi space straight x space minus space 4 πt minus straight pi over 4 close parentheses space space space space space space space space... left parenthesis 1 right parenthesis
    The standard wave equation can be written as,
              straight y space equals space straight a space sin space left parenthesis kx minus ωt plus straight ϕ right parenthesis                         ...(2)
    where straight alpha is amplitude, k the propagation constant and straight omega the angular frequency,
    comparing the Eqs. (i) and (ii), we have 
                           straight k space equals space 0.5 space straight pi comma space space space straight omega space equals space 4 straight pi
    therefore    Speed of transverse wave 
                                             straight v equals straight omega over straight k equals fraction numerator 4 straight pi over denominator 0.5 straight pi end fraction
space equals space 8 space straight m divided by straight s

    Question 1938
    CBSEENPH11020802

    The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room A?

    • 2

    • 4

    • 1 half
    • 1

    Solution

    A.

    2

    Sabine's formula for reverberation time is
                              straight T space equals space fraction numerator 0.16 space straight V over denominator begin display style sum for blank of end style as end fraction
    where V is volume of hall in m3.
    sum from blank to blank of as space equals space straight a subscript 1 straight s subscript 1 plus straight a subscript 2 straight s subscript 2 plus.......... space equals space total absorption of the hall (room)
    Here s1, s2, s3 ...... are the surface areas of the absorbers and a1, a2, a3.......are their respective absorption coefficients.
    therefore space space space space space space space fraction numerator straight T apostrophe over denominator straight T end fraction space equals space fraction numerator straight V apostrophe over denominator straight s apostrophe end fraction cross times straight s over straight V
                      equals space fraction numerator left parenthesis 2 right parenthesis cubed over denominator left parenthesis 2 right parenthesis squared end fraction equals space 8 over 4 space equals space 2
    Hence comma space space space straight T apostrophe space equals space 2 straight T space equals space 2 cross times 1 space equals space 2 straight s

    Question 1939
    CBSEENPH11020803

    Which one of the following statements is true?

    • Both light and sound waves in air are transverse

    • The sound waves in air are longitudinal while the light waves are transverse

    • Both light and sound waves in air are longitudinal

    • Both light and sound waves can travel in vacuum

    Solution

    B.

    The sound waves in air are longitudinal while the light waves are transverse

    In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.

    Question 1941
    CBSEENPH11020862

    A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by

    • πrρv

    • πrρv2

    • πr2ρv

    • πr2ρv2

    Solution

    D.

    πr2ρv2

    Cross-sectional area A = πr2

    Volume of liquid flowing per second = AV = πr2v

    Mass of the liquid flowing out per second = πr2

    Initial momentum of liquid per second = mass of liquid flowing x speed of liquid

    = πr2vρ x v = πr2v2ρ

    Since the liquid does not rebound after impact, the momentum after impact is zero.

    Therefore, the rate of change of momentum = πr2v2ρ

    According to Newton's second law, the force exerted on wall = rate of change of momentum

    =  πr2ρv2

    Question 1942
    CBSEENPH11020868

    A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litre per second. On increasing the pressure. this increases to two litres per second. The firman has now to

    • push forward twice as hard

    • push forward four times as hard

    • push backward four times as hard

    • push backward twice as hard

    Solution

    B.

    push forward four times as hard

    The rate of change of mass, dm/dt = avρ

    Therefore, F =vdm/dt

    = (avρ)v = av2ρ

    A volume of liquid flowing per second = av

    This is proportional to velocity. By doubling this volume of liquid flow per second the force becomes four times. when liquid flows forward, the hosepipe tens to come backwards. So to keep it intact, it should be pushed forward. Thus, the hosepipe should be pushed forward four times.

    Question 1943
    CBSEENPH11020872

    A liquid is allowed into a tube of truncated cone shape. Identify the correct statement from the following.

    • The  speed is high at the wider end and low at the narrow end

    • The speed is low at the wider end and high at the narrow end.

    • The speed is same at both ends in a streamline flow.

    • The liquid flows with the uniform velocity in the tube.

    Solution

    B.

    The speed is low at the wider end and high at the narrow end.

    For an incomressible liquid equation of continuity.

    AV = = constant or A ∝(1/V)

    Therefore, at the wider end speed below and at the narrow end speed will be high.

    Question 1945
    CBSEENPH11020876

    Two identical glass spheres filled with air are connected by a horizontal glass tube. The glass tube contains a pellet of mercury at its mid-point. Air in one sphere is at 0°C and the other is at 20°C. If both the vessels are heated through 10°C, then neglecting the expansions of the bulbs and the tube.

    • the mercury pellet gets displaced towards the sphere at a lower temperature.

    • The mercury pellet gets displaced towards the sphere at a higher temperature

    • The mercury pellet does not get displaced at all.

    • the temperature rise causes the pellet to expand with any displacement.

    Solution

    C.

    The mercury pellet does not get displaced at all.

    Let n1 and n2 be the number of moles in the bulbs at 0°C and 20°C, respectively, then,

    pV = 273n1R = 293n2R

    n1n2 = 293273 ... (i)

    when the vessels are heated, let the volume of low temperature be V1 and that of the other be V2. Since pressure are same, hence

    V1p = 283 n1R and V2p = 303 n2R

    V1V2 = 283 n1303 n2 ... (ii)

    From Eqs. (i) and (ii), we get 

    V1V2 = 283303 x 293273 = 1

    Thus, the mercury pellet remains at the same position.

    Question 1946
    CBSEENPH11020893

    A girl is swinging on a swinging plate form in sitting position when the girl stands up then the period of swinging?

    • Will not change

    •  Will depend on height of girl

    • Will be longer

    •  Will be shorter

    Solution

    D.

     Will be shorter

    When the girl stands up, the centre of mass gets higher which effect on point of oscillation. As a result effective length will decrease. The period of will varies with square root of the effective length, time period will decrease.

    we know this equation,

    T=2πLg

    Question 1947
    CBSEENPH11020894

    The displacement of the motion of a particle is represent by a equation  Y=ASinωt+BCosωt. The motion of the particle is

    • SHM with amplitude A2+B2 

    • SHM with amplitude A+B

    • SHM with amplitude A

    • Oscillatory but not in SHM

    Solution

    A.

    SHM with amplitude A2+B2 

    A linear combination of sine and cosine functions

            y=Asinωt+Bcosωt             ....1

      Let   A=αcosϕ                                 ....2         B=αsinϕ                                  ....3              

    then equation (1) becomes

                  y=acosϕsinωt+asinϕcosωt      y=asin(ωt+ϕ)                           ..sin(α+β)=sinαcosβ+cosαsinβ

    it is clear that the equation number (2) in simple harmonic motion with the amplitude α squaring and adding (2) and(3),we get

              A2+B2α2(cos2ϕ+sin2ϕ)   α=A2+B2    

    Question 1948
    CBSEENPH11020895

    A ball falling in a lake depth 200 m shows a decrease of 0.1 % in its volume at the bottom then the bulk modulus of the material of the ball is

    • 19.6×108 N/m2

    • 19.2×107N/m2

    • 24×109 N/m2

    • 24×105 N/m2

    Solution

    A.

    19.6×108 N/m2

    The pressure exerted by water on the ball p=hρg

     Given:- ρ=1000 kgm3(density of water) ,

                 h=200m

                 p=103×200×9.8=19.6×105

                 VV=0.1%=0.1100

     Bulk modulus B=-pVV=hρgVV=19.6×1050.1×10-2     

    The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs.If p is positive, ∆V is negative.Thus for a system in equilibrium, the value of bulk modulus B is always positive. So B is positive

                         B  =19.6×108 Nm2 

    Question 1949
    CBSEENPH11020896

    A fixed volume of iron is drawn into a wire of length l. The extension x produced in this wire by a constant force F is proportional to

    • l

    • 1l

    • l2

    • 1l2

    Solution

    C.

    l2

    Young's mosulous=stressstrain=FAll   ...(l-length ,l-change in length)

     l=FYA l   V=Al  A=Vll2=FYVl2l1Vl2 l2             ....F =constant ;Y= constant 

    Graph between l and lis straight line.

    Question 1950
    CBSEENPH11020897

    Two bars A and B of circular cross-section and same volume made of same material are subjected under same tension. If the diameter of A is half that of B. The ratio of extension of A to B will be

    • 16

    • 2

    • 14

    • 4

    Solution

    A.

    16

    Y=stressstrain=FAlll=FAY lBut V =πr2lso    L=Vπr2=FVπr22Y1r4lAIB=rBrB4=rBrB24=16

    Question 1952
    CBSEENPH11020899

    A capillary tube of radius r can support a liquid of weight 6.28 x 10-4 N. If the surface tension of the liquid is 5 x 10-2N/m. The radius of capillary must be

    • 2.0×10-3m

    • 2.0×10-4m

    • 1.5×10-3m

    • 12.5×10-4m

    Solution

    A.

    2.0×10-3m

    Given:- weight =6.24×10-4N,

                 W=T.2πr

             r=W2πT=6.28×10-42×3.14×5×10-2

              r = 2×10-3m                   

    Question 1953
    CBSEENPH11020900

    The soap bubbles have radii in the ratio 2: 1 ratio of excess pressure in side there is

    • 1:4

    • 4:1

    • 2:1

    • 1:2

    Solution

    D.

    1:2

    P1rhence ,P1P2=1r11r2=r2r1=12        ...since r1:r2=2:1                   P1:P2=1:2         

    Question 1954
    CBSEENPH11020901

    Standing wave are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m/s. The frequency is

    • 10Hz

    • 5Hz

    • 4Hz

    • 2Hz

    Solution

    B.

    5Hz

    n=p2lTm    =2l               where l is the length of the stretched string   =5×202×10   n=5Hz

    Question 1955
    CBSEENPH11020902

    A steel wire 0.5 m long has a total mass 0.01 kg and stretched with a tension of 800 N. The frequency with which it vibrates in its fundamental note will be

    • 4Hz

    • 2Hz

    • 200Hz

    • 160Hz

    Solution

    C.

    200Hz

    Given  length of wire L=0.5m ,  tension on wire t= 800Nm=ML=0.010.5kg/mFundamental frequency is n =12LTm=12×0.5800×0.50.01 n  =200Hz

    Question 1956
    CBSEENPH11020903

     A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is

    • 48N

    • 81N

    • 24N

    • 16N

    Solution

    B.

    81N

    String is vibrating in fundamental mode,it's frequency

    'n' is given by ,

    12lTmTherefore the frequency of the fork n will beN= 2n=12lTmIf number of loops formed is P thenN=PlTmby squaring on both sideN2=p2l2×Tm p2T =N2l2mn, l and m are constant

    PT=constanthence,  P1T1=P2T2          6×36 =4T2               T2=9             T2 =81N

    Question 1957
    CBSEENPH11020904

    The equation of travelling wave is y= 60 cos (1800 t-6x) where y is in micron and t in second and x in metre. The ratio of maximum particle velocity to velocity of wave propagation is

    • 3.6

    • 3.6×10-4

    • 3.6×10-11

    • 3.6×10-6

    Solution

    B.

    3.6×10-4

    y(x,t)=A sin(kx-ωt+ϕ)the equation isy =60 cos(1800t-6x)Velocity of the wave is given by  ν=ωk=18006=3.6×10-4Maximum particle velocity isumax= =60×10-6×1800 m/s           (as t in second, x in meter and y is in micron so A=60×10-6)The ratio is umax=60×10-6×1800300=3.6×10-4

    Question 1958
    CBSEENPH11020905

    In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.170.The frequency of the wave will be

    • 2.94Hz

    • 0.73Hz

    • 0.36Hz

    • 1.47Hz

    Solution

    D.

    1.47Hz

                            

    Maximum displacement from mean occurs atT4.   

    For a motion in SHM particle taken 14th of periodic time to reach maximum displacement.

    frequency n =1T=10.68=1.47Hz

    Question 1960
    CBSEENPH11020907

    22 g of CO2 at 270 C is mixed with 16 g of O2 at 370C. If both gases are considered as ideal, then temperature of mixture is

    • 30.50C

    • 370C

    • 270C

    • 29.30C

    Solution

    D.

    29.30C

    22 g of CO2 at T1 =27oC = 27+273=300k is mixed with 16 g of O2 at  T2=370C=37+273=310k molar mass of carbon =12g/mol and molar mass of oxygen =16g/molby addition of both 12+16×2= 44g/molsimillarly , O2 16×2 =32 g/molT=n1Cv1T1+n2Cv2T2n2Cv1+n2Cv2  =2244×3R×300+1632×52R×3102244×3R+1632×52R  =302.3k =29.30C

    Question 1961
    CBSEENPH11020908

    At a given temperature the pressure of an ideal of density p is proportional to

    • ρ2

    • ρ

    • 1ρ2

    • 1ρ

    Solution

    B.

    ρ

    P=13ρc2 ρTherefore at given temperature the presure of an ideal gas is proportional to ρ.

    Question 1962
    CBSEENPH11020909

    Air is expanded from 50 litres to 150 litres at atmosphere the external work done is (1 atmospheric pressure =1×105N/m2)

    • 2×10-8J

    • 2×104J

    • 200J

    • 2000J

    Solution

    B.

    2×104J

    Work done is W= PV                             =2×105×(150-50)×10-3                             =2×104J

    Question 1963
    CBSEENPH11020910

    Helium at 270C has a volume 8 litres. it is suddenly compressed to a volume of 1 litre. The temperature of the gas will be V= 53

    • 93270C

    • 12000C

    • 9270C

    • 1080C

    Solution

    C.

    9270C

    TV1γ-1=TV2γ-2T2=V1V2T1T2=V1V253-2×300T2=8123×300=4×300=1200kT2=1200-273=9270C

    Question 1965
    CBSEENPH11020912

    A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take

    • 50 sec

    • 90 sec

    • 60 sec

    • 48 sec

    Solution

    D.

    48 sec

    The rate of cooling α excess of temperature

    Newton's of cooling 

    dQdt=KT2-T1

    800-600600=K800+6002-300   13=K4600-500t=K600+5002-30010t=K25Dividing equation (1) by (2), we gett30=4025 or t=4025×30=48 sec

    Question 1966
    CBSEENPH11020929

    A cylinder contains 10kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the cylinder, if final pressure is 2.5×106 N/m2 will be( temperature of gas is constant)

    • 15.2 kg

    • 3.7 kg

    • zero

    • 7.5 kg

    Solution

    D.

    7.5 kg

    Initial mass of the gas m1=10kg

    Initial pressure P1=10N/m2

    Final pressure P2 =2.5×106 N/m2

    According to kinetic theory of gases

           PV=13mc-2or pmHence, P1P2=m1m2or      1072.5×16=10m2or  m2 =2.5 kgHence quantity of gas taken out from the cylinder is 10-2.5=7.5 kg

    Question 1967
    CBSEENPH11020932

    A resonance air column of length 40 cm resonates with tunning fork of frequency 450Hz. Ignoring end corrections , the velocity of sound in air will be

    • 1020m/s

    • 920m/s

    • 820m/s

    • 720m/s

    Solution

    D.

    720m/s

    We know that the minimum length of the air column tube corresponds to the fundamental mode of vibrations. Hence by ignoring end corrections

     Length of column =40cm=0.4m

    Frequency of tuning fork = 450 Hz

    Using the formula n=ν4l

    or  v= 450×4×0.4=720m/s

    Question 1968
    CBSEENPH11020933

    If vibration of string is to be two times, then tension in the string must be made

    • Eight times

    • Four times

    • Twice

    • Half

    Solution

    B.

    Four times

    Here initial vibration n1=n , final vibration n2 =2n

    Initial tension T1 =T

    Vibration frequency of string

    n=12lTmnTHence, n1n2=T1T2or           T1T2=n1n22=n2n2=12or   T2=4T1

    Question 1969
    CBSEENPH11020934

    A siren emitting sound of frequency 500 Hz is going away from a stationary listener with a speed of 50 m/s, the frequency of sound heard directly from the siren is 

    • 286.5 HZ

    • 481 Hz

    • 434.2 Hz

    • 580 Hz

    Solution

    C.

    434.2 Hz

    The frequency of siren =500HzThe speed of siren νs =50m/sUsing Doppler's principlen' =νν+νsnn'=330330+50×500n'=330×500380=434.2Hz

    Question 1970
    CBSEENPH11020938

    The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k = 8.31 J/mole K)

    • 1.7×103 J

    • 10.2×103 J

    • 3.4×103 J

    • 6.8×103 J

    Solution

    C.

    3.4×103 J

    According to kinetic theory, K.E of 1g-mole of an ideal gas

    E=32RT 

    where R is universal gas constant, T and n are the temperature of the gas and the number of moles respectively. 

    Hence K.E at normal temperature 0o C =273 K

    E= 32×8.31×273= 3.4×103 J

    Question 1971
    CBSEENPH11026110

    The radius R of the soap bubble is doubled under isothermal condition. If T be the surface tension of soap bubble. The work done in doing so is given by

    • 32 πR2T

    • 24 πR2T

    • 8 πR2T

    • 4 πR2T

    Solution

    A.

    32 πR2T

    Surface tension is force per unit length ( or surface energy per unit area) acting in the plane of interface between the plane of the liquid and any other of the substance. 

     Initial surface energyω1 = 2×4π R2 TFinal surface energyω2 = 2×4π 2R R2 T       =32 πR2T

    Molecules inside the liquid. Forces on a molecule due to others are shown above.

    Question 1972
    CBSEENPH11026116

    An ice-cube of density 900 kg /m3 is floating in water of density 1000 kg / m3. The percentage of volume of ice-cube outside the water is

    • 20%

    • 35%

    • 10%

    • 25%

    Solution

    C.

    10%

    Density is quantity of mass per unit volume. 

    ρ =mV

    A liquid is largely incompressible and its density is nearly constant at all pressures The percentage of the volume  of ice cube outside the water

    =ρwater - ρiceρwater×100=1000-9001000×100=10%

    Question 1973
    CBSEENPH11026123

    A tank is filled with water upto height H. When a hole is made at a distance h below the level of water. What will be the horizontal range of water jet?

    • 2 h H - h

    • 4 h H + h

    • 4 h H - h 

    • 2 h H + h

    Solution

    A.

    2 h H - h

    Applying Bernoulli's theorem

    The velocity of water at point ν = 2 ghTime take to reach at point C is tso,  H - t =12gt2t =2H - hgNow horizantol rangeR = νt  =2 gh × 2H -hg=2H - hh

    Question 1974
    CBSEENPH11026183

    A charged oil drop is suspended in uniform field of 3 x 104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 x 10-15 kg and g = 10 m/s2 )

    • 3.3 × 10-18 C

    • 3.2 × 10-18 C

    • 1.6 × 10-18 C

    • 4.8 × 10-18 C

    Solution

    A.

    3.3 × 10-18 C

    In steady state, 

    electric force on drop = weight of drop 

     q E =  mg

      q = mgE          = 9.9 × 10-15 × 103 × 104  q  = 3.3 ×10-18 C

    Question 1975
    CBSEENPH11026188

    A frame made of metallic wire enclosing a surface area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by

    • 100%

    • 75%

    • 50%

    • 25%

    Solution

    C.

    50%

    Surface energy = surface tension × surface area

                      E = T × 2A

    New surface energy

               E1 = T × 2 A2        = T × A

    % decrease in surface energy

                   =E1 - E2E × 100= 2TA -TA 2TA× 100

    % decrease in surface energy = 50%

    Question 1976
    CBSEENPH11026190

    The potential energy of a molecule on the surface of a liquid compared to one inside the liquid is

    • zero

    • lesser

    • equal

    • greater

    Solution

    D.

    greater

    When the surface area of a liquid is increased, molecules from the interior of the liquid rise to the surface. As these molecules reach the surface work is done against the cohesive force. This work is stored in the molecules in the form of potential energy. Thus, the potential energy of the molecules lying in the surface is greater than that of the molecules in the interior of the liquid.

    The above figure is an intermolecular cohesive force in a bulk of liquid with a free surface.

    Question 1977
    CBSEENPH11026193

    Aerofils are so designed that the speed of air

    • on top side is more than on lower side

    • on top side is less than on lower side

    • is same on both sides

    • is turbulent

    Solution

    A.

    on top side is more than on lower side

    The aerofoils are so designed that

    Pupper side < Plower side

    So that the aerofils get a lifting force in upward direction

    According to Bernoulli's theorem, where the pressure is large, the velocity will be minimum or vice-versa.

    Thus vupper side > vlower side

    Question 1978
    CBSEENPH11026211

    Two glass plates are separated by water. If surface ension of water is 75 dyne/cm and area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is

    • 102 dyne

    • 104 dyne

    • 105 dyne

    • 106 dyne

    Solution

    C.

    105 dyne

    The shape of water layer between the two plates is shown in figure

                    

    Thickness d of the film = 0.12 mm
                                      = 0.012 cm 

    Radius R of the cylindrical face = d2 

    Pressure difference across the surface 

                          = TR = 2Td

    Area of each plate wetted by water = A 

    Force F required to separate the two plates is given by

    F =  pressure difference × area

        =  2Td A

    Putting the given values, we get

    F = 2 × 75 × 80.012

    F = 105 dyne

    Question 1979
    CBSEENPH11026219

    The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is

    • 30 N

    • 150 N

    • 300 N

    • 600 N

    Solution

    C.

    300 N

    Pressure at neck of bottle

       P1F1A1

           = F1π r12

    Similarly, the pressure at bottom of bottle

        P2F2A2

        P2F2π r22

    According to Pascal's law, liquids transmits pressure equal in all directions.

         F1πr12 = F2π r22

        12π 32 = F2π 152

         F2    = 12 × 15 × 153 × 3

          F2 = 300N

    Question 1980
    CBSEENPH11026220

    A square wire frame of size L is dipped in a liquid. On taking out a membrane is formed. If the surface tension of liquid is T, then force acting on a frame will be

    • 2T L

    • 4T L

    • 8T L

    • 16T L

    Solution

    C.

    8T L

    Since, the wire frame is dipped in liquid, therefore its membrane has two free surfaces. 

    Total length of square wire frame in contact of membrane

                           = 2 x perimeter of square

                            = 2 × 4L

                              = 8L

    Hence, force acting on a frame
                        F = T l

                          = T x SL

                          F = 8TL

    Question 1981
    CBSEENPH11026221

    64 spherical rain drops of equal size are falling vertically through air with a terminal velocity 1.5 ms-1 .If these drops coalesce to form a big spherical drop, then terminal velocity of big drop is

    • 8 ms-1

    • 16 ms-1

    • 24 ms-1

    • 32 ms-1

    Solution

    C.

    24 ms-1

    Volume of big drop = 64 x volume of a small drop

          43π R3 = 64 × 43πr3

          ⇒ R = 4 r

    The terminal velocity of spherical rain drop

           ν = 2 r2 ρ - σ9 η

           ⇒ v ∝ r2

         v1v2 = rR2       = 142

           v1v2 = 116

    ∴     v2 = 16 v1

              = 16 × 1.5

           v2 = 24 ms-1

    Question 1982
    CBSEENPH11026222

    The equation of a wave is y = 5 sin t0.04- x4 ; where x is in cm and t is in second. The maximum velocity of the wave will be

    • 1 ms-1

    • 2 ms-1

    • 1.5 ms-1

    • 1.25 ms-1

    Solution

    D.

    1.25 ms-1

    Equation of wave 

              y = 5 sin t0.04 - x4

    The standard equation of a wave in the given form

              y =α sin ωt - 2πxλ

    Comparing the given equation with the standard equation, we get

          α = 5 

    and  ω = 10.04

            ω = 25

     Therefore, maximum velocity of particles of the medium,

        vmaxαω

                 = 5 × 125 cm s-1

         vmax = 1.25 ms-1   

    Question 1983
    CBSEENPH11026230

    If there were no gravity, which of the following will not be there for a fluid?

    • Viscosity

    • Surface tension

    • Pressure

    • Archimedes' upward thrust

    Solution

    D.

    Archimedes' upward thrust

    Archimedes' upward thrust will be absent for a fluid, if there were no gravity.

    Question 1984
    CBSEENPH11026250

    The cylindrical tube of a spray pump has a cross-section of 8 cm2 , one end of which has 40 fine holes each of area 10-8 m2. If the liquid flows inside the tube with a speed of 0.15 m min-1 the speed with which the liquid is ejected through the holes is

    • 50 m/s

    • 5 m/s

    • 0.05 m/s

    • 0.5 m/s

    Solution

    B.

    5 m/s

    One of the fundamental principles used in the analysis of uniform flow is known as continuity of flow. This principle is derived from the fact that mass is always conserved in fluid systems regardless of the pipeline complexity or direction of flow.

    According to the equation of continuity,

              Q = A1 V1 = A2 V2

                      a v = constant                       ... (i)

    Where Q is the volumetric flow rate

             A is the cross-sectional area of flow

              V is the mean velocity

    Area of cross secrction of the spray pump,

    a1 = 8 cm2 = 8 × 10-4 m2

    number of holes n = 40

              v1 = 0.15 m min-10.1560

             ∴   For tube, 

                8 × 10-4 × 0.1560 = a1 v1

        For holes,

                 40 × 10-8 × v   = a2 v2 

    According to law of continuity from equation (i)

    ∴                a2 v = a1 v1

    ∴         40 × 10-8 × v  = 8 × 10-4 × 0.1560

    ⇒                   v = 8 × 10-4 × 0.1540 × 10-8 × 60

    ⇒                    v = 5 m/s

    Question 1985
    CBSEENPH11026256

    Horizontal tube of non-uniform cross-section has radii of 0.1 m and 0.05 m respectively at M and N. For a streamline flow of liquid, the rate of liquid flow is

                  

    • changing continuously with time

    • greater at M than at N

    • greater at N than at M

    • same at M and N

    Solution

    C.

    greater at N than at M

    The mass flow per unit time passing through each cross-section does not change, even if the pipe diameter changes. This is the law of conservation of mass.

    A continuity equation in physics is an equation that describes the transport of some quantity. It is perticularly simple and powerful when applied to conserved quantity. 

    According to the equation of continuity

                ρ A v  = constant

     If the fluid is incompressible e.g water, with ρ being effectively constant, then

                 A v = constant

    Where A is the area of cross-section, v is the velocity and ρ is the density of a fluid.

    Av gives the volume flux or flow rate remains constant throughout the pipe of flow.      

    The velocity of flow will increase if cross-section decreases and vice-versa.

     

    i.e                         A1 V1 = A2 V2

    ⇒                            Av = constant

    Therefore, the rate of liquid flow will be greater at N than at M.

    Question 1986
    CBSEENPH11026259

    If two soap bubbles of different radii are connected by a tube, then

    • air flows from bigger bubble to the smaller bubble till sizes become equal

    • air flows from bigger bubble to the smaller bubble till sizes are interchanged

    • air flows from smaller bubble to bigger

    • there is no flow of air

    Solution

    C.

    air flows from smaller bubble to bigger

    The excess pressure inside the soap bubble is inversely proportional t radius of soap bubble i.e 

                         ρ  1r

    where r is the radius of the bubble.

    It flows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus if these two bubbles are connected by a tube, air will flow from smaller bubble to a bigger bubble and the bigger bubble grows at the expense of the smaller one. Air flows from smaller bubble to bigger as the pressure in a smaller bubble is higher.

    Question 1987
    CBSEENPH11026263

    A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If  T  is the surface tension of soap solution, the energy spent in the blowing

    • 3 πTr2

    • 6πTr2

    • 12πTr2

    • 24 πTr2

    Solution

    D.

    24 πTr2

    Initially area of soap bubble

               A1 = 4π r2

    Under isothermal condition radius becomes 2 r, Then, area

           A24π 2r2

                = 4 π. 4r2

            A2 = 16 πr2

    Increase in surface area

           A = 2 A2 - A1

                   = 2 16πr2 - 4πr2

             A = 24 π r2

    Energy spent

             W = T × ΔA

                  = T . 24π r2

             W = 24 πT r2 J

    Question 1988
    CBSEENPH11026280

    A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water 

    • V2

    • V3

    • V4

    • V

    Solution

    A.

    V2

    This difference in pressure between the top and the bottom of the object produces an upward force on it. This is called upthrust.

    According to Archimedes Principle, the upthrust on an object in a fluid is equal to the weight of the fluid displaced, so the volume of the object multiplied by the density of the fluid.

    When a body (sphere) is half immersed, then 

       Upthrust = weight of sphere

           V2 × ρliq × g = v  ×  ρ × g

    ∴                ρ = ρliq2

    When body (sphere) is fully immersed then,

    Upthrust = weight of sphere + weight of water pourd in sphere

               V × ρliq × g = V × ρ × g + V' × ρliq × g

                   V' = V2

    Question 1989
    CBSEENPH11026283

    The speed of air flows on the upper and lower surfaces of a wing of an aeroplane are v1 and v2 respectively. If A is the cross section area of the wing and ρ is the density of air, then the upward life is

    • 12ρ Av2 - v1

    • 12ρ Av1 + v2

    • 12 ρ A  v12 - v22 

    • 12ρ A v12  +  v22 

    Solution

    C.

    12 ρ A  v12 - v22 

    Due to the specific shape of wings when the aeroplane runs air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called dynamic lift act on the plate .

    Upward lift = pressure difference × area of wing

                   = 12 ρA v12 - v22

    Question 1990
    CBSEENPH11026286

    The work done in increasing the size of a soap film for 10cm × 6cm  to 10cm × 11cm is 3  × 10-4 J. The surface tension of the film is

    • 1.0  × 10-2 N/m

    • 6.0  × 10-2 N/m

    • 3.0  × 10-2 N/m

    • 1.5  × 10-2 N/m

    Solution

    C.

    3.0  × 10-2 N/m

    Given:-

    Increasing size ΔA = (10 × 10-2 × 11 × 10-2 - 10 × 10-2 × 6 × 10-2 )

                            ΔA = ( 10 × 11 - 10 × 6 ) 

    Work done = area increased  × surface tension

          surface tension 

                  ST3 × 10-42 × 10 × 11 - 10 × 6  × 10-4          

    (since soap film has two surfaces)

                   ST = 3 × 10-2 N/m

    Question 1991
    CBSEENPH11026294

    A balloon contains 500 m3 of helium at 27o C and 1 atmosphere pressure. The volume of the helium at - 3° C temperature and 0.5 atmosphere pressure will be

    • 1000 m3

    • 900 m3

    • 700 m3

    • 500 m3

    Solution

    B.

    900 m3

    According to Boyle's law

       V ∝ TP

    The volume of a given amount of gas is proportional to the ratio of its Kelvin temperature and its pressure

       PVT = C

    As the temperature goes up and vice versa.

        P1 V1T1 = P2V2T2etc   

    Using gas equation law

             P1 V1T1 = P2 V2T2

    ⇒   1 × 500300 = 0.5 × V2270

    ⇒    V2 = 900 m3

    Question 1992
    CBSEENPH11026298

    An ice-cube of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of volume ofice-cube outside the water is

    • 20%

    • 35%

    • 10%

    • 25%

    Solution

    C.

    10%

    The percentage of volume of ice cube outside the water is

           = ρwater - ρiceρwater

        = 1000 - 9001000×100 

      = 10 %

    Question 1993
    CBSEENPH11026335

    The lower end of capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?

    • 5.9

    • 4.9

    • 2.9

    • 1.9

    Solution

    C.

    2.9

        hHg - height of the mercury

         hw - height of water     

            hHg2 THg cosθHgg r ρHg

    and    hw2Tw cosθHg g r ρw

    where, the symbols have their usual meanings

     Now, 

               hwhHg = hwhHg × ρHgρw × cos θwcos θHg

                      = 0.0750.465 × 13.6 × cos 0ocos 40o

               hwhHg = 2.9

    Question 1994
    CBSEENPH11026369

    A capillary tube of length L and radius r is connected with another capillary tube of the same length but half the radius in series. The rate of steady volume flow of water through first capillary tube under a pressure difference of p is V. The rate of steady volume flow
    through the combination will be (the pressure difference across the combination is p)

    • 17 V

    • 1617 V

    • V17

    • 1716 V 

    Solution

    C.

    V17

    Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is potential difference is replaced by Δp and electrical resistance

            R18 η Lπ r4 = R

    and    R28 ηLπ r24

                   = 16 R

    Electric current is replaced by rate of volume flow V'

           p = VR1 = VR                      .....(i)

            p' = p = V' Req

                      = V' (R1 + R2 )  

                       = 17 V'R

          From Eqs. (i) and (ii), we get

             VR = 17 V'R

    ⇒         V' = V17    

    Question 1995
    CBSEENPH11026374

    A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

    • dHdt  1r5

    • dHdt  r4

    • dHdt  1r4

    • dHdt  r5

    Solution

    D.

    dHdt  r5

    Viscous force acting on spherical drop

          Fv = 6 πηrv

    ∴  Terminal velocity v = 29r2 σ - ρη

    where, η = coefficient of viscosity of liquid 

      σ = density of material of spherical drop

       ρ = density of liquid

    Power imparted by viscous force = Rate of production of heat

         P = dHdt

            = Fv . v

             = 6 πηrv2

              = 6 πηrv . 29 r2 σ - ρ gη2

     ⇒   dHdt ∝ r5

    Question 1996
    CBSEENPH11026389

    A solid sphere of volume V and density p floats at the interface of two immiscible liquids of densities and p, respectively. If ρ1 < ρ < ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquid is

    • ρ - ρ1ρ2 - ρ

    • ρ2 - ρρ - ρ1

    • ρ + ρ1ρ + ρ2

    • ρ + ρ2ρ + ρ1

    Solution

    B.

    ρ2 - ρρ - ρ1

    V = Volume of solid sphere

    Let V1 = Volume of the part of the sphere immersed in a liquid of density p1 and 

    V2 = Volume of the part of the sphere immersed in liquid of density p2

    The law of flotation says that  for a floating object the weight of the object equals the weight of the fluid.

    According to law of flotation,

            Vρ g = V1 pg + V2 pg             ....(i)

           V1 ( ρ - ρ1 ) g = V2 ( p- ρ )

    ⇒     V1V2 = ρ2 - ρρ - ρ1

    Question 1997
    CBSEENPH11026410

    Assertion:  A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink. 

    Reason:  The buoyancy of an object depends both on the materials and shape of the object.

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.

    Solution

    C.

    If assertion is true but reason is false.

    When a needle is placed carefully on the surface of water,  it floats on the surface of water due to the surface tension of water,  which does not allow the needle to sink.

    In case of a steel ball, the surface tension of the water is not sufficient to keep it floating, so it sinks down.

    Question 1998
    CBSEENPH11026419

    A cylindrical tank is filled with water to level of 3 m. A hole is opened at height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. The square of the speed with which water is coming out from the orifice is (Take g = 10 m s-2)

    • 50 m2 s-2

    • 40 m2 s-2

    • 51.5 m2 s-2

    • 50.5 m2 s-2

    Solution

    A.

    50 m2 s-2

    Suppose A be the area of cross section of tank, a be the area of hole, ve be the velocity of
    efflux, h be the height of liquid above the hole, 

         A

    Let v be the speed with which the level decreases in the container. Using equation of
    continuity, we get

            ave = Av

    ⇒       v = aveA 

    Using Bernoulli's theorem, we have

          P0 + hρg + 12 ρν2 = P012 ρ ve2

    ρ = density of fluid

    P = pressure

    ⇒    hρg  + 12 ρ aveA2  =  12 ρve2

    ⇒                    ve2 2hg1 - a2A2

                               = 2 ×  3 - 0.525 1 - 0.12

    ⇒                      ve2 = 50 m2 s-2

    Question 1999
    CBSEENPH11026441

    Viscous force exerted by the liquid flowing between two plates in a streamline flow depends upon the

    • velocity gradient in the direction perpendicular to the plates

    • area of plate

    • co-efficient of viscosity of the liquid

    • all of these

    Solution

    D.

    all of these

    Viscous force exerted by the liquid flowing between two plates is given by

              F = - η Advdy

    So Viscous force exerted by the liquid flowing between two plates in a streamline flow depends upon the all the three factors mentioned in option.

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