Science Chapter 2 Is Matter Around Us Pure
  • Sponsor Area

    NCERT Solution For Class 9 About 2.html

    Is Matter Around Us Pure Here is the CBSE About 2.html Chapter 2 for Class 9 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 9 About 2.html Is Matter Around Us Pure Chapter 2 NCERT Solutions for Class 9 About 2.html Is Matter Around Us Pure Chapter 2 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 9 About 2.html.

    Question 1
    CBSEENSC9004743
    Question 11
    CBSEENSC9004753

    Write the steps you would use for making tea. Use the words—solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

    Solution

    Step for making tea:

    1) Take 100ml of water as solvent in a kettle and boil it.
    2) when solvent boil add milk, tea leaves, sugar. these are solute. now again boil the solution for few mintues. sugar is soluble in water thus it will be disslove.
    3) colour of tea leaves goes into solution as filtrate. The remaining tea leaves being insoluble remains as residue. 
    4). filter the solution, enjoy your tea.




    Question 13
    CBSEENSC9004755

    Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

    Substance Dissolved

    Temperature in K

    283

    293

    313

    333

    353

    Potassium nitrate

    21

    32

    62

    106

    107

    Sodium chloride

    36

    36

    36

    37

    37

    Potassium chloride

    35

    35

    40

    46

    54

    Ammonium chloride

    24

    37

    41

    55

    66



    Solution

    As solution cools, potassium chloride gets crystallized or precipitated. Because at 353K, solubility of potassium chloride is 54 g per 100 g of water. Whereas at room. temperature (20°C) it is 35 g. So excess of potassium chloride dissolved gets precipitated.

    Question 14
    CBSEENSC9004756

    Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

    Substance Dissolved

    Temperature in K

    283

    293

    313

    333

    353

    Potassium nitrate

    21

    32

    62

    106

    107

    Sodium chloride

    36

    36

    36

    37

    37

    Potassium chloride

    35

    35

    40

    46

    54

    Ammonium chloride

    24

    37

    41

    55

    66



    Solution
    The solubility of each salt at 293 K in 100g of water:

    Potassium nitrate =32
    sodium chloride    = 36
    potassium chloride=35
    ammonium chloride = 37

    since ammonium chloride is maximum dissolve in 100g of water therefore the solubility of ammonium chloride is maximum at 293K
    Question 16
    CBSEENSC9004758

    Explain the following giving examples:

    Saturated solution

    Solution

    Saturated solution. A solution in which no more of the solid (solute) can be dissolved at a given temperature is called a saturated solution. Suppose 40 g of a solute is the maximum amount that can be dissolved in 100 g water at 298 K. Then (100 + 40) g. of solution so obtained is the saturated solution at 298 K.

    Question 17
    CBSEENSC9004759

    Explain the following giving examples:

    Pure substance

    Solution

    Pure substance. The substances, which always have the same colour, taste or texture at a given temperature and pressure are pure substances. A pure substance has a characteristic melting point or boiling point at a given pressure. For example, pure water is always colourless, odourless and tasteless and boils at 373 K at normal atmospheric pressure.

    Sponsor Area

    Question 18
    CBSEENSC9004760

    Explain the following giving examples:


    Colloid

    Solution

    Colloid: The particles of a colloid are uniformly spread throughout the solution. A colloid has the following characteristics:

    (i) It is a heterogeneous mixture.

    (ii) The size of particles of a colloid lies between 1–100 nm and are not visible to naked eye.

    (iii) The particles of a colloid can scatter a beam of light.

    (iv) The particles of a colloid cannot be separated from the mixture by filtration.

    (v) The particles of a colloid do not settle when left undisturbed. Examples are smoke, milk, fog, cloud.

    Question 19
    CBSEENSC9004761

    Explain the following giving examples:


    Suspension

    Solution

    A suspension is a heterogeneous mixture in which the solute particle do not dissolve but remain suspended throughout the bulk of the medium. A suspension has following characteristics: 

    (i) It is a heterogeneous mixture.

    (ii) The particles of a suspension are greater than 100 nm and are visible to naked eye.

    (iii) The particles of a suspension can scatter a beam of light.

    (iv) The particles of a suspension settle down when left undisturbed for some time.

    (v) The particles of a suspension can be separated from its mixture by filtration. Examples—Fine clay particles in water.

    Question 20
    CBSEENSC9004762

    Classify each of the following as a homogeneous or heterogeneous mixture : soda water, wood, air, soil, vinegar, filtered tea.

    Solution

    Homogeneous mixture : Soda water, air, vinegar, filtered tea.

    Heterogeneous mixture : Wood, soil.

    Question 21
    CBSEENSC9004763

    How would you confirm that colourless liquid given to you is pure water?

    Solution

    Every liquid has a characteristic boiling point at 1 atmospheric pressure. Boiling point of pure water is 373K at 1 atomspheric pressure. If the given sample is colourless liquid and boils at exactly 373K at 1 atmospheric pressure, then it is pure water. If the boiling point is even slightly above 373K, then it is contaminated.

    Question 22
    CBSEENSC9004764

    Which of the following matter falls in the category of a ‘pure substance’?

    (a) Ice
    (b) Milk
    (c) Iron
    (d) Hydrochloric acid
    (e) Calcium oxide
    (f)Mercury
    (g) Brick
    (h) Wood
    (i) Air

    Solution

    These are ‘pure substance’
    (a) Ice-A Compound
    (c) Iron-An Element
    (d) Hydrochloric acid-A Compound
    (e) Calcium oxide-A Compound
    (f)Mercury-An Element


    Question 24
    CBSEENSC9004766

    Which of the following will show Tyndall effect?

    Salt solution, Milk, Copper sulphate solution, Starch solution.

    Solution

    Milk and Starch solution are colloids and will show Tyndall effect.

    Question 25
    CBSEENSC9004767

    Classify the following into elements, compounds and mixtures.

    Sodium, Soil, Sugar solution, Silver, Calcium carbonate, Tin, Silicon, Coal, Air, Soap, Methane, Carbon dioxide and Blood.

    Solution

    Elements:Sodium, Silver, Tin, Silicon.

    Compounds:Calcium carbonate, Methane, Carbon dioxide.

    Mixtures:Soil, Sugar solution, Coal, Air, Soap, Blood.

    Question 26
    CBSEENSC9004768
    Question 27
    CBSEENSC9004769

    What is chemically a pure substance?

    Solution

    A pure substance or chemical substance is a material that has constant composition and has consistent properties throughout the sample. For example sodium chloride, Diamond etc.

    Question 28
    CBSEENSC9004770

    Are the naturally occurring materials in nature chemically pure substances? Explain.

    Solution

    Most of the naturally occurring material in nature are not chemically pure substances. They are purified by chemical or physical process. For example the minerals are available mixed with other material or at least mixed will dust particles. 

    Question 29
    CBSEENSC9004771

    How does the meaning of pure substance different for a common man than to a chemist? Explain.

    Solution

    A pure element or compound contains only one substance, with no other substances mixed in it.

    For a common man, pure substance means that it is not adulterated. For example, for a common man, pure milk means the liquid material that is given by cow or buffallo and it is not mixed with water or other materials to thicken it. But for a chemist, any milk is not a pure substance. Milk is a mixture of water, fats and proteins. These materials are different in milk obtained from different sources. Similarly pure fruit juice for a common man means is that it is not mixed with fruit juice of other fruits or water or salt. But for a chemist any fruit juice is a mixture of water and several other substances. For a common man pure gold means gold of a given composition. But a chemist knows that no ornaments can be made from pure gold. All ornaments of gold contain gold and copper.

    Question 31
    CBSEENSC9004773

    How would you test the purity of a given substance?

    Solution

    A pure substance or chemical substance is a material that has constant composition and has consistent properties throughout the sample. it has a fixed melting point or boiling point at constant pressure. For example, pure water boils at 373K at 1 atmospheric pressure. But water containing some invisible dissolved substance boils at a temperature above 373K.

    Question 32
    CBSEENSC9004774

    State why we cannot call any vegetable or fruit a pure substance?

    Solution

    A pure substance is a material that has constant composition and has consistent properties throughout the sample. A fruit or a vegetable does not always have the same colour, taste or texture due to the presence of various constituents in different proportions.

    Question 33
    CBSEENSC9004775

    Define and explain the term:Mixture.

    Solution

    Mixtures are constituted by more than one kind of pure form of matter, known as a substance. In a mixture the constituents are not chemically combined and can be separated into pure substances even by physical processes. Sea water is a mixture of dissolved salts and water. On heating water evaporates and solid salt is left behind. The composition of a mixture can be varied and the properties of a mixture depends on its composition. For example, various types of stainless steel with distinct properties can be prepared by mixing steel with different amounts of chromium and nickel.


    Question 34
    CBSEENSC9004776

    Explain how are mixtures classified on the basis of their physical properties.

    Solution

    Mixtures are classified as homogeneous and heterogeneous on the basis of their physical properties:

    Homogenous mixture

    Heterogeneous mixture

    A homogenous mixture has no visible boundaries of separation between the various constituents. It has a uniform composition throughout its mass. All homogenous mixture is called solution.

    A heterogeneous mixture has visible boundaries of separation between the various constituents. It does not have a uniform composition throughout its mass.

    A mixture of sugar in water is a solution. It has same sugar- water composition throughout and all parts of it have the same sweetness. A solution has only one phase and there is no boundary of separation as between sugar and water particles in a sugar solution.

    The mixture of sugar and sand is heterogeneous mixture. There are distinct boundaries of separation between sugar and particles. Further different parts of this mixture will have different sugar and sand compositions, i.e., there will be more of sugar particles in some parts of the mixture while other parts may have more of sand particles.

    A mixture of two miscible liquids is also a homogeneous mixture. e.g., a mixture of water and alcohol. Other examples are air, alloys like brass, amalgam etc.

     

    Other example are suspension of solids in liquid and mixture of immiscible liquids like water and oil.

    Question 35
    CBSEENSC9004777

    You are given two liquids A and B. One is a compound or a pure substance and the other is a mixture (solution). How will you select the compound?

    Solution

    The liquid compound would boil off at a constant temperature whereas the liquid mixture (solution) would boil off at two different temperatures. Thus the liquid which boils throughout at a constant temperature is a compound or pure substance.

    Question 36
    CBSEENSC9004778

    When iron filings and sulphur are mixed in a particular fashion, the following observations are made. No iron gets attracted to a magnet placed in the resultant matter.A foul smelling gas is released when the mixture is treated with dilute sulphuric acid. Giving reasons, state whether the resultant substance is a mixture or a compound.

    Solution

    The resultant substance is not a mixture as the iron and sulphur are mixed in a fixed proportion and also the properties of the constituents of the mixture are not shown. That is, no iron filings get attracted to a magnet. Dilute sulphuric acid with iron sulphide (a compound) gives a foul smelling gas. Thus, the resultant substance is a compound of iron and sulphur.

     

    Question 37
    CBSEENSC9004779

    How do you establish that air is a mixture and not a pure substance?

    Solution

    Air is a homogeneous mixture of several gases.
    mostly of nitrogen (~78%) and oxygen(~20%). There are other gases as well; argon slightly less than 1%, and of course, carbon dioxide,  (~.04%). Therefore it can be established as follows:

    (i) The amounts of gases present in air at different places varies only marginally.

    (ii) No definite formula can be assigned to air.

    (iii) The different gases present in air are not separable by a physical process : fractional distillation of liquid air.

    (iv) The properties of air are the properties of its constituents. Air supports combustion because of oxygen present in it. Air turns lime water milky due to the presence of carbon dioxide in it.
    Question 38
    CBSEENSC9004780

    Give the distinguishing features of a pure substance or a compound and a mixture.

    Solution

    Various points of distinction between a chemical compound and a mixture are summarised below:

     

    Characteristics

    Pure substance or compound

    Mixture

    1. Composition

     

    The elements in a compound are present in definite proportion by weight.

     

    The ingredients of a mixture may be present in varying ratio.

    2. Homogeneity

     

    A compound is always homogeneous.

     

    A mixture may be homogeneous (solution) or heterogeneous. The constituents of a mixture can be separated by simple methods.

     

    3. Properties

    .A compound has entirely different properties from those of its constituents.

     

    Properties of a mixture are an average of the properties of its constituents.

     

    4. Separation

     

    The constituents of a compound cannot be separated by simple separation techniques

    The constituents of a heterogeneous mixture can be separated by simple physical separation techniques.

    5. Energy changes

     

    Energy in the form of heat, light or electricity is either evolved or absorbed when a compound is formed.

    There is generally no energy change when a mixture is formed from its constituents.

    6. Volume change

    At constant temperature and pressure, formation of a compound may involve either no change or large change in volume.

    At constant temperature and pressure, formation of a mixture involves either very little or no change in volume.

    Question 39
    CBSEENSC9004781

    List the various types of homogeneous and heterogeneous mixtures with examples.

    Solution

    (i) Homogeneous mixtures

    (a) Solid solution. Alloys, e.g., brass, steel, bronze.(b) Liquid solution. Sugar solution in water, sea water.
    (c) Gaseous solution. Air, natural gas.

    (ii) Heterogeneous mixtures:

    (a) Solid + Solid : sugar and sand, iron filing and sulphur, copper sulphate and potassium permanganate.
    (b) Solid + Liquid : Pastes, suspensions etc.
    (c) Solid + Gas : Smoke.
    (d) Liquid + Liquid : Milk, emulsions.
    (e) Liquid + Gas : Soap bubbles.


    Sponsor Area

    Question 40
    CBSEENSC9004782

    What is meant by a pure substance?

    Solution

    A pure substance or chemical substance is a material that has constant composition and has consistent properties throughout the sample. For example sodium chloride. Diamond etc.

    Question 41
    CBSEENSC9004783

    List the point of difference between homogeneous and heterogeneous mixtures.

    Solution

    Differences between homogeneous and heterogeneous mixtures :

    Homogeneous mixture

    Heterogeneous mixture

    (i) It has no visible boundary or boundaries of separation between its constituents.

    (i) It has visible boundary/boundaries of separation between its constituents.

    (ii) It has a uniform composition throughout its mass.

    (ii) The composition of the mixture varies from mass.

    For example, sugar in water, salt in water, Sulphur in carbon disulphide, water and alcohol etc.

    For example, sand and salt, sugar and salt, wood blood, water in oil etc.

    Question 42
    CBSEENSC9004784

    Define the following terms:

    True solution

    Solution

    True solution is defined as a homogeneous mixture of two or more substances. The term homogeneous means that every portion of the solution has the same properties, e.g., sugar dissolved in water, iodine in ethyl alcohol (tincture). In molecular term, it also means that unit volume of the solution has the same number of atoms/molecules or ions.

    Question 43
    CBSEENSC9004785

    Define the following terms:

    Solute

    Solution

    Solute. The substance (s) present in smaller proportion in a solution is called solute.

    Question 44
    CBSEENSC9004786

    Define the following terms:

    Solvent

    Solution

    Solvent. The substance present in larger proportion in a solution is called solvent.

    Question 45
    CBSEENSC9004787

    Define the following terms:

    Solubility

    Solution

    The maximum amount of the solid that can be dissolved in a given amount of the solvent (water) at room temperature is termed its solubility.

    Question 46
    CBSEENSC9004788

    Differentiate between a saturated solution and an unsaturated solution. How will you test whether a given solution is saturated or not?

    Solution

    Saturated solution. A solution in which no more of the solid (solute) can be dissolved at a given temperature and pressure is called a saturated solution.

    Unsaturated solution. A solution in which more of the solid can be dissolved at the given temperature and pressure is called an unsaturated solution.

    If we consider both solution as a salt solution then If to a given solution, more salt is added and it dissolves, then it is unsaturated. If the salt added settles down,  it is saturated.
    Question 47
    CBSEENSC9004789

    How will you prepare a saturated solution of copper sulphate in water at 50°C? What will happen if this solution is allowed to cool?

    Solution

    Take water in a beaker. Heat the water gently to 50°C and slowly add copper sulphate powder, stirring the solution continuously with the help of a glass rod. Continue adding the compound, keeping constant  temperature of solution 50°C, till some compound remains undissolved and settles down. Then quickly filter the solution. The solution so obtained is a saturated solution of copper sulphate at 50°C. On cooling, the solubility of copper sulphate will decrease and it will start settling down at the bottom of the beaker.

    Question 48
    CBSEENSC9004790

    Given a solution of substance A, how will you test whether it is saturated or unsaturated with respect to A at the prevailing temperature? What is observed when a hot saturated solution of a substance is allowed to cool?

    Solution

    To given solution add a small amount of substance A to its given solution. If A dissolves and clear solution is formed, then it is an unsaturated solution of substance A. If the substance A added does not dissovle and settles down, then the solution is saturated. 


    When a hot saturated solution of a substance is allowed to cool, some of the salt reappears. 

    Question 49
    CBSEENSC9004791

    When a ray of light passes through a solution, it is not visible. Explain.

    Solution

    The particles of a solution are smaller than 10 metre in diameter. such small particles do not scatter a beam of light passing through them. So, the path of light is not visible in a solution. 

    Question 50
    CBSEENSC9004792

    What would happen if a saturated solution at a given temperature is allowed to cool?

    Solution

    A saturated solution at a given temperature contains the maximum amount of solute that can be dissolved at this temperature. on cooling or at lower temperature, the solubility of the solute decreases. So on cooling a saturated solution, solid solute starts reappearing or settles down.

    Question 51
    CBSEENSC9004793

    An alloy cannot be separated into its constituents by physical methods but still it is considered a homogeneous mixture. Explain.

    Solution

    Alloys are considered as homogeneous mixtures because,

    (i) The composition of an alloy is uniform throughout its mass.
    (ii) It shows the properties of its constituents and can have variable  composition of its constituents.

    Question 52
    CBSEENSC9004794

    What makes water as a universal solvent?

    Solution

    Water acts as a solvent due to:

    (i) The polar nature of its molecules

    (ii) Its ability to produce soluble salt on reaction with a large number of substances.

    Question 53
    CBSEENSC9004795

    A saturated solution becomes unsaturated on heating, Comment.

    Solution

    Kinetic energy of solution particle increases with increasing temperature. It allows the solvent molecules to more effectivley break apart the solute molecules that are held together by intermolecular attraction, hence more solute will be dissovled in solution. Thus, it is correct to say that a saturated solution becomes unsaturated on heating.

    Question 54
    CBSEENSC9004796

    Which is the difference between aqueous and non-aqueous solutions? Give one example of each.

    Solution

    Solution in which water is the solvent are called aqueous solution. For example, when sugar is dissolved in water, the solution obtained is called an aqueous solution of sugar.

    The solution obtained by dissolving a solute in any solvent other than water are called non aqueous solution. For example, if a solute is dissolved in any solvent such as benzene, alchol, ether, carbon, disulphide, carbon tetrachloride, acetone,etc., 
    For example, solution of Iodine in alchol.

    Question 55
    CBSEENSC9004797

    How do you express concentration of a solution?

    Solution

    Concentration of a solution may be expressed in terms of percentage by mass of solute per 100 gram of the solution.

    percent sign space solute space equals space fraction numerator mass space of space solute over denominator mass space of space solution space end fraction cross times 100

    Thus, if a solution is 5%, it means it is 5 gram of a solute dissolved in 100 gram of the solution or it contains 5 gram solute and 95 gram solvent.

    Concentration of a solution may also be expressed as mass of solute dissolved in 100 cm3 of the solution. Thus, 10% sugar solution by volume means 10 g sugar dissolved in 100 cm3 of solution.

    Question 56
    CBSEENSC9004798

    40 g of sugar is dissolved in 250 ml of solution. Calculate the concentration of the solution in terms of mass by volume percentage of solution.

    Solution

    we have given

    Mass of Sugar(solute) = 40g
    Vol. of solution = 250 ml

    Therefore,    

    Mass by volume percentage fo solution is given by:
    Mass of solute Volume of solution x 100

    Thus, 
    40250 x 100 = 16%


    Question 57
    CBSEENSC9004799

    What is a suspension?

    Solution

    A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of a suspension are visible to the naked eye. 

    Properties of a Suspension:

    1.) Suspension is a heterogeneous mixture.
    2.) The particles of a suspension can be seen by the naked eye.
    3.) The particles of a suspension scatter a beam of light passing through it and make its path visible.
    4.) The solute particles settle down when a suspension is left undisturbed, that is, a suspension is unstable. They can be separated from the mixture by the process of filtration.

    Question 58
    CBSEENSC9004800

    What are colloidal solutions? Give examples.

    Solution

    A colloidal solution is said to be intermediate state of the solution and the suspension. It is neither a true solution nor a suspension. In colloidal solution, the particle size is between 1 nm to 100 nm. For example, a solution of soap in water, milk, blood, writing inks are some of the colloidal solutions. Though these particles are 10 to 1000 times the size of a single small molecule, these are not visible to naked eyes. But these are large enough to scatter the light that passes through the dispersion medium. The medium of colloidal solution is known as dispersion phase (dispersion medium) and the particles dissolved are called dispersed phase. 

    Properties of a colloid:

    1) A colloid is a heterogeneous mixture. The size of particles of a colloid is too small to be individually seen by naked eyes.

    2) Colloids are big enough to scatter a beam of light passing through it and make its path visible.

    3) They do not settle down when left undisturbed, that is, a colloid is quite stable.

    4)They cannot be separated from the mixture by the process of filtration. But, a special technique of separtion known as centrifuagtion can be used to separate the colloidal particels. 

     

    Question 59
    CBSEENSC9004801

    16. In what respect does a true solution differ from a colloidal solution and a suspension? How will you test whether a given solution is a colloidal solution or a suspension?

    Solution

    Comparison of properties of solution, colloidal solution and suspension.

    Property

    Solution

    Colloidal solution

    Suspension

    (1) Appearance

     

    Homogeneous and transparent

     

    Heterogeneous and transluscent

    Heterogeneous and opaque

    (2) Particle size

     

    < 1 nm (10–7 cm)

     

    1 nm—100 nm (10–7 cm—10–5 cm)

    > 100 nm (10–5 cm)

     

    (3) Visibility

     

    Particles are not visible even with a powerful microscope

     

    Particles can be seen with a high power microscope

     

    Particles can be seen with naked eyes

     

    (4) Stability

     

    Stable

    Stable

     

    Unstable

    (5) Diffusion

    Diffuse rapidly NaCl dissolved in water

    Diffuse slowly Blood

    Do not diffuse Mud water


    Test of a colloidal solution or a suspension.

    (i) A colloidal solution is turbid and particles settle down on adding a salt whereas no change is noticed when a salt is added to a true solution.

    (ii) If the particles in a heterogeneous and opaque solution can be seen with naked eyes and get settled on keeping, then it is a suspension.

    Question 60
    CBSEENSC9004802

    What are colloids? State their characteristics.

    Solution

    If the size of the particles of the solute (also called dispersed phase) is between 1 nm and 100 nm, then these are called colloids. Their solutions are known as colloidal solutions. The colloids exhibit the following characteristic properties:

    (i) Brownian motion. Colloidal particles move at random in zig-zag paths like gas particles. This is called Brownian motion. This type of motion is caused due to the collisions between the particles of the dispersion medium and the colloidal particles.

    (ii) Tyndall effect. When strong beam of light is passed through a true solution taken in a beaker placed in a dark room, the path of light through the solution is dark. But if the light is passed through a colloidal solution under conditions as above, the path of light through the colloidal solution becomes visible. This is called Tyndall effect. The colloidal particles become illuminated because they scatter the light falling on them in all directions.


    Fig. Tyndall effect. 

    (iii) Electrophoresis: When an electric current is passed through a colloidal solution, the particles move either towards the positive or negative electrode. This phenomenon is known as electrophoresis. By the direction of the movement of colloidal particles, we can know the nature of the charge on them.
    Electrophoresis 

    Question 61
    CBSEENSC9004803

    The solubility of urea is 109.4 g/100 g in water at 20° C. A solution is formed by dissolving 500 g urea in 1 kg water. State the nature of the solution formed.

    Solution

    The solution contain more amount of solid in water in case of ((109.4 g/100 g) at 20°C. On other hand solution formed by 500g urea in 1 kg is contains lesser amount of the solid in water. Hence, solution formed is unsaturated.


    Question 62
    CBSEENSC9004804

    Explain what happens when:

    (i) a beam of light is passed through a colloidal solution.

    (ii) an electric current is passed through a colloidal solution.

    Solution

     

    i) When a beam of light is passed through a colloidal solution the path of the light is illuminated. This is because colloidal particles scatter the light in the all directions. 

    ii) When an electric current is passed through a colloidal solution, colloid particles being either positively or negtively charged, starts move towards the oppositely charged electrode.

    Question 63
    CBSEENSC9004805

    Classify the following mixtures into true solution, colloid and suspension:

    (i) Mud water, (ii) sand water, (iii) lime water, (iv) aerated water, (v) milk, (vi) sugar in water, (vii) common salt in benzene, (viii) writing ink, (ix) blood, (x) jellies, (xi) butter, (xii) gold ornaments, (xiii) vinegar, (xiv) aluminium paint, (xv) mist, (xvi) lemonade, (xvii) coke, (xviii) wheat flour in water.

    Solution

    True solutions. Lime water, aerated water, sugar in water, gold ornaments, vinegar, lemonade, coke.

    Colloids. Milk, writing ink, blood, jellies, butter, aluminium paint, mist.

    Suspensions. Mud water, sand water, common salt in benzene, wheat flour in water.

    Question 64
    CBSEENSC9004806

    How does a true solution differ from a colloidal solution? Differentiate on the basis of respective particle size.

    Solution

    Property

    Solution

    Colloidal solution

    (1) Appearance

     

    Homogeneous and transparent

     

    Heterogeneous and transluscent

    (2) Particle size

     

    < 1 nm (10–7 cm)

     

    1 nm—100 nm (10–7 cm—10–5 cm)

    (3) Visibility

     

    Particles are not visible even with a powerful microscope

     

    Particles can be seen with a high power microscope

     

    (4) Stability

     

    Stable

    Stable

     

    (5) Diffusion

    Diffuse rapidly NaCl dissolved in water

    Diffuse slowly Blood

    Question 66
    CBSEENSC9004808

    Given below the list of some colloids. Make a list selecting them on the basis of type of dispersing medium. Also give the nature of dispersed phase in each.

    fog, clouds, shaving cream, milk, face cream, milk of magnesia, foam, rubber, sponge, jelly, cheese, butter, paint, writing ink, smoke, blood, coloured gemstone.

    Solution

    Selection of colloids on the basis of type, dispersed phase and dispersion medium:

    Dispersing medium

    Colloids

    Dispersed phase

    Type

    1. Gas

    fog, clouds, mist smoke, automobile exhaust

    liquid solid

    aerosol

    2. Liquid

    shaving cream,

    milk, face cream, writing

    ink, blood, paint

    milk of magnesia, mud

    gas liquid

     

     

     

     

    solid

    foam emulsion

     

     

     

     

    sol

    3. Solid

    foam, rubber, sponge, pumice
    jelly, cheese, butter, coloured gemstone

    gas



    liquid


    solid

    foam gel



    solid sol

    Question 67
    CBSEENSC9004809

    Giving example, state the use of a solution, a colloid and a suspension in our daily life.

    Solution

    Solutions: When two reacting substances are kept in contact, the reaction does not take place or is very slow. But when their solutions are mixed, the reaction takes place very fast. This happens because in solutions, reacting substances come in contact with each other at molecular level. When solid washing soda is poured over clothes, it does not wash but when clothes are mixed with solution of washing soda, clothes are easily cleaned.

    Colloids: Some insoluble substances particularly in pharmaceuticals become more effective when administered in colloidal form. Many medicines are insoluble in water but are used in the form of colloids.

    Suspensions: Many substances insoluble or sparingly soluble in dispersing medium become useful for analytical purposes when dispersed temporarily. For example, sparingly soluble barium sulphate dispersed in water is an opaque medium and is used for diagnostic X-rays.

    Question 68
    CBSEENSC9004810

    Give four properties each of a solution, a colloid and a suspension.

    Solution

    Property

    Solution

    Colloidal solution

    Suspension

    (1) Appearance

     

    Homogeneous and transparent

     

    Heterogeneous and transluscent

    Heterogeneous and opaque

    (2) Particle size

     

    < 1 nm (10–7 cm)

     

    1 nm—100 nm (10–7 cm—10–5 cm)

    > 100 nm (10–5 cm)

     

    (3) Visibility

     

    Particles are not visible even with a powerful microscope

     

    Particles can be seen with a high power microscope

     

    Particles can be seen with naked eyes

     

    (4) Stability

     

    Stable

    Stable

     

    Unstable

    (5) Diffusion

    Diffuse rapidly NaCl dissolved in water

    Diffuse slowly Blood

    Do not diffuse Mud water

    Question 69
    CBSEENSC9004811

    A mixture is homogeneous and light is not visible on passing through it. State what can you infer about the nature of the mixture and size of its particles.

    Solution

    The mixture is homogeneous, it must be a  true solution. The particles of a true solution are smaller than 1 nm diameter and cannot scatter beam of light.

    Question 70
    CBSEENSC9004812

    The particles of a mixture are approximately 40–50 nm in diameter. Can the constituents of the mixture be separated by centrifugation or filtration.

    Solution

    Since the size of the particles are approximately 40-50 nm in diameter. Therefore it is colloidal solution. 
    The colloid can be separated by centrifugation not by filtration.

    Question 71
    CBSEENSC9004813

    A mixture is heterogeneous. How will you know whether it is a solution, a colloid or a suspension.

    Solution

    The given mixture cannot be a solution because a solution is always homogeneous. If this mixture is left undisturbed for some time and no solid settles down, then it is colloid otherwise it is suspension.


    Question 72
    CBSEENSC9004814

    Why does the beam of light illuminated when it enters a room through a small hole?

    Solution

    When a beam of light entering through a small hole appears illuminatied because the light beam is scattered by randomly moving dust particles.

    Question 73
    CBSEENSC9004815

    What is Tyndall effect? Why is Tyndall effect observed when sun light passes through dense forest.

    Solution

    The illumination of beam of light due to scattering on collision with  tiny particle (which act as particles of colloid dispersed)  is called Tyndall effect.

    In dense forest, particularly in the early morning, there is fog, i.e., tiny water droplets suspended in air. These droplets are large enough to scatter beam of light. This is why Tyndall effect can be observed when sunlight passes through dense forest.

     

    Question 74
    CBSEENSC9004816

    To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

    Solution

    Mass of sodium chloride = 36 g
    Mass of water = 100 g
    Mass of solution = 100+36 = 136 g


    Concentraction of solution =Mass of soluteVolume of solution 

    Thus,

    36136 x 100  =26.47%

    Question 75
    CBSEENSC9004817

    Why is the separation of a mixture done?

    Solution

    Separation of components of a mixture is done for the following purposes:

    (i) To remove harmful components such as insects from grains or insects and other soluble impurities from water.

    (ii) To remove an undesirable component such as stones etc. from wheat, rice.

    (iii) To obtain a useful component such as butter from milk or gur from sugarcane.

    (iv)To obtain a pure sample of a substance.

    Question 76
    CBSEENSC9004818

    What is the principle of separation?

    Solution

    Principle of separation depend on the difference in physical or chemical properties of constituents of a mixture. Depending upon the kind of constituents and their respective properties, different techniques are used.

    Question 77
    CBSEENSC9004819

    What are the commonly used techniques used by farmers in villages to purify food grains?

    Solution

    Farmers generally use the following techniques:

    (i) Winnowing: It is the method of separating lighter husk from grain with the help of air.

    (ii) Hand-picking. The method of hand-picking is used to separate undesirable components from the mixture such as small pieces of stones, husk, grain from wheat, rice and pulses.

    (iii) Sieving. It is the processes of separating particles of different sizes by allowing them to pass through sieve having different hole sizes.

    Question 78
    CBSEENSC9004820

    Name the physical processes commonly used in laboratories, industries to separate mixtures.

    Solution

    The physical process commonly used in laboratories are:

    (i) Evaporation. It is the process of removing water from solution by leaving in sun or heating on water bath.

    (ii) Sedimentation. It is the process in which heavy solid particles are allowed to settle at the bottom of the vessel and then followed by decantation of the clear liquid.

    (iii) Magnetic separation. It is the process in which magnetic substances are separated from non-magnetic substances with the help of a magnet.

    (iv) Centrifugation. It is the process in which lighter suspended particles in a liquid are made to settle down by rotating the mixture at high speed.

    (v) Filtration. It is the process of separating solid particles from liquid by passing the liquid mixture through a filter paper.

    (vi) Crystallisation. It is the process of getting crystals of pure substance from its solution.

    (vii) Sublimation. In this process a solid substance from a mixture of solids is directly converted into vapours (followed by condensation) by heating without passing through liquid state.

    (viii) Distillation. It is the process in which one liquid is separated from its mixture by converting it into vapours and then cooling them.

    Question 79
    CBSEENSC9004821

    Give two examples of solutions containing a non-volatile component and a volatile component. How will you obtain non-volatile component from such a solution?

    Solution

    (a) Solution of sugar (non volatile component) in water (volatile component).
    (b) Solution of ink (non-volatile component) in water (volatile component).

    Non-volatile component from a solution can be obtained by the process of evaporation.

    Experiment. Take a beaker half filled with water. Put a china dish or watch glass on the mouth of the beaker. Pour some drops of solution in the watch glass. Heat the beaker. Your will find that water from solution starts evaporating. Continue heating till evaporation goes on and only solid material is left behind. Solid material left behind is the non-volatile component. The volatile component, i.e., water gets evaporated.

    Fig. 2.2. Evaporation.

     

    Sponsor Area

    Question 80
    CBSEENSC9004822

    What is centrifugation. How it is done?

    Solution

    Centrifugation is the process of separating suspended particles from a liquid like colloids by churning the liquid at a high speed. The principle is that denser particles are forced to the bottom and lighter stay at the top when spun rapidly.

    The mixture is taken in a closed bottle and rotated at a high speed. The heavy particles settle at the bottom while light particles remain behind. For example, to separate cream from milk, milk is churned for 2-3 minutes. Cream collects at the centre and being lighter than milk floats at the top of the mixture.


    Question 81
    CBSEENSC9004823

    Give any four applications of centrifugation.

    Solution

    Centrifugation is a process which involves the application of the centripetal force for the sedimentation of heterogeneous mixtures with a centrifuge.
     Applications of centrifugation:

    (a) Used in dairies and home to separate cream from milk or butter from cream.

    (b) Used in washing machines to squeeze out water from clothes.

    (c) Used in laboratories to separate colloidal particles from their solutions.

    (d) Used in diagnostic labs for blood and urine test.

     

    Question 82
    CBSEENSC9004824

    Discuss the use of separating funnels.

    Solution

    Separating funnels are used to separate two liquids which do not form homogeneous mixture on mixing but form two distinct layers. The principle is that immiscible liquids separate out in layers depending on their densities. The immiscible liquids like water and mustard oil when mixed form two separate layers, hence they can be separated using separating funnels.

    Experiment. Take the mixture of water and mustard oil in a separating funnel. Allow the mixture to stand undisturbed for sometime. When separate layers of oil and water are formed, oil being lighter makes the upper layer. Open the stopcock of the separating funnel and pour out the lower layer of water into the beaker carefully. Oil remaining behind is then collected in another beaker or flask.

    Fig. (a) Mixture (b) pure liquids.

    Question 83
    CBSEENSC9004825

    Define and explain distillation.

    Solution

    Distillation is a process to obtain pure liquid from its solution. Distillation can be defined as the conversion of impure liquid into vapours by evaporation and then condensation (cooling) of the vapours into the pure liquid.

    Impure liquid or mixture is taken in a flask, fitted with a condenser as shown in Fig. When heated, liquid gets evaporated and the vapours pass to the delivery tube where they condense into liquid. The pure condensed liquid is collected in receiver. The solid remains behind in the flask, hence gets separated. This method is generally used for the separation of components of a mixture containing two miscible liquids that boil without decomposition and have sufficient difference in their boiling points.

    Fig. Distillation.

    Question 84
    CBSEENSC9004826

    What types of mixtures can be separated by simple distillation ?

    Solution

    Simple distillation can be used to separate the following mixtures:

    (a) It is used for the separation of components of a mixture containing two miscible liquids that boil without decomposition and have sufficient difference in their boiling points.

    (b)This method is used for the separation of different gases from air, different factions from petroleum products etc.

    Question 85
    CBSEENSC9004827

    How will you separate a mixture of ethyl alcohol (boiling point 78°C) and water (boiling point 100°C) ? Describe the process.

    Solution

    Ethyl alcohol and water are completely miscible liquids and these have sufficient difference (22°C) in their boiling points. So these can be separated by the process of simple distillation.

    Experiment. Arrange the apparatus as shown in Fig. Take the mixture of ethyl alcohol and water in a distillation flask. Heat slowly till the liquid starts boiling and the temperature of the liquids starts rising. At 78°C, vapours of alcohol will be formed and get collected as liquid in the receiver after passing through the condenser. Temperature will remain constant till all the alcohol in the mixture is evaporated. When the temperature starts rising, stop heating. Pure water is left in the distillation flask and alcohol collects in the receiver.

    Question 86
    CBSEENSC9004828

    A mixture of acetone and toluene was distilled in a distillation flask. On slow heating it was found that temperature became constant at 56°C for some time and the residual liquid left was toluene. When the heating was continued, again the temperate became constant at 80°C.

    Answer the following questions.

    (a) What is the boiling point of acetone and toluene ?

    (b) Why was it possible to separate acetone from toluene ?

    (c) Why does temperature remain constant for some time at 56°C even though heating was continued.

    Solution

    (a) The boiling point of acetone and toluene are 56°C and 80°C respectively.

    (b) The mixture of acetone and toluene could be separated by simple distillation because these two miscible liquids have a large difference in their boiling points.

    (c) The temperature at 56°C remains constant for some time because heat given at this temperature first gets converted into latent heat of vaporisation of acetone.

    Question 87
    CBSEENSC9004829

    What is the advantage of distillation over evaporation?

    Solution

    Distillation is more advantageous than evaporation because, in evaporation liquid is lost where as in distlliation liquid can be recoverd.

    Question 88
    CBSEENSC9004830

    What is chromatography? State its applications.

    Solution

    Chromatography is the technique whereby different soluble components of a mixture are separated due to their differential movement over a stationary phase under the influence of mobile phase, i.e., solvent. If hexane is the solvent then the component that is more soluble inhexane will move faster. In this way the different components get separated.

    The technique of chromatography is used to separate

    (a) components of dyes

    (b) pigments from natural colours

    (c) amino acids

    (d) sugars from urine samples

    (e) drugs in blood.

    Question 89
    CBSEENSC9004831

    What is blue/black ink that we use in our daily life?

    Solution

    Blue/black ink is a solution of dye in water. This dye is a mixture of two or more coloured components. For example black ink is used in printing news paper, books, etc.

    Question 90
    CBSEENSC9004832

    Describe the process to show that the dye used in blue/black ink is a mixture of two or more components.

    Solution
    Take a strip of a special type of filter paper. Draw a line with the help of a pencil approximately one inch above one of the edges. Put a small spot of ink on the centre of line with the help of a thin capillary tube. This strip of paper is then suspended in a beaker or jar containing water so that the line is a little above water level. Here water acts as the mobile phase and strip of paper is the stationary phase. You will find that water rises up the paper by capillary action and flows over the spot. As the water rises up, it takes along with it the dye particles also. The spots of separated coloured compounds would be visible at different heights from the position of initial spot. The number of components in ink dye is equal to the number of spots. Usually a dye is a mixture of two or more coloured compounds.



    Question 91
    CBSEENSC9004833

    What is the important criteria to use chromatography for separating components of a mixture?

    Solution

    Chromatography is the process of separation of the individual components of a mixture based on their relative affinities towards stationary and mobile phases.

    Question 92
    CBSEENSC9004834

    What is a fractionating column? Explain its advantage.

    Solution

    When the boiling points of the two liquids are quite close (difference less than 20-25°C) the separation is not successfully affected by using simple distillation process but is affected by fitting the flask with a fractionating column.
    A fractionating column is a long tube provided with obstructions or packed with glass beads. The obstructions provided slows down the passage of vapours upwards and that of the liquid downwards.

    Fig. Fractionating columns. 

    Advantages. Suppose you have a mixture of liquids (say A with boiling point 80°C and B with boiling point 90°C). Set up the apparatus as shown in Fig. 2.9. On heating the flask the vapours obtained consist more of A and less of B. As the vapours rise up the fractionating column, they condense partially and the condensed liquid flows down. Vapours of B having higher boiling point will condense more rapidly than the vapours of A having lower boiling point. As a result the vapours moving up get richer in A. The condensed liquid flowing down meets the ascending stream of vapours and in the process takes away more of high boiling liquid B. The process is


    Fig. Fractional distillation.

    repeated throughout the length of fractionating column. Consequently, the vapours which escape from the top of the fractionating column into the condenser consist of almost pure A. Thus the distillate received is pure A whereas the liquid left behind in the flask is rich in B. Pure B can then be obtained by simple distillation. The process of fractional distillation with fractions A and B are repeated if the boiling point difference is very low.

    Question 93
    CBSEENSC9004835

    Give two important industrial applications of fractional distillation.

    Solution

    Two important industrial applications of fractional distillation:

    (a) Separation of petroleum products-petrol, diesel and other organic compounds are obtained from crude petroleum.

    (b) Separation of gases like oxygen, nitrogen, argon from air.


    Question 94
    CBSEENSC9004836

    How is air separated into its constituents?

    Solution

    Air is a homogeneous mixtures of several gases which have boiling points much below room temperature. Air is first of all brought to liquid form under higher pressure and reduced temperatures. Then at constant high pressure, it is slowly warmed up in fractionating column. Different fractions of air are then obtained in gaseous form at different temperatures. The gas with lowest boiling point is obtained first and so on.

    Question 95
    CBSEENSC9004837

    Oxygen, argon and nitrogen boil at –182°C, –186°C and –196°C respectively.

    Answer the following questions:

    (a) Arrange the gases in the increasing value of boiling point.

    (b) Which gas forms the liquid first as the air is cooled?

    (c) Which gas is obtained first when liquid air boils up?

    Solution

    (a) In the increasing value of boiling point.
    Nitrogen < argon < oxygen

    (b) Oxygen forms liquid first as the air is cooled.

    (c) Nitrogen has least boiling point. hence it obtained first.

    Question 96
    CBSEENSC9004838

    Diesel oil, kerosene, petrol and petroleum ether are different fractions of crude petroleum. These have boiling point ranges as 525–670K, 453–550K, 343–473K and 303–363K respectively. When crude petroleum is heated on a fractionating column, arrange the fractions in the order these are collected in the distillate.

    Solution

    Fractional distlillation is based on the difference in boiling point is less than 25K. Since the boiling point difference is more than 25 K . Therefore the order of distillate is petroleum ether, petrol, kerosene, diesel oil.

    Question 97
    CBSEENSC9004839

    Name the process used to obtain pure solid from its impure sample. Describe the process taking impure sample of sugar as an example.

    Solution

    The process of crystallization is used to purify a solid from its impurities.

    Description of crystallization:
    i)  Take about 5 gm of a sample of impure sugar dissolved in minimum amount of water.

    ii) Filter the impurities. Saturate the solution by heating in a china dish.

    iii) when some solid starts appearing.

    iv) Cover the solution with filter paper and leave it undisturbed to cool slowly for a day.

    The crystals of pure sugar so obtained are separated either by decantation or filtration.

    Question 98
    CBSEENSC9004840

    What are the advantages and disadvantages of crystallization over evaporation?

    Solution

    Crystalization seperate as solid crystals and if we evapourated its being lost and both on heating.

    Advantages of crystallization:
    (i) Fine crystals of the substance are obtained.
    (ii) Solids which decompose on heating can be safely purified.

    Disadvantages of crystallization:

    (i) Some substance remains with the liquid/solvent.(ii) Liquid needs to be purified for further use.



    Question 99
    CBSEENSC9004841

    What is decantation? Explain.

    Solution

    Decantation is the process of separation of insoluble solids from liquids. A suspension of solid particles in a liquid is allowed to stand for sometimes. Insoluble solid particles settle down at the bottom due to their weight. This is called sedimentation. The clear liquid is then transferred into another container, without disturbing the settled particles. In other words clear liquid is decanted and separated from solid.

    Question 100
    CBSEENSC9004842

    Explain the methods by which you would separate camphor from sodium chloride.

    Solution

    Camphor would separate from sodium chloride by sublimation or Filtration.

    (i) Sublimation: Mixture of camphor and sodium chloride is heated in a china dish covered with an inverted funnel closed from the lower end. Camphor sublimes and collects on the inner wall of the funnel. Sodium chloride is left as residue in china dish.

    (ii) Filtration. Dissolve the mixture in water. Sodium chloride dissolves in water while camphor remains as residue. It is then filtered. Residue left on filter paper is camphor. Sodium chloride is then recovered from water by evaporation.

    Question 101
    CBSEENSC9004843

    You are given a mixture of sand, water and mustard oil. How will you separate the components of the mixture.

    Solution

    Take the mixture of sand, water and mustard oil. Let the solution to stand undisturbed for about half an hour. The heavy sand particles settle down at the bottom of the container. Now transfer the liquid to another beaker in such a way that the sand remains in the first beaker.

    Take the remaining mixture of mustard oil and water in a separating funnel. Let the funnel to stand undisturbed for half an hour. Water and mustard oil forms two distinct layers. Oil being lighter forms upper layer while the water forms lower layer. Open the tap of the separating funnel and let the lower layer to collect down in the beaker. Close the tap when lower water layer is completely separated. Take out the oil layer in another beaker. Thus the components of mixture are separated.

    Question 102
    CBSEENSC9004844
    Question 103
    CBSEENSC9004845

    How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25 °C), which are miscible with each other?

    Solution

    As both kerosene and petrol do not decompose on heating and the boiling points difference between kerosene and petrol is more than 250C. Thus, these can be separated by using simple distillation technique. 

     

    Question 104
    CBSEENSC9004846

    Name the technique to separate:

    (i) butter from curd
    (ii) salt from sea-water 
    (iii) Camphor from salt.

    Solution

    Techinque to separate:
    (i) Butter from curd : Centrifugation
    (ii) Salt from sea-water : Evaporation
    (iii) Camphor from salt : Sublimation

    Question 105
    CBSEENSC9004847

    What type of mixtures are separated by the technique of crystallization?

    Solution

    Crystallization is used to:

    (i) Purify salt that we get from sea water.

    (ii) Obtain pure substances from their impure samples. For example separation of crystal of alum (phitkari) from impure samples.

    Question 106
    CBSEENSC9004848

    Define physical and chemical changes.

    Solution

    Physical change.: Physical change is a temporary change in which the composition and chemical nature of the substance remains the same but the state of the substance and physical properties changes. For example, when ice changes into water and water changes into vapours. All have different appearance and display different physical properties but they are chemically same.

    Chemical change. A chemical change is a permanent change in which not only physical properties but chemical properties also change. This is because new substance/s is formed in a chemical change. Burning of candle is a chemical change. Here both the physical and chemical changes are taking place and new substances are formed.

    Question 107
    CBSEENSC9004849

    Give reason for each of the following:

    (i) Ripening of a fruit is a chemical change.

    (ii) Crystallization of a substance from its solution is a physical change.

    (iii) Formation of curd from milk is a chemical change.

    (iv) Separation of gases from air is a physical change.

    Solution

    (i) Composition and properties of fruit initially are entirely different than those after ripening of fruit. It cannot be reversed. Hence, it is a chemical change.

    (ii) Crystallization of a substance is a physical change because there is no change in composition and chemical properties of the substance. The crystallized substance can again be made into solution.

    (iii) Curd and milk have entirely different properties and curd cannot be changed to milk. Hence it is a chemical change.

    (iv) Separation of gases from air is a physical change because separated gases can again be mixed.

    Question 108
    CBSEENSC9004850

    Giving reason, explain which of the following is a physical change.

    (i) Electric current is passed through water and hydrogen and oxygen gas are formed.

    (ii) Electric current is passed through a bulb and it glows.

    Solution

    (i) When electric current is passed through water, it break water molecule in hydrogen and oxygen gas. Hence new products are formed whose properties is different from the water, thus it is a chemical change.

    (ii) Electric current is passed through a bulb and it glows—is a physical change because when current is stopped, bulb does not glow and no new substance is formed.

    Question 109
    CBSEENSC9004851

    Giving reason explain burning of oil is a chemical change.

    Solution

    Burning of oil is chemical change because during the buring oil react with oxygen and produce new substances, whose properties are different from the oil.

    Question 110
    CBSEENSC9004852

    Does the energy changes take place both in physical and chemical changes?

    Solution

    Yes, energy changes take  place in both physical and chemical changes. For example evaporation of liquid is a physical change and energy is absorbed in this case. Burning of candle is chemical change and energy is evolved. Hence energy is involve in both case.

    Question 111
    CBSEENSC9004853

    Differentiate between physical and chemical changes.

    Solution

    Given below is a comparison of two types of changes:

    Physical changes

    Chemical changes

    1.Changes take place only in properties such as colour, physical state, density etc.

    1.Change result in the formation of new chemical substance.

    2.Change is temporary.

    2.Change is permanent.

    3.Original substance(s) can be obtained back easily.

    3.Original substance cannot be obtained back easily.

    4.Chemical properties of the substance remain unchanged even after  the physical change.

    4.New substance with different properties are formed.

    Question 112
    CBSEENSC9004854

    Give some examples of physical and chemical changes.

    Solution

    Physical changes:
    Examples of physical changes are: breaking of glass tumbler, drying of clothes, cutting a fruit with knife, breaking of a brick with hammer, changing of visible shapes of moon, melting of wax.

    Chemical changes:
    Examples of chemical changes are: burning of paper, cooking of food, rusting of iron, making of ice-cream, burning of wax, eating of sugar.

    Question 113
    CBSEENSC9004855

    Classify the following as a chemical or physical change.

    (i) Cutting of trees; (ii) Melting of butter in a pan; (iii) An almirah gets rusted; (iv) Water boils to form steam; (v) Electric current is passed through water and it is broken down into hydrogen and oxygen gas; (vi) Dissolving salt in water.

    Solution

    (i) Cutting of trees–Physical Change

    (ii) Melting of butter in a pan–Physical change

    (iii) An almirah gets rusted–Chemical change

    (iv) Water boils to form steam–Physical change

    (v) Electric current is passed through water and it is broken down into hydrogen and oxygen–Chemical change

    (vi) Dissolving salt in water–Physical change

    Question 114
    CBSEENSC9004856

    Name and explain any five physical changes.

    Solution

    1. Change of physical state: Solids can be converted into liquid and then to a gas by heating or vice-versa.

    2. Glowing of an electric bulb: When a bulb is switched on, its filament made of tungsten starts glowing. When the bulb is switched off, the bulb stops glowing and there is no change in its filament.

    3. Dissolution of salts: When common salt is added to water, a solution is formed. When the solution is heated, water evaporates and white solid of common salt is left behind. This white solid has the same properties as the original substance.

    4. Mixing of iron filings and sulphur: If the iron filings are separated from sulphur with the help of a magnet, the properties of the substances recovered are the same as the original substances, i.e., iron filings and sulphur.

    5. Sublimation of naphthalene: When naphthalene is heated, it gets converted into vapours which are sublimed (condensed) into a white solid, which has the same properties as the original substance.

    Question 115
    CBSEENSC9004857

    Melting of wax is a physical change but burning of wax is a chemical change. Explain.

    Solution

    (i) Melting of wax changes the state of wax from solid to liquid. No new substance is formed. The chemical nature of solid wax and liquid wax is the same. Hence melting of wax is a physical change.

    (ii) Burning of wax gives new gaseous products.  Hence burning of wax is a chemical change.

    Question 116
    CBSEENSC9004858

    Classify the following as chemical or physical changes:

    • cutting of trees

    • melting of butter in a pan

    • rusting of almirah

    • boiling of water to form steam

    • passing of electric current through water and the water breaking down into hydrogen and oxygen gases

    • dissolving common salt in water

    • making a fruit salad with raw fruits

    • burning of paper and wood.

    Solution

    • Cutting of trees—physical change

    • melting of butter in a pan—physical change

    • rusting of almirah—chemical change

    • boiling of water to form steam—physical change

    • passing of electric current through water and the water breaking down into hydrogen and oxygen gases—chemical change

    • dissolving common salt in water—physical change

    • making a fruit salad with raw fruits—physical change

    • burning of paper and wood—chemical change

    Question 117
    CBSEENSC9004859

    Try segregating the things around you as pure substances or mixtures.

    Solution

    Pure substances—copper wire, water, sugar, salt, etc.

    Mixtures—fruits, furniture, paper, clothes, food, butter, shoes, vegetables, toothpaste, etc.

    Question 118
    CBSEENSC9004860

    Define the term—Element. Give an illustration in support of your definition.

    Solution

    Element: An element is a substance which contains same kind of atom. It cannot be converted into anything visibly simpler than itself. For example, element mercury is composed of only one kind of atoms (with atomic no. 80). Gold, copper, hydrogen are other examples of elements. After the discovery of isotopes, the definition of an element is modified. It is more correct to say that element is a substance made up of atoms, all having the same atomic number.

    ILLUSTRATION

    Take about 5.0 g of mercuric (II) oxide (red oxide of mercury) in a hard glass test tube fitted with cork and a delivery tube as shown in Fig. 2.10. Heat the tube over a bunsen flame, first slowly and then strongly. After sometime you would observe a gas coming out of the capillary tube. If you bring a glowing matchstick near the mouth of the tube, the stick starts burning with a flame. That is due to evolved oxygen which is supporter of combustion. Continue heating the glass tube. After sometime, no more gas is evolved and a shining liquid is left at the bottom of the glass tube. That is mercury.

     

    Conclusion. Red mercuric oxide on heating breaks up into two simpler substances, mercury and oxygen which are not further broken. Thus mercury and oxygen are elements.

    Question 119
    CBSEENSC9004861

    Name a substance that cannot be decomposed under ordinary circumstances either by a physical process or a chemical process.

    Solution

    A substance that cannot be decomposed by a physical or a chemical process is known as an element. For example, oxygen, silver etc.

    Sponsor Area

    Question 120
    CBSEENSC9004862

    Define the term compound.

    Solution

    A compound is a substance which is formed by the combination of two or two more elements in a fixed proportion by weight. The properties of a compound are entirely different from its constituents. A compound can be decomposed into two or more simpler substances. For example, water is a compound formed by combination of 89% oxygen and 11% hydrogen by weight irrespective of its source. Its properties namely, density, physical state, reactivity etc. are quite different from the properties of hydrogen and oxygen. Water can be easily decomposed into hydrogen and oxygen. Carbon dioxide is made of carbon and oxygen, salt is made of sodium and chlorine and sugar is made of carbon, oxygen and hydrogen having their constituent elements in fixed proportions. Carbon dioxide, salt and sugar are compounds.

    Question 121
    CBSEENSC9004863

    Give the difference between an element and a compound.

    Solution

    Element

    Compound

    1. It cannot be broken into more simpler form by chemical methods.

    1. It can be decomposed into its constituents by chemical methods.

    2. It is composed of atoms of one kind only.

    2. It is composed of atoms of different elements in a fixed proportion by weight.

    For Example Copper, iron, etc.

    For example H2O, methane etc.

    Question 122
    CBSEENSC9004864

    Classify each of following as element, compound or mixture.

    (i) Sugar (ii) Iron (iii) Air (iv) Water

    (v) Milk (vi) Phosphorus (vii) Urea (viii) Carbon dioxide

    (ix) Diamond (x) Brass.

    Solution

    Classification of following substance as:

    Elements : Iron, phosphorus, diamond.

    Compounds : Sugar, water, carbon dioxide, urea.

    Mixture : Air, milk, brass.

    Question 123
    CBSEENSC9004865

    Which one of the following processes give a compound?

    (i) Sodium is burnt in air;

    (ii) Chalk and marble chips are ground together;

    (iii) Mixing of sulphur with sodium chloride;

    (iv) Addition of alcohol in water.

    Solution

    (i) Sodium is burnt in air to give a compound because sodium reacts with oxygen of air to give a new compound, sodium oxide.

    Question 125
    CBSEENSC9004867

    Classify the following substances as elements, compounds or mixtures:

    (i) Sodium, (ii) Hydrogen, (iii) Brass, (iv) Carbon dioxide, (v) Gun powder, (vi) Soda water, (vii) Water, (viii) Chlorine, (ix) Air, (x) Common salt, (xi) Stainless steel, (xii) Iron filings, (xiii) Oxygen, (xiv) Milk, (xv) Blood, (xvi) Ammonia, (xvii) Glass, (xviii) Marble, (xix) Emulsion, (xx) Wood.

    Solution

    Classification of substances:

    Elements—(i) Sodium, (ii) Hydrogen, (viii) Chlorine, (xii) Iron filings, (xiii) Oxygen.

    Compounds—(iv) Carbon dioxide, (vii) Water, (x) Ammonia, (xvi) Common salt, (xviii) Marble.

    Mixtures—(iii) Brass, (v) Gun powder, (vi) Soda water, (ix) Air, (xi) Stainless Steel, (xiv) Milk, (xv) Blood, (xvii) Glass, (xix) Emulsion, (xx) Wood.

    Question 126
    CBSEENSC9004868

    Why is hydrogen considered an element?

    Solution

    Hydrogen is considered an element because:

    (i) Hydrogen cannot be broken into new substances by ordinary physical or chemical means i.e., by applying heat, light or energy.

    (ii) Hydrogen has one kind of atoms.

    Question 127
    CBSEENSC9004869

    Why is water considered a compound?

    Solution

    Water is considered as compound because:

    (i) Water is composed of two elements:hydrogen and oxygen.

    2H2 +O2 -->2H2O

    (ii) The ratio of hydrogen and oxygen by mass in any sample of pure water is the same.

    (iii) The properties of water are different from its constituent elements: hydrogen and oxygen.

    (iv) Water can be decomposed by chemical means (e.g., electrolytically) into hydrogen and oxygen.

    Question 128
    CBSEENSC9004870

    Name two:(i) gaseous elements, (ii) two liquid elements and (iii) two solid elements.

    Solution

    (i) Gaseous elements: Oxygen and Nitrogen

    (ii) Liquid elements: Mercury and Bromine

    (iii) Solid elements: Gold and Silver

    Question 129
    CBSEENSC9004871

    Name the elements present in each of the following compounds.

    (i) Magnesium carbonate

    (ii) Sulphuric acid

    (iii) Potassium nitrate.

    Solution

    Elements present in each of the compounds are:

    (i) Magnesium carbonate consists of Magnesium, Carbon and Oxygen.

    (ii) Sulphuric acid consists of Hydrogen, Sulphur and Oxygen.

    (iii) Potassium nitrate consists of Potassium, Nitrogen and Oxygen.

    Question 130
    CBSEENSC9004872

    A pure substance on heating leaves behind a white residue and gives out a colourless, odourless gas. Is the given substance an element, compound or mixture ?

    Solution

    (i) The given substance cannot be an element because on heating it breaks up into a gas and a white residue.

    (ii) Since the substance is pure, it cannot be a mixture. Therefore, the given substance is a compound.

    Question 132
    CBSEENSC9004874

    Giving reasons, choose the elements from the following substances:

    (i) Stainless steel, (ii) Diamond, (iii) Graphite,

    (iv) Iron filings, (v) Soda water, (vi) Common salt.

    Solution

    Element are made of only one kind of substance. i.e., 
    (iv) Iron filings is an element as it is made up of only iron (Fe) atoms. (ii) Diamond and (iii) graphite are also composed of carbon atoms.

    Other substances are not elements, since these are made of different substance.

    (i) Stainless steel is a mixture of varying amounts of iron and carbon.

    (v) Soda water is a mixture of carbon dioxide in water.

    (vi) Common salt (NaCl) is a compound of sodium and chlorine.

    Question 133
    CBSEENSC9004875

    Which of the following are elements:

    Sulphur, ice, smoke, manganese, glucose.

    Solution

    Sulphur and manganese are elements.

    Question 134
    CBSEENSC9004876

    Give three reasons to support that brass is a mixture, not a compound.

    Solution

    Composition of different substance in random amount is known as mixture.

    Brass is a mixture because:

    (i) Different samples of brass may have variable composition.

    (ii) Different constituents of brass are separated by fractional melting.

    (iii) The properties of brass depend on the percentage of each pure substance in it.

    Question 135
    CBSEENSC9004877

    Name the types of elements.

    Solution

    Elements are roughly divided into

    (i) Metals  : For example Sodium, Iron, etc.
    (ii) Non-metals : For example Nitrogen, oxygen, etc.
    (iii) Metalloids: For example Boron, silicon, etc.

    Question 136
    CBSEENSC9004878

    Write physical properties of metals?

    Solution

    Physical properties of Metals:

    (i) State: Metals are amorphous solids. Only mercury is an exception which is not a solid. It is a liquid. They retain their shapes under normal conditions.

    (ii) Hardness: Metals are very hard. Sodium is an exception which is not very hard. It can be cut into pieces easily. Hardness varies from metal to metal.

    (iii) Metallic lustre: Most of the metals have a shining surface. They have a metallic lustre. Their shining can be intensified by polishing.

    (iv) Malleability: Metals can be beaten into sheets. For example aluminium is beaten into sheet for different purpose.

    (v) Ductility: Metals are ductile because they can be drawn into wires of desired diameter. metals like copper and aluminium used in electrical wires.

    (vi) Conductivity. Metals are good conductors of heat and electricity.

    (vii) Sonorousity. Metals are sonorous. They produce a ringing sound when struck.

    (viii) High melting point. Metals have very high melting point. gallium is an exception which melt on your hand.

    (ix) Tensile strength. Metals have a tensile strength. They can be stretched up to some degree without fracturing.

    Question 137
    CBSEENSC9004879

    Write some physical properties of non-metals.

    Solution

    Physical Properties of non-metals:

    (i) State: Non-metals are crystalline solids, liquids or gases.

    (ii) Hardness: Non-metals are less hard than metal.

    (iii) Lustre: Non-metals have a dull lustre. They do not reflect light well.

    (iv) They are poor conductors of heat and electricity.

    (v) They are non-malleable and non-ductile. Solid non-metals are brittle. They can be broken into pieces when hammered.

    (vi) They have low boiling and melting points.

    (vii) They have low densities.

    Question 138
    CBSEENSC9004880

    What are metalloids? Give examples.

    Solution

    Metalloids are elements which have properties intermediate between those of metals and non-metals. Examples are silicon, germanium etc.

    Question 139
    CBSEENSC9004881

    Give an account of elements known to us.

    Solution

    There are 115 elements known to us. 92 are natural elements. Rest 23 elements are man-made. 11 elements are gases, 2 elements are liquids and rest are solids at room temperature.

    Question 140
    CBSEENSC9004882

    How many elements are non-metals? Out of these how many are solids, liquids and gases?

    Solution

    There are 22 non-metals:

    10 non-metals are solids. Important solid non-metals are boron, carbon, silicon, phosphorus, arsenic, sulphur and iodine.

    11 non-metals are gases. Some of these are hydrogen, nitrogen, oxygen, fluorine, chlorine. 1 non-metal is a liquid. For example, bromine.

    Question 141
    CBSEENSC9004883

    Give some examples where the property-malleability and ductility of metals are used in our daily life.

    Solution

    Malleability means that metals can be hammered into sheets and foils. For example, aluminium foils are used for wrapping food stuffs, silver foils are used for decorative purposes on sweets and fruits.

    Ductility means that metals can be drawn into wires. Gold and silver wires are used in ornaments. Aluminium and copper wires are used for conduction of electric current.

    Question 142
    CBSEENSC9004884

    Give an example of a metal which:

    (a) is liquid at room temperature (b) can be easily cut with a knife (c) is the best conductor of heat (d) is the poorest conductor of heat

    Solution

    (a) Mercury is liquid at room temperature.

    (b) Sodium can be easily cut with knife.

    (c) Silver is the best conductor.

    (d) Lead is the poorest conductor of heat.

    Question 143
    CBSEENSC9004885

    Mention the metals which are used for the following purposes :

    (a) Domestic utensils and factory equipment

    (b) Jewelleries

    (c) Packaging food materials

    (d) Atomic energy and space research

    (e) Transport of electricity

    Solution

    (a) Iron, aluminium and copper can be used for domestic utensils and factory equipment.

    (b) Gold and silver used for making jewelleries.

    (c) Thin foils of aluminium used for packing food material.

    (d) Titanium, Zirconium and Uranium can be used in atomic energy and  space research.

    (e) Copper and aluminium can be used to conduct electricity.

    Question 144
    CBSEENSC9004886

    From amongst the list, choose the metals and non-metals, state the property on the basis of which you make your choice.

    Chlorine, Aluminium, Sulphur, Sodium, Silver, Helium, Carbon

    Solution

    Chlorine is a non-metal. It is a gas at room temperature and is brittle.

    Aluminium is a metal. It is good conductor of heat and electricity.

    Sulphur is a non-metal. It is brittle and non-malleable or ductile.

    Sodium is a metal. It shows lustre and malleable.

    Silver is a metal. It shows lustre and is the best conductor of heat.

    Helium is a non-metal. It is a gas and does not conduct electricity.

    Carbon is non-metal. It is non-malleable and non-ductile.

    Question 145
    CBSEENSC9004887

    State two properties of sodium in which it shows behaviour which is not expected from its classification as a metal.

    Solution

    (a) Metals are generally hard but sodium is soft.

    (b) Metals are sonorous but sodium is not.

    Question 146
    CBSEENSC9004888

    State two properties of carbon which are not expected from its classification as a non-metal.

    Solution

    (a) Non-metals do not conduct heat and electricity but carbon in the form of graphite is a good conductor of electricity.

    (b) Non-metals are brittle and soft but carbon in the form of diamond is very shining or hard substance.

    Question 147
    CBSEENSC9004889

    Distinguish between metals and non-metals.

    Solution

    Metal

    Non-Metal

    1. Metals are solids at room temperature.

    1. Non-metals may be a gas, a liquid or a solid at room temperature.

    2. Metals have a shining metallic luster. Their shining can be intensified by polishing.

     

    2. Non-metals have a dull non-metallic luster.

     

    3. Metals are hard substances.

    3. Non-metals have hardness of varying category.

     

    4. Metals are malleable.

    4. Non-metals are non-malleable.

    5. Metals are ductile and can be drawn into wires

    5. Non-metals are not ductile and cannot be drawn into wires.

    6. Metals have high specific gravity.

     

    6. Non-metals have low specific gravity.

    7. Metals are good conductors of heat and electricity.

    7. Non-metals are bad conductors of electricity, except graphite which is a good conductor of electricity.

     

    8. Metals form alloys with other metals.

    8. Non-metals do not form alloys with other non-metals.

    Example of metal silver, sodium, iron, copper, gold, etc.

    Example of non-metal Iodine, carbon, oxygen, nitrogen, etc.

    Question 148
    CBSEENSC9004890

    Write characteristics of compounds.

    Solution

    Characteristics of compounds:

    (ii) Compounds are the substances formed by chemical combination of two or more elements.

    (ii) The constituent elements are present in a fixed ratio. For example water is compound and compose in 1:2 such that 1 part of oxygen and 2 part of hydrogen.

    (iii) A chemical reaction takes place during the formation of a compound.

    (iv) Properties of a compound are different to those of its elements.

    (v) Constituent elements cannot be separated by physical processes.

    Question 149
    CBSEENSC9004891

    How can matter be classified on the basis of its chemical constitution?

    Solution

    Matter can be classified on the basis of its chemical constitution. such as element, compound or mixture.

    Question 150
    CBSEENSC9004892

    What process is used to separate husk from corn?

    Solution

    Husk can be separate from corn by winnowing.

    Question 151
    CBSEENSC9004893

    Name three materials that are used as filters.

    Solution

    Cotton, filter paper, sand can be used for Filtration.

    Question 152
    CBSEENSC9004894

    Name the method to separate iron from waste material.

    Solution

    Iron can be separated by magnetic separation.

    Question 153
    CBSEENSC9004895

    What method will you use to separate copper sulphate from its solution?

    Solution

    Copper sulphate can be separated by crystallization.

    Question 154
    CBSEENSC9004896

    Is the rusting of iron in air a physical change or chemical change?

    Solution

    Rusting of iron is a Chemical change.

    Question 159
    CBSEENSC9004901
    Question 164
    CBSEENSC9004906
    Question 165
    CBSEENSC9004907

    Name two elements which are liquids at room temperature.

    Solution

    Mercury and bromine are liquid at room temperature.

    Question 166
    CBSEENSC9004908

    Name two elements that become liquids at a temperature just above 30°C.

    Solution

    Gallium and cesium become liquid at 30°C.

    Question 167
    CBSEENSC9004909

    How many elements are gaseous?

    Solution

    11 elements are gas : H, He, N, O, F, Ne, Cl, Ar, Kr, Xe, Rn.

    Question 168
    CBSEENSC9004910
    Question 169
    CBSEENSC9004911

    Name a non-metal which conducts heat and electricity.

    Solution

    Carbon in the form of graphite can conduct heat and electricity.

    Question 170
    CBSEENSC9004912
    Question 175
    CBSEENSC9004917

    How will you obtain pure copper sulphate from an impure sample?

    Solution

    Pure copper sulphate obtained by crystallisation from an impure sample.

    Question 177
    CBSEENSC9004919

    When you slowly heat a liquid, at some point the temperature of the liquid becomes constant. What is this point?

    Solution

    When liquid is slowly heated, at some point the temperature of the liquid becomes constant because the liquid changes its state to gas state and it is known as boiling point of the liquid.

    Question 178
    CBSEENSC9004920

    What should be the minimum difference in temperature between the boiling points of two miscible liquids that these can be separated by simple distillation method?

    Solution
    The minimum difference in the boiling point of two miscible liquid is btween 20K to 25K.
    Question 179
    CBSEENSC9004921

    What is the process called in which pigments of natural colours can be separated?

    Solution

    The pigments of natural colours can be separated by Chromatography.

    Question 180
    CBSEENSC9004922

    Which technique is used in dairies and home to separate butter from cream?

    Solution

    Butter is separated from cream by centrifugation.

    Question 181
    CBSEENSC9004923
    Question 182
    CBSEENSC9004924

    Take the odd substance out :

    Camphor, naphthalene, sodium chloride, ammonium chloride

    Solution

    Sodium chloride is odd substance of the given list because they are separted by sublimation but sodium chloride is not separted by this method.

    Question 186
    CBSEENSC9004928
    Question 187
    CBSEENSC9004929
    Question 189
    CBSEENSC9004931
    Question 190
    CBSEENSC9004932
    Question 192
    CBSEENSC9004934
    Question 193
    CBSEENSC9004935
    Question 196
    CBSEENSC9004938

    Impure salts can be purified by the process of __________.

    Solution

    crystallisation

    Impure salts can be purified by the process of..................

    Question 197
    CBSEENSC9004939
    Question 199
    CBSEENSC9004941
    Question 200
    CBSEENSC9004942
    Question 201
    CBSEENSC9004943
    Question 202
    CBSEENSC9004944
    Question 203
    CBSEENSC9004945

    Most elements are ________.

    Solution

    solids

    ,

    Most elements are..................

    Question 205
    CBSEENSC9004947
    Question 209
    CBSEENSC9004951
    Question 210
    CBSEENSC9004952
    Question 211
    CBSEENSC9004953
    Question 213
    CBSEENSC9004955

    Which one of the following is a chemical change:

    • evaporation of spirit

    • freezing of water

       

    • heating of copper and sulphur

    • mixing of H2 and O2.

    Solution

    C.

    heating of copper and sulphur

    Which one of the following is a chemical change :

    Question 214
    CBSEENSC9004956
    Question 215
    CBSEENSC9004957
    Question 222
    CBSEENSC9004964
    Question 227
    CBSEENSC9005851

    An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

    Solution

    Yes, even when the net external force is zero, an object may travel with non-zero velocity. 

    A rain drop falls downwards with a constant velocity. The net force on the drop is zero because the weight of the drop is balanced by the upthrust and the viscosity of air. 

    Question 228
    CBSEENSC9005852

    When a carpet is beaten with a stick, dust comes out. Explain why?

    Solution

    When a carpet is beaten with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet. 

    Question 229
    CBSEENSC9005853

    Why it is advised to tie any luggage kept on the roof of a bus with a rope?

    Solution

    When a bus takes sharp turn on the road, a sudden jerk is felt. As a result of the sudden jerk, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. In order to avoid this, the luggage is tied with a rope on the roof. 

    Question 233
    CBSEENSC9005857

    A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40,000 N. If the track offers a friction force of 5,000 N, then calculate:

    (a) the net acceleration force;

    (b) the acceleration of the train; and

    (c) the force of wagon 1 on wagon 2.

    Solution

    Total mass of engine and 5 wagons, m = 8,000 + 5 x 2,000 = 18,000 kg
    (a) The net accelerating force is given by, 
    F = Engine Force  - Friction Force
       = 40,000 - 5,000
       = 35000 N. 
    (b) The acceleration of the train is given by, 
    space straight a space equals space straight F over straight m space equals space 35000 over 18000 space equals space 35 over 18 space equals space 1.94 space ms to the power of negative 2 end exponent. space space
    (c) The force of wagon 1 on wagon 2
     = Net accelerating force - mass of wagon x acceleration 
    equals space 35000 minus space 2000 space cross times space 35 over 18
equals space 35000 minus 3888.8

equals space 31 comma 111.2 space straight N. space

     

     

     

    Question 234
    CBSEENSC9005858

    An automobile vehicle has a mass of 1,500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms–2?

    Solution

    Here,
    Mass of the vehicle, m = 1500 kg
    Acceleration, a = -1.7 ms-2
    Force, F = ma = 1500 x (-1.7) = - 2,550N

    The force between the vehicle and the road is 2,550 N.
    The negative sign implies thatforce is acting in a direction opposite to the direction of the vehicle.

    Question 236
    CBSEENSC9005860

    Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at constant velocity. What is the friction force that will be exerted on the cabinet?

    Solution

    The cabinet will move with constant velocity only when the net force on it zero.

    Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet. 

    Question 237
    CBSEENSC9005861

    Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms–1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

    Solution

    Here,  
    Mass of two objects are equal but their motion is in opposite direction. 
    That is,
    m1 = m2 = 1.5 kg
    Initial velocity, u1 = 2.5 ms-1, u2 = -2.5 ms-1.
    Let v be the velocity of the combined object after the collision.
    By conservation momentum,
    Total momenta after collision = Total momenta before collision
    i.e., (m1 + m2) v = m1u1 + m2u2

    rightwards double arrow(1.5 + 1.5)v = 1.5 x 2.5 + 1.5 x (-2.5)
    rightwards double arrow 3.0 v =  0
     So, v = 0 ms-1
    Hence, the objects will have zero velocity after collision. 

    Question 238
    CBSEENSC9005862

    According to third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

    Solution

    Action and reaction can never cancel each other because they always acts on two different bodies. 
    The force of friction between the truck's tyres and the road is very large. Also, the weight of the truck is massive. So, when we push the truch it does not move.



    Question 239
    CBSEENSC9005863

    A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

    Solution

    Here,    
    Mass of the hockey ball, m = 200 g = 0.2 kg
    Initial velocity of the ball, u = 10 ms-1
    Final velocity, v = -5 ms-1
    So, change in momentum = m(v - u)
                                            = 0.2(-5 - 10)
                                            = -3 kg ms-1.

    Question 240
    CBSEENSC9005864

    A bullet of mass 10 g travelling horizontally with a velocity of 150 ms–1 strikes a stationary wooden block and comes to at rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

    Solution

    Here,
    Mass of the bullet, m = 10 g = 0.01kg
    Velocity with which the bullet is fired, u = 150 ms-1,
    Final velocity, v = 0 m/s
    Time taken for the bullet to strike the object, t = 0.03 s 
    Now, using the formula,
            straight a space equals space fraction numerator straight v minus straight u over denominator straight t end fraction space equals space fraction numerator 0 minus 150 over denominator 0.03 end fraction space equals space minus 5 comma 000 space m s to the power of negative 2 end exponent 
    Acceleration is negative. 
    The distance of penetration of the bullet into the block is given by the second equation of motion, 

    Question 241
    CBSEENSC9005865

    An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1 collides with, and sticks to, a stationary wooden block of mass 5 kg then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also calculate the velocity of the combined object.

    Solution

    Here,  
    Mass of the object, m1 = 1 kg
    Velocity of the object, u1 = 10ms-1
    Mass of the wooden block, m2 =5 kg
    Velocity after collision, u2 =0 m/s
    Let v be the velocity of the combined object after the collision
    That is, 
    Total momentum just before the impact is, 
      m1u1 + m2u2 = 1 x 10 + 5 x 0 = 10 kg ms-1 
    Total momentum just after the impact is, 
         (m1 + m2) v = (1+5)v = 6v kg ms-1 

    Using the law of conservation of momentum, 
                  6v = 10 
    rightwards double arrow            v = 10/6 = 5/3 = ms-1 
     
    Therefore, 
    Total momentum just after the impact= 6 x 5/3
                                                              = 10 ms-1.

    Question 245
    CBSEENSC9005869

    The following is the distance-time table of an object in motion

    Time in seconds

    Distance in meters

    0

    0

    1

    1

    2

    8

    3

    27

    4

    64

    5

    125

    6

    216

    7

    343

    (a) What conclusion can you draw about the acceleration? Is it constant? Increasing? Decreasing? Or Zero? 

    (b) What do you infer about the forces acting on the object?

    Solution

    (a) The above table shows that the motion is accelerated and acceleration is increasing uniformly with time.
    (b) As the acceleration is increasing uniformly, the force acting on the body is also increasing uniformly with time.

    Question 246
    CBSEENSC9005870

    Two persons manage to push a motorcar of mass 1,200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms–2. With what force does each person push the motorcar? (Hint : All persons push the motorcar with same muscular effort.)

    Solution

    Here,  
    Mass of the motor-car, m = 1200 kg
    Acceleration, a = 0.2 s-1. 
    Clearly, the unbalanced force is exerted when the third person applies the force.
    Therefore,
    Force applied by the third person = ma
                           
                                                      = 1200 x 0.2
                                                      = 240 N
    As all the three persons push the motorcar with the same muscular effort, force exerted by each person = 240 N. 

    Question 248
    CBSEENSC9005872

    A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to halt after that. If the collision lasts for 1 s:

    (a) Which vehicle experiences the greater force of impact?

    (b) Which vehicle experiences the greater change in momentum?

    (c) Which vehicle experiences the greater acceleration?

    (d) Why is the car likely to suffer more damage than the truck?

    Solution

    (a) Both vehicles experience equal forces of action and reaction because both the objects are moving.

    (b) Since, both truck and car are moving with the same magnitude of velocity v, momentum change for both vehicles is same. 

    (c) Mass and acceleration are inversely proportional to each other. Therefore, car experiences greater acceleration due to its smaller mass.

    (d) Due to its smaller mass or opposition to the force exerted on it, the car is likely to suffer more damage than the truck. 

    Question 254
    CBSEENSC9005878

    SI unit of force is

     

    • kg-m/s

    • Newton

       

    • dyne

    • kg-wt

    Solution

    B.

    Newton

     

    Question 264
    CBSEENSC9005888

    A bullet of mass ‘a’ and velocity ‘b’ is fired into a large block of mass ‘c’. The final velocity of the system is


      Solution

      Here,
      Mass of the bullet, m = 10 g = 0.01 kg
      Initial velocity, u = 150 ms-1
      Final velocity, v =0 m/s
      Time taken for the bullet to reach the target, t = 0.03 s
      The distance of penetration of the bullet into the block, 
      straight s space equals space ut plus 1 half at squared space
space space space equals space 150 space cross times 0.03 space plus space 1 half cross times left parenthesis negative 5 comma 000 right parenthesis space cross times space left parenthesis 0.03 right parenthesis squared
space space space equals 4.5 space minus space 2.25 space equals space 2.25 space straight m. space space 

      Question 265
      CBSEENSC9005889
      Question 266
      CBSEENSC9005890
      Question 273
      CBSEENSC9005897

      How can we change the state of motion of objects.


      Solution

      We can bring objects into motion by pushing, hitting and pulling them, as shown in the Fig.
      When we apply an unbalanced force on the particle, the body is set into motion. 




      Question 274
      CBSEENSC9005898

      Define the term force.

      Solution

      Force is defined as a push or a pull which tends to change the state of rest or uniform motion or direction of motion of a body.

      For instance, force exerted by the engine makes the train move from its actual position of rest while the force exerted by the brakes slows down or stops the moving train.

      Question 275
      CBSEENSC9005899

      State the various effects produced by a force.

      Solution

      The various effects produced by force are :

      (i) An applied force can change the speed of the object making it move faster or slower. 

      (ii) Force can change the direction of motion of the object.

      (iii) Force can change the shape of the object. 

      Question 276
      CBSEENSC9005900

      Give an illustration to show that force can change the speed of an object.

      Solution
      Examples for force can change the speed of an object:

      1. Let us consider the case of a horse which exerts force on a cart, pulls it from rest. Subsequently, exerting larger force, the horse makes the cart move with larger speed.

      2. A player can stop a moving ball by exerting force on the ball in the opposite direction of its motion.

       
      Question 277
      CBSEENSC9005901

      Give an illustration to show that a force can change the direction of motion of an object.

      Solution

      An applied force can change the direction of motion of an object. This can be illustrated from the following example given below. 

      When a ball at the end of a string is rotated in a circular motion, the hand exerts a force radially towards the centre of the circular path. This force continuously changes the direction of motion of the ball, thus enabling the ball to move along a circular path.

      Question 278
      CBSEENSC9005902

      Give an illustration to show that a force can change the shape of an object.

      Solution

      Consider a spring attached to a rigid support at its one end, as shown in Fig.(a).

      Fig.(a): A spring expands on application of force; (b) A spherical rubber ball becomes oblong as we apply force on it.

      If we pull the free end of the spring, it gets elongated. Thus, on appying a force, a spring expands.
      Similarly, if we hold a rubber ball between our palms and push the two palms against each other, we find that the ball is no longer round but oblong. See fig. (b). The oblongation of the ball is due to the application of the force.

      Question 279
      CBSEENSC9005903

      Define resultant force.

      Solution

      The forces that are acting simultaneously on a body producing the same effect on a body as all the forces together produce is called as the resultant force. 

      Question 280
      CBSEENSC9005904

      What are balanced forces? Give examples.

      Solution

      When the resultant of several forces acting on a body is zero, the forces are said to be ‘balanced forces’.
      Balanced forces do not change the speed but changes the shape of the object. 

      Examples of balanced forces:
      (i) If a bag full of books is held in hand steadily at a certain height from the ground, the resultant force on the bag is zero and so its position does not change.
      (ii) Consider a wooden block lying on a table as shown in Fig.(a). 

      It has strings tied to its two opposite faces. /the wooden block move towards left, if the string is pulled at point A. If we pull at point B, it begins to move towards right.
      But, if we pull from both, the sides with equal effort (and hence equal force), the block does not move.
      The two forces have now balanced each other.
      iii) In a tug-of-war, the rope does not move in any direction if the two teams pull the rope with equal effort.
      The forces exerted by the two teams are equal and opposite and so get balanced. 
      This is illustrated as shown in fig. (b).


      Question 281
      CBSEENSC9005905

      Give examples to show that balanced forces change the shape of the object on which they are applied.

      Solution

      When balanced force is applied on any object, it changes the shape of the object. 
      From the examples, we have 


      Fig (a). Balanced force cause change in shape. 

      (i) Consider a spring as shown in Fig (a).
      If the string is pulled at it's two ends, the spring gets elongated. And if we push the spring at its two ends, it gets compressed. In both cases the two forces being equal and opposite, balance each other but change the shape of the spring.

      (ii) When a rubber ball or an inflated balloon is pressed between the two palms, its shape changes. The forces exerted by the two palms are equal and opposite and thus balance each other. Hence, the shape of the ball changes. 

      Question 282
      CBSEENSC9005906

      What are unbalanced forces. Give examples.

      Solution

      When the resultant of several forces acting on a body is not zero, the forces are said to be unbalanced forces. 
      A change in the state of rest or uniform motion of a body is produced by unbalanced forces. 

      Below are given some examples of unbalanced forces:

      (i) In a tug-of-war, when one of the two teams pulls the rope with a larger force, they are able to pull the weaker team towards it. Here the two forces are not balanced. Therefore, it results in the motion of the weaker team towards the larger force along the rope. 

      (ii) When we stop peddaling a bicycle, it begins to slow down. Enough force is to be applied to overcome the frictional force between the road and the tyres of the bicycle.

      Therefore, we can say that objects continue to be in motion or rest, with the same velocity unless acted upon by unbalanced forces. 

      Question 283
      CBSEENSC9005907

      What were the views of Greek philosopher Aristotle about the state of motion of objects?

      Solution

      The Greek philosopher Aristotle (384–324 B.C.) asserted that the natural state of a body is the state of rest. Every single body in motion tends to slow down and comes to rest. An external force is necessary to maintain its motion.

      For example, a cart on a road has to be constantly pushed to keep it in motion. A single push will not take it far.

      With such observations, Aristotle concluded that an external force is necessary to keep a body moving.

      However, these views were proved wrong by Galileo Galilei (1564–1642) later on.

      Question 284
      CBSEENSC9005908

      Describe Galileo’s experiments with inclined planes and state the conclusion.

      Solution

      Galileo deduced that objects move with constant speed when no force acts on them, by observing the motion of objects on inclined planes.

      i) Galileo studied the motion of marbles on an inclined plane, in his first experiment.
      He observed that when a marble rolls down an inclined plane, its velocity increases, as shown in Fig.9.6 (a).
      Here the marble falls under the unbalanced force of gravity.
      The velocity of the marble decreases when it rolls up the inclined plane (against the force of gravity), as shown in Fig.9.6(b).
      From these observations, Galileo argued that the velocity of a marble rolling on flat horizontal surface should remain constant.

      Fig. Motion of a marble (a) down the inclined plane and (b) up the inclined plane.

      The above experiments suggest that an unbalanced force (external force) is required to change the motion of an object while no unbalanced force is needed to keep an object moving with a constant velocity.
      From the observations above, 
      Galileo concluded the following law of inertia:

      1. A body moving with a certain speed along a straight line path will continue to move with the same speed along the same straight line path, unless acted upon by an external force. 

      Question 285
      CBSEENSC9005909

      State Newton’s three laws of motion.

      Solution

      The three Newton's laws of motion are: 

      First Law: An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.

      Second Law: The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

      Third Law: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction.

       

      Question 286
      CBSEENSC9005910

      State Newton’s first law of motion. Explain it with the help of suitable examples. 

      Solution

      Newton's first law of motion states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. 

      (i) First part, says that a body at rest continues in its state of rest. For example, when a bus suddenly starts moving forward, the person falls backward. This happens because, as the bus moves the lower part of his body begins to move along with the bus while the upper part of his body continues to remain at rest due to inertia. Hence, a person falls backward when the bus starts.

      (ii) Second part, says that a body in uniform motion continues moving in straight line path with a uniform speed. For example, when a moving bus suddenly stops, a person sitting inside the bus falls forward. As the bus stops, the lower part of his body comes to rest along with the bus while upper part of his body continues to remain in motion due to inertia and so he falls forward.

      (iii) Third part, says that a body moving with a uniform speed in a straight line cannot change itself its direction of motion. For example, when a bus takes a sharp turn, a person sitting in the bus gets a force acting away from the centre of the curved path due to his tendency to move in the original direction. He has to hold on to a support to prevent himself from swaying away in the turning bus. 

      Question 287
      CBSEENSC9005911

      Define the term inertia, what are the different types of inertia?

      Solution

      Inertia of a body is the natural tendency of a body to resist any change in its state of rest or of uniform motion along a straight line.

      For example, a book lying on a table will continue to remain there until an external force is applied on it to remove or displace it from that position.

      There are three types of inertia, namely: 

      (i) inertia of rest,
      (ii) inertia of motion and
      (iii) inertia of direction. 

      Question 288
      CBSEENSC9005912

      What is relationship between mass and inertia? Give the SI unit of mass and inertia.

      Solution

      Mass and inertia are correlated. Mass of a body is a measure of its inertia. Both of them are directly related. More massive is the body, larger is the inertia or opposition offered by a body to change its state of motion. 

      SI unit of mass is kilogram (kg)
      SI unit of inertia is kilogram (kg)


      Question 290
      CBSEENSC9005914

      Define inertia of rest and give some examples.

      Solution

      The tendency of a body to remain in its position of rest unless acted by an external force is called ‘inertia of rest’. 

      Examples:

      (i) Dust is removed from a hanging carpet by beating it with a stick. As the carpet is beaten, it suddenly moves forward while the dust particles tend to remain at rest due to inertia and so fall off.

      (ii) When we shake a tree, the fruits or dry leaves fall down from the tree due to inertia of rest.

      Question 291
      CBSEENSC9005915

      Define the term inertia with respect to motion. Give some examples.

      Solution

      The tendency of a body to remain in its state of uniform motion in a straight line, unless acted upon by an external force is called ‘inertia of motion’.

      Examples:

      (i) An athelete runs for certain distance before taking a jump so that inertia of motion may help to take a longer jump. 

      (ii) If a horse running fast suddenly stops, the rider is thrown forward if he is not firmly seated. 

      Question 292
      CBSEENSC9005916

      Define inertia of direction. Give some examples of inertia of direction.

      Solution

      The inability of a body to change by itself its direction of motion is called Inertia of direction.

      Examples of inertia of direction:

      i) A dog chases a hare, the hare runs along a zig-zag path. It becomes more difficult for the dog to catch the hare because dog has more inertia of direction than that of the hare. 

      (ii) Consider a stone being rotated at the end of a string in a circle. If the string is suddenly released, the stone flies off tangentially to its circular path.

      (iii) As a bicycle moves, the water drops sticking to its tyre start leaving it tangentially.

      Question 293
      CBSEENSC9005917

      Why is the Newton’s first law of motion also called ‘law of inertia’?

      Solution

      Newton’s first law of motion states that, a body by itself cannot change its state of rest or of uniform motion. This property of a body is also called inertia.

      That is why the Newton’s first law of motion is also called ‘law of inertia’.

      Question 294
      CBSEENSC9005918

      Explain the law of inertia.

      Solution

      According to the law of inertia, a body continues to be in its state of rest or of uniform motion unless an external force acts on it to change that state.

      For example, a table lying on the floor will remain in its position unless an external force acts on it to displace it from that position.

      Question 295
      CBSEENSC9005919

      Explain how does Newton’s first law of motion define force?

      Solution

      According to Newton’s first law of motion, a body at rest or in uniform motion will remain at rest or in uniform motion unless acted upon by an external force.

      Therefore, force is  defined as an external agency which changes or tends to change the state of rest or of uniform motion of an object. 

      Question 296
      CBSEENSC9005920
      Question 297
      CBSEENSC9005921

      In the following example, try to identify the number of times the velocity of ball changes:

      “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards the player of his own team”.

      Solution
      In the given situation, velocity of ball changes three times. 

      First time, when the football player of one team kicks the ball, velocity changes. 
      Second time, the velocity changes when another player of same team kicks the football.
      Third time, the velocity changes when the goalkeeper of the opposite team kicks the football.

       
      Question 298
      CBSEENSC9005922

      Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

      Solution

      Initially, the tree and the leaves are in a state of rest. When the branches are shaken vigorously, they come into motion while, the leaves tend to remain in a state of rest due to inertia of rest. Therefore, the leaves get detached from the branches and fall down. 


      Question 299
      CBSEENSC9005923

      Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

      Solution

      When a moving bus brakes to a stop, the lower part of our body which is in contact with the bus comes to rest while the upper part of our body tends to keep moving due to inertia of motion. Hence, one tends to fall forward. When the bus accelerates from rest, the lower part of our body comes into motion alongwith the bus while the upper part of body tends to remain at rest due to inertia of rest. Therefore, we fall backwards. 

      Question 300
      CBSEENSC9005924

      Write the name of physical quantity which is the combined effect of mass and velocity.

      Solution

      Momentum is the combined effect of mass and velocity. 

      Momentum = mass * velocity

      Question 301
      CBSEENSC9005925

      Taking suitable examples, develop the concept of momentum.

      Solution

      The concept of momentum can be developed by considering the following example. Consider a tennis ball which hits a player. The tennis ball does not harm her. But, at the same time when a fast moving cricket ball hits a spectator, it may hurt him. 
      Similarly, a truck at rest does not harm anyone when parked along a roadside. But a truck moving even at a low speed of 5 ms–1 may kill a person standing in its path.

      From the above examples, we can say that the impact produced by a moving body or the effect of motion depends both on both mass and velocity. 
      The product of mass and velocity of a body is called its momentum. 


      Question 302
      CBSEENSC9005926

      Define momentum of a body. Is it a scalar or vector quantity?

      Solution

      Momentum of a body is the quantity of motion possessed by the body. Momentum is equal to the product of the mass and velocity of the body and is denoted by p.
      Momentum = Mass x Velocity
      rightwards double arrow   p = mv

      Momentum is a vector quantity and its direction is the same as the direction of velocity of the object.

      Question 303
      CBSEENSC9005927

      What is the SI unit of momentum?

      Solution

      The SI unit of momentum is kilogram metre per second (kg ms–1).

      Question 304
      CBSEENSC9005928

      State and explain Newton’s second law of motion.

      Solution

      Newton's second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. 

      The law can be divided into two parts:

      (i) Rate of change of momentum is directly proportional to the applied force. Larger the force acting on a body, greater is the change in its momentum. Change in momentum is equal to the product of mass and the change in velocity. Mass of the body remains constant. Therefore, rate of change of momentum is directly proportional to the rate of change of velocity i.e., acceleration. 
      So, Force is directly proportional to mass and acceleration.

      (ii) The change of momentum occurs in the direction of the force.
      If a body is at rest, a force will set it in motion. If a body is moving with a certain velocity, a force will increase or decrease this velocity accordingly as the force acts in its same or opposite direction.

      Question 305
      CBSEENSC9005929

      Explain how Newton’s second law can be used to define the unit of force. Define the SI unit of force.

      Solution

      Measurement of force from Newton’s second law.
      Consider a force F acts on a body of mass m and changes its velocity from u to v in t seconds.
      Then
      Initial momentum of the body, p1 = mu
      Final momentum of the body, p2 = mv
      Change of momentum = p2 - p1
                                         = mv - mu
                                         = m(v - u)
       Time taken = t
      therefore Rate of change of momentum = 
      fraction numerator Change space of space momentum over denominator Time space taken end fraction space equals space fraction numerator straight m left parenthesis straight v minus straight u right parenthesis over denominator straight t 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 equals space ma space open square brackets because space space fraction numerator straight v minus straight u over denominator straight t end fraction space equals space straight a close square brackets

      According to Newton’s second law, the rate of change of momentum is directly proportional to the applied force.
      So, 
      space space space space space space straight F space proportional to space ma space
rightwards double arrow space straight F space equals space kma
      Here,
       k is a constant.
      The unit of force is so chosen that k is equal to one.
      If m = 1, a = 1 and F = 1, then k = 1. 
      Therefore, F = ma  
      rightwards double arrow    Force = Mass x Acceleration
      SI unit of force is Newton.
      One newton is that force which produces an acceleration of 1 m/s2 in a body of mass 1 kg.  
      1 newton = 1kg x  1 m/s or  1N = 1kg m/s2

      Thus, force is measured using the second law of motion.
      If the mass and acceleration of a body are known, we can determine the force acting on it.

      Question 306
      CBSEENSC9005930

      Give some examples to illustrate Newton’s second law of motion.

      Solution

      Examples of Newton's second law of motion:
      (i) We have seen a cricket player lowers his hands while catching a ball. By lowering the hands, he increases the time of catching the ball. As a result, the rate of change of momentum decreases and by Newton’s second law, the force exerted on his hands is less. So he is less likely to hurt.

      (ii) Chinawares are wrapped in straw paper before packing. The straw paper between the chinawares increases the time of experiencing the jerk during transportation. Hence, they strike against each other with a less force and are less likely to be damaged.

      (iii) When we jump on a muddy floor, than on a hard floor we are less hurt. On a muddy floor, the floor is carried in the direction of the jump and the time interval for which force acts is increased. This decreases the rate of change of momentum and hence the force of reaction. Thus, the person is hurt less. 

      Question 307
      CBSEENSC9005931

      Show that Newton’s first law of motion is contained in the second law.

      Solution

      First law of motion as a special case of second law.
      Newton’s first law of motion can be deduced from second law.
      space According space to space second space law comma space space space straight F space equals space ma space equals space straight m open parentheses fraction numerator straight v minus straight u over denominator straight t end fraction close parentheses
space or space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space Ft space equals space straight m left parenthesis straight v minus straight u right parenthesis
      That is,
      when F = 0, v = u for whatever time t is taken because m ≠ 0.
      This means that the object will continue moving with uniform velocity u throughout the time t. If u is zero then v will also be zero. That is the object will remain at rest. This is the same statement as Newton's law of motion. 
      Hence first law of motion is contained in the second law.

      Question 308
      CBSEENSC9005932

      What is the momentum of a man of mass 75 kg, when he walks with a uniform velocity of 2 m/s?

      Solution

      Here,
      Mass of man, m = 75 kg
      Man is moving with a uniform velocity, v = 2m/s
      Momentum,  p = mv
                             = 75 kg x 2m/s
                             = 150 kg m/s. 

      Question 309
      CBSEENSC9005933

      By how much does the momentum of a body of mass 5 kg change when its speed

      (i) decreases from 20 m/s to 0.20 m/s; and

      (ii) increases from 30 m/s to 40 m/s?

      Solution

      (i) Here,
      Mass of the body, m = 5 kg
      Initial velocity, u = 20 m/s
      Final velocity, v = 0.20 m/s
      Change in momentum = m(v - u)
                                         = 5(0.20 - 20) kg m/s
                                         = -99 kg m/s
      The negative sign shows that the momentum decreases. 
      (ii) Here, 
      Mass, m = 5 kg
      Initial velocity, u = 30 m/s
      Final velocity, v = 40 m/s 
      Change in momentum = m(v - u)
                                         =5(40-30)
                                         = 50 kg m/s
      Here, the momentum of the body decreases, when it's speed increases. 

      Question 310
      CBSEENSC9005934

      A man throws a ball of mass 0.5 kg vertically upwards with a velocity of 10 m/s. What will be its initial momentum? What would be its momentum at the highest point of its reach?

      Solution

      Mass of the ball, m = 0.5 kg
      Initial velocity, u = 10 m/s
      Initial momentum  = mu = 0.5 x 10 = 5 kg m/s
      At the highest point, velocity of ball is zero.
      So, momentum of the ball 0.5 x 0 = 0. 

      Question 311
      CBSEENSC9005935

      Calculate the force required to produce an acceleration of 5 ms 2 in a body of mass 6 kg.

      Solution

      Here,
      Mass of the body, m = 6 kg
      Rate of acceleration, a = 5 ms-2
      Therefore,
      Required force,  F = ma
                                 = 6 x 5  
                                 = 30 N.  

      Question 312
      CBSEENSC9005936

      Calculate the mass of the body, when a force of 525 N, produces an acceleration of 3.5 mJs2.

      Solution

      We have, 
      Force of the body, F = 525 N
      Acceleration, a = 3.5 m/s2
      Therefore,
      Mass of the body, m = F/a
                                     = 525/3.5
                                     = 150 Kg.

      Question 313
      CBSEENSC9005937

      What would be the force required to produce an acceleration of 2 m/s2 on a body of mass 12 kg? What would be its acceleration if the force was doubled?

      Solution

      Case I:
      Acceleration, a = 2 m/s2 
      Mass of the body, m = 12 kg
      Therefore,
      Force ,   F = ma = 12 x 2 = 24 N. 
      Case 2:
      Mas of the body, m = 12 kg
      When force is doubled on the body,
      Force acting, F = 48 N
      Therefore,
      Acceleration,  a = F/m = 48/12 = 4m/s2  

      Question 314
      CBSEENSC9005938

      Which would require a greater force—accelerating a 2 kg mass at 5 ms –2 or a 4 kg mass at 2 ms–2.

      Solution

      Here,  
      Mass of first body, m1 = 2 kg
      Body is accelerating at a rate of, a1 = 5ms-2
      Mass of second body, m2 = 4kg
      Rate of acceleration, a2 = 2ms-2
      Force acting on the first body,
      F1 = m1a1 = 2 x 5  = 10 N
      Force acting on the second body, 
      F2 = m2a2 = 4 x 2 = 8 N
      Clearly,   F1 > F2

      Thus, the body which is accelerating at a rate of 5 ms-2  would require a greater force. 

      Question 315
      CBSEENSC9005939

      What will be acceleration of a body of mass 5 kg if a force of 200 N is applied on it?

      Solution

      Here,
      Mass of the body, m = 5 kg
      Force acting on the body, F = 200 N
      Acceleration of the body is given by, a = F/m
                                                               = 200/5
                                                               = 40 m/s2.  

      Question 316
      CBSEENSC9005940

      A bullet of mass 10 g is fired with a rifle. The bullet takes 0.003 s to move through its barrel and leaves it with a velocity of 300 m/s. What is the force exerted on the bullet by the rifle?

      Solution

      Here,
      Mass of the bullet, m = 10 g = 0.010 kg
      Initial velocity of the bullet, u = 0
      Velocity of bullet attained, v = 300 m/s
      time taken by the bullet to move through the barrel, t = 0.003 s
      Therefore, 
      Force is given by,
        straight F equals straight m open parentheses fraction numerator straight v minus straight u over denominator straight t end fraction close parentheses space

space space equals space 0.010 open parentheses fraction numerator 300 minus 0 over denominator 0.003 end fraction close parentheses

space space space equals space 1000 space straight N.

      Question 319
      CBSEENSC9005943

      A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers in the motorcar is 1000 kg.

      Solution

      Here,
      Initial velocity of motorcar, u = 108 km h-1 
                                                  equals space 108 space cross times space 5 over 18 space
equals space 30 space ms to the power of negative 1 end exponent
      The brakes are aplied and motorcar comes to rest.
      So,
      Final velocity= 0
      Time taken for the motorcar to stop, t = 4s
      MAss of the motorcar, m = 1000 kg
      therefore  Force, 
      straight F equals space straight m open parentheses fraction numerator straight v minus straight u over denominator straight t end fraction close parentheses space equals space 1000 space open parentheses fraction numerator 0 minus 30 over denominator 4 end fraction close parentheses space equals space minus 750 space straight N.

      The negative sign shows that the force exerted by the brakes is opposite to the direction of motion of motorcar. 

      Question 320
      CBSEENSC9005944

      A force of 5 N gives a mass m1 and acceleration of 10 ms–2 and a mass m2 and acceleration of 20 ms–2. What acceleration would it give if both the masses were tied together?

      Solution

      Given, 
      Force acting on the body, F = 5 N 
      Acceleration, a  = 10 ms-2
      So, mass of first body, m1straight F over straight a subscript 1 space equals space 5 over 10 equals space 0.5 space k g
      Rate of acceleartion of body = 20 ms-2 
      MAss of second body, m2straight F over straight a subscript 2 =  5 over 20 space equals space 0.25 space k g
      When the two masses are tied together, the total mass is, 
      m = m1+m2 = 0.50 +0.25 = 0.75 kg 
      Acceleration produced in the combined mass by the force of 5 N is, 
      a =  straight F over straight m space equals space fraction numerator 5 over denominator 0.75 end fraction space equals space 6.67 space m s to the power of negative 2 end exponent 

      Question 321
      CBSEENSC9005945

      A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force applied by the player to stop the ball?

      Solution

      Given, 
      Mass of the cricket ball, m = 70 g
      Velocity of the ball, u = 0.5 m/s 
      Final velocity of the ball, v = 0 m/s
      Time taken, t = 0.5 secs 
      Force acting on the ball, F =  straight m space open parentheses fraction numerator straight v minus straight u over denominator straight t end fraction close parentheses space equals space 0.070 open parentheses fraction numerator 0 minus 0.5 over denominator 0.5 end fraction close parentheses space equals space minus 0.07 space straight N
      Negative sign shows the retarding nature of the force. 

      Question 322
      CBSEENSC9005946

      A car of mass 1000 kg moving with a velocity of 45 km/h collides with a tree and comes to a stop in 5 s. What will be the force exerted by the car on the tree?

      Solution

      Here, 
      Mass of car, m = 1000 kg
      Initial velocity, u = 45 km/hr =  fraction numerator 45 cross times 1000 over denominator 3600 end fraction space equals space 25 over 2m/s
      Final velocity = 0 m/s
      Time taken for car to come to a halt = 5 s
      Therefore, force is given by
      F =  m open parentheses fraction numerator v minus u over denominator t end fraction close parentheses space equals space 1000 open parentheses fraction numerator 0 minus begin display style bevelled 25 over 2 end style over denominator 5 end fraction close parentheses space equals space minus 2500 space N
      The negative sign shows the retarding nature of force. 

      Question 323
      CBSEENSC9005947
      Question 324
      CBSEENSC9005948

      For how long should a force of 100 N act on a body of mass 20 kg so that it acauires a velocity of 100 m/s?

      Solution

      Given,
      Force, F = 100 N
      Mass of the body, m = 20 kg
      therefore Acceleration is given by, 
        Acceleration comma space straight a space equals space straight F over straight m space equals space 100 over 20 space equals space 5 straight m divided by straight s squared
Now comma

I n i t i a l space v e l o c i t y comma space straight u space equals space 0

F i n a l space v e l o c i t y comma space straight v space equals space 100 space straight m divided by straight s

Acceleration comma space straight a space equals space 5 straight m divided by straight s squared

therefore space Time space for space which space the space force space is space applied comma space straight t space equals space fraction numerator straight v minus straight u over denominator 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 space space space space space space space space space space space space space space space space space space space space space space space space space 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 100 minus 0 over denominator 5 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 space space space space space space space space space space space space space space space space space space space space equals space 20 straight s space space space space space space

      Question 328
      CBSEENSC9005952

      A driver accelerates his car first at the rate of 1.8 m/s2 and then at the rate of 1.2 m/s2. Calculate the ratio of the force exerted by the engine in the two cases.

      Solution

      Initial rate of acceleration = 1.2 m/s2
      Rate of acceleration = 1.2 m/s2
      Let m be the mass of the car. 
      Then,
      Ratio of the force exerted by the engine is, 
       space space straight F subscript 1 over straight F subscript 2 space equals space ma subscript 1 over ma subscript 2 space equals space fraction numerator 1.8 over denominator 1.2 end fraction space equals space 18 over 12 space equals 3 over 2.


      Therefore, the ratio is 3:2. 

      Question 329
      CBSEENSC9005953

      A man in a circus show jumps from a height of 10 m and is caught by a net spread below him. The net sags down 2 m due to his impact. Find out the average force exerted by the net on the man to stop his fall. Take the mass of the man to be 60 kg and consider the value of acceleration during his free fall as 10 m/s2.

      Solution

      Given, the man jumps from a height of 10 m.
      Initial velocity, u = 0

      Distance travelled by the mn, s = 10 m
      Final velocity, v =?
      Acceleration, a = 10 m/s2. 
      The velocity of the man after he has descended 10 m vertically downward is given by,
      space space space space space space space space space space straight v squared minus straight u squared space equals space 2 a s space space left square bracket space Third space equation space of space motion right square bracket

rightwards double arrow space straight v squared minus space left parenthesis 0 right parenthesis squared space equals space 2 cross times 10 cross times 10 space space
rightwards double arrow space space space space space space space space space space space space space space straight v space equals space 10 square root of 2 space straight m divided by straight s

      Now, the acceleration produced when the net sags by 2m is given by the third equation of motion, 

                          straight v squared minus straight u squared space equals space 2 a s 
      F i n a l space v e l o c i t y comma space straight v space equals space 0 space straight m divided by straight s space
I n i t i a l space v e l o c i t y space comma straight u equals 10 square root of 2 space straight m divided by straight s
D i s t a n c e space b y space w h i c h space t h e space n e t space s a g s space straight s space equals space 2 straight m 
      T h e r e f o r e comma space

space space space space space space space left parenthesis 0 right parenthesis squared minus space 200 space equals space 2 cross times straight a cross times 2 space 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 a space equals space minus 200 over 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 space minus 50 space space straight m divided by straight s squared

Negative space sign space implies space that space there space is space retardation. space

So comma space

Average space force space exerted space by space the space net space on space the space
man space to space stop space his space fall space is comma space

straight F space equals space ma space

space space space equals space 60 cross times 50 space equals space 300 space straight N 

                  

       

       

       



       

       

       

      Question 330
      CBSEENSC9005954

      The velocity-time graph of a ball of 20 g moving along a straight line on a long table is given in Fig. 


      How much force does the table exert on the ball to bring it to rest?

      Solution

      Given,
      Mass of the body, m = 20 g =  0.02 kg
      From the given graph, we note that
      Initial velocity, u = 20 cm/s = 0.20 m/s
      Final velocity, v = 0 cm/s
      Time taken, t = 10 - 0  = 10s
      The force exerted on the ball is,
      therefore space space space space straight F space equals space straight m open parentheses fraction numerator straight v minus straight u over denominator straight t end fraction close parentheses space
space space space space space space space space space space space equals space 0.02 space open parentheses fraction numerator 0 minus 0.20 over denominator 10 end fraction close parentheses space
space space space space space space space space space space space equals space minus 0.0004 space straight N. space
                        
      The negative sign shows that the friction force exerted by the table is opposite to the direction of motion of the ball.

      Question 331
      CBSEENSC9005955

      The velocity-time graph of a car is shown in Fig a. The car weighs 1000 kg.
      (a) What is the distance travelled by the car in the first two seconds?
      (b) What is the braking force applied at the end of 5s to bring the car to a stop within one second?

      Solution


      Fig. (a): Velocity-time graph of the car.
      left parenthesis straight a right parenthesis

space D i s t a n c e space t r a v e l l e d space b y space t h e space c a r space i n space f i r s t space 2 straight s space equals space A r e a space o f space increment A B M
space space space 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 A M space cross times space B M
space space space 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 left parenthesis 2 minus 0 right parenthesis space cross times space left parenthesis 15 minus 0 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 space equals space 15 space straight m
      (b) It is clear from the graph that the velocity at the end of 5 s is 15 m/s.

      So,
      I n i t i a l space v e l o c i t y comma space straight u space equals space 15 space straight m divided by straight s

Time space taken comma space straight t space equals space 1 space straight s

Final space velocity comma space straight v space equals space 0

T h e r e f o r e comma space
A c c e l e r a t i o n comma space straight a space equals space fraction numerator straight v minus straight u over denominator straight t 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 equals space fraction numerator 0 minus 15 over denominator 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 equals space minus 15 space straight m divided by straight s squared

R e t a r d a t i o n space equals space 15 space straight m divided by straight s squared

B r a k i n g space f o r c e comma space F space space equals space m 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 equals space 1000 space cross times space 15 space 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 space equals space 15000 space straight N.

                            

      Question 332
      CBSEENSC9005956

      Plot the distance-time graphs for an object which starts from rest, in the following cases:
      (a) a = 2 m/s2; and (b) a = 8 m/s2.

      Solution

      A table of distance-time graph is made in both the cases. 
      In both cases, 
      Initial velocity, u = 0
      therefore space space space space straight s space space equals u t plus 1 half a t squared space
space space space space space space space space space space space equals space 0 space plus space 1 half a t squared space
space space space space space space space space space space space equals space 1 half a t squared


       
      The distance-time graph is as shown in fig. 9.14 and 9.15:



      Question 333
      CBSEENSC9005957

      State and explain Newton’ s third law of motion.

      Solution

      Newton’s third law of motion.
      According to Newton's third law, to every action, there is an equal and opposite reaction.
      When one objects exerts a force (action) on another object, then the second object also exerts a force (reaction) on the first.
      These two forces are always equal in magnitude but opposite in direction.

      Fig. Action of Newton’s third law.

      As shown in fig.,
      If straight F with rightwards arrow on top subscript AB is the force exerted by body A on B and straight F with rightwards arrow on top subscript BA is the force exerted by B on A, then according to Newton's third law,
                          straight F with rightwards arrow on top subscript BA space equals space minus straight F with rightwards arrow on top subscript AB space
      We can see that the forces are equal and opposite. 
      According to the third law of motion, single force can never exist.
      The forces always exist in pairs.
      The two opposing forces are known as action and reaction forces.
      But the forces of action and reaction always act on two different bodies. 

      Question 334
      CBSEENSC9005958

      Give a simple experiment to demonstrate that action and reaction forces are equal and opposite.

      Solution

      Consider two spring balances connected together.
      The fixed end of B is attached to a rigid support as shown in the fig. below.

      Fig. Action and reaction forces are equal and opposite.
      i) When a force is applied through the free end of spring balance A, it is observed that both the spring balances show the same readings on their scales.
      ii) This means that the force (action) exerted by the spring balance A on balance B is equal but opposite in direction to the force (reaction) exerted by the balance B on A. This proves that action and reaction are equal and opposite forces.

      Question 335
      CBSEENSC9005959

      Give some examples to illustrate Newton’s third law of motion.

      Solution

      Illustration's of Newton's third Law of motion: 
      (i) While walking we press the ground with our feet in the backward direction (action), as shown in Fig. 9.18.
      The ground exerts an equal and opposite force on us. The horizontal component of this force of reaction enables us to move forward. 

      Fig. Action and reaction force acting while we walk on the ground. 

      (ii) While swimming, a person pushes water with his hands and feet in the backward direction (action). Water pushes the person with an equal force in the forward direction, which is the reaction force. 
      (iii) When a gun is fired, the exploding gases, which come out because of a chemical reaction, exert a forward force on the bullet. The bullet exerts an equal and opposite reaction force on the gases. This results in the recoil of the gun, as shown in fig. below



      Fig. A forward force on the bullet and recoil of gun.

      (iv) When a sailor jumps out of a rowing boat, the boat moves backwards. As the sailor jumps forward, he applies a forward force (action) on boat and the boat moves backwards due to the force of reaction.

      Fig. 9.21. As the sailor jumps in forward direction, the boat moves backwards.

      Question 336
      CBSEENSC9005960

      State law of conservation of momentum. Deduce this law from Newton’s second law of motion.

      Solution

      Law of conservation of momentum.
      The law states that if a number of bodies are interacting with each other, that is, the bodies are exerting forces on each other, their total momentum remains conserved before and after the interaction, provided there is no external force acting on them.

      Derivation using Newton’s second law of motion: 
      Let p1 and p2 represent the sum of momenta of a group of objects before and after the collision, respectively.
      Let t be the time elapsed during the collision.

      According to Newton’s second law of motion,

      External force = Rate of change of momentum
      rightwards double arrow space space straight F space equals space fraction numerator straight p subscript 2 minus straight p subscript 1 over denominator straight t end fraction
      I f space t h e r e space i s space n o space e x t e r n a l space f o r c e comma space t h a t space i s space straight F space equals space 0 comma space t h e n

space space space space space space space space fraction numerator straight p subscript 2 minus straight p subscript 1 over denominator straight t end fraction space equals space 0 space space
rightwards double arrow space space space space space space space space space space space space space space straight p subscript 2 space equals space straight p subscript 1.

      Hence, in the absence of an external force, the total momentum of a group of objects remains unchanged or conserved during the collision.
      This is the law of conservation of momentum.

      Question 337
      CBSEENSC9005961

      Deduce law of conservation of momentum from Newton’s third law of motion.

      Solution

      Consider two bodies A and B of masses mA and mB moving in the same direction along a straight line.
      The bodies are moving with velocities uA and uB respectively, such that uA > uB.
      They collide for time t.
      After collision, let their velocities become v1 and v2
      straight F subscript A B end subscript space equals space R a t e space o f space c h a n g e space o f space m o m e n t u m space o f space straight A space
space space space space space space space equals space fraction numerator straight m subscript straight A left parenthesis straight v subscript straight A minus straight u subscript straight A right parenthesis over denominator straight t end fraction
      Force space exerted space by space straight B space on space straight A space is comma

straight F subscript B A end subscript space equals space R a t e space o f space c h a n g e space o f space m o m e n t u m space o f space straight B
space space space space space space space space equals space fraction numerator straight m subscript straight B left parenthesis straight v subscript straight B space minus space straight u subscript straight B right parenthesis over denominator straight t end fraction
      Now, according to Newton's third law of motion, 
      The forces are equal and opposite. 
      space space 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 subscript A B end subscript space equals space minus straight F subscript B A end subscript

rightwards double arrow space space fraction numerator straight m subscript straight A left parenthesis straight v subscript straight A space minus space straight u subscript straight A right parenthesis over denominator straight t end fraction space equals space minus fraction numerator straight m subscript straight B left parenthesis straight v subscript straight B minus straight u subscript straight B right parenthesis over denominator straight t end fraction

rightwards double arrow space space straight m subscript straight A straight v subscript straight A space minus space straight m subscript straight A straight u subscript straight A space equals space minus straight m subscript straight B straight v subscript straight B space plus space straight m subscript straight B straight u subscript straight B

rightwards double arrow space straight m subscript straight A straight u subscript straight A space plus space straight m subscript straight B straight u subscript straight B space equals space straight m subscript straight A straight v subscript straight A space plus space straight m subscript straight B straight v subscript straight B
      That is,
      Total momentum before collision = Total momentum after collision.
      Thus, total momentum of the two bodies is conserved provided no external force acts on them.
      This proves the law of conservation of momentum.

       

      Question 338
      CBSEENSC9005962

      Give some examples to illustrate the law of conservation of momentum.

      Solution

      The law of conservation of momentum can be illustrated using the following examples: 

      (i) Recoil of the gun

      The bullet and the gun, both are at rest before firing and initial momentum of the system is zero. When the bullet is fired, it moves forward with a large velocity, As per the conservation of momentum, the gun must moves backward with a velocity such that the final mometum is zero.

      (ii) Flight of rockets and jet engines:

      When a rocket is fired, it throws a stream of burnt gases in the vertically downward direction with a large velocity. So, the outgoing gases have large momentum in the downward direction. According to the law of conservation of momentum, the rocket gets an equal and opposite momentum which causes it to fly up. 

      Question 339
      CBSEENSC9005963

      If action is always equal to the reaction, explain how a horse can pull a cart.

      Solution

      The horse pulls the cart with a force (action) in the forward direction. The cart also pulls the horse with an equal force (reaction) in the backward direction. This way, the two forces get balanced. The horse pushes the  ground with its feet in the backward direction, while pulling the cart. So, the reaction of earth makes it move in the forward direction alongwith the cart. 

      Question 340
      CBSEENSC9005964

      Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

      Solution

      Water is ejected with a large forward force (action) and due to reaction, the fireman experiences a large backward thrust. Therefore, the force experienced by him makes it difficult in holding the hose. 

      Question 341
      CBSEENSC9005965

      From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms–1. Calculate the initial recoil velocity of the rifle.

      Solution

      Mass of bullet, m1 = 50g  = 0.05 kg
      Mass of rifle, m2 = 4 kg
      Initial velocity of bullet, u1 = 0
      Initial velocity of rifle, u2 = 0
      Final velocity of rifle, v1 = 35 ms-1

      Final velocity of rifle, v2 = ?
      According to the law of conservation of momentum,
      Total momenta after the fire = Total momenta before the fire.
              m1v1+ m2v2 = m1u1+ m2u2

                  0.05 x 35 + 4v2 = 0+0
      rightwards double arrow space space straight v subscript 2 space equals space minus fraction numerator 0.05 space cross times space 35 over denominator 4 end fraction
space space space space space space space space space space equals space minus 7 over 16 space
space space space space space space space space space space equals space minus 0.44 space ms to the power of negative 1 end exponent. 
      The negative sign indicates that the direction in which the pistol would recoil is opposite to that of the bullet.

      Question 343
      CBSEENSC9005967

      A bullet of mass 20 g is horizontally fired with a horizontal velocity 150 ms–1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

      Solution

      For bullet:
      m1 = 20 g = 0.02kg
      u1 = 0 m/s
      v1 = 150 ms-1

      For pistol:
      m1 = 2kg
      u1 = 0 m/s
      v2 = ?
      According to the law of conservation of momentum,
      Total momenta after the fire = Total momenta before the fire.
      That is, 
      m1v1 + m2v2 = m1u1 + m2u2

      0.20 x 150 + 2 x v2 = 0.02 x 0 + 2 x 0
      rightwards double arrow       2v2 = -3
      rightwards double arrow         v2 = -1.5 ms-1
      Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet.

      Question 344
      CBSEENSC9005968

      A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block?

      Solution

      Let the final velocity of the bag along with the bullet embedded in it be v.
      therefore space space For space bullet colon thin space m subscript 1 space equals space 10 space straight g space equals space 0.01 space kg semicolon space space straight u subscript 1 space equals space 400 space straight m divided by straight s semicolon space space straight v subscript 1 space equals space straight v space straight m divided by straight s

space space space space space space For space bag colon space space space space straight m subscript 2 space equals space 900 space straight g space equals space 0.9 space kg semicolon space space straight u subscript 2 space equals space 0 semicolon space space space straight v subscript 2 space equals straight v space straight m divided by straight s
      Now, According to the law of conservation of momentum,
      Total momentum before collision = Total momentum after collision
      That is, 
      space space space space space space space space space space space space space space space space straight m subscript 1 straight u subscript 1 space plus space straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 plus straight m subscript 2 straight v subscript 2

rightwards double arrow space space space space space space space space space space space space straight m subscript 1 straight u subscript 1 plus space straight m subscript 2 straight u subscript 2 space equals space left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v space space space space space space space space space space open square brackets because space space straight v subscript 1 space equals straight v subscript 2 space equals space straight v close square brackets

rightwards double arrow space space space space space space space space space space space space space space space space space space space straight v space equals space fraction numerator straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 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 equals fraction numerator 0.01 space cross times 400 space plus 0.9 space cross times 0 over denominator 0.01 space plus 0.9 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 equals fraction numerator 4 over denominator 0.91 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 equals 400 over 91 equals 4.4 space straight m divided by straight s.

      Question 345
      CBSEENSC9005969

      A target of mass 400 g moving with a horizontal speed of 10 m/s is hit by a bullet of mass 0.01 kg moving in opposite direction. If both the bullet and the target come to rest after the collision, calculate the velocity of the bullet at the time of striking the target.

      Solution

      Mass of the target, m1 = 400 g = 0.4 kg
      Initial velocity of the target, u1 = 10m/s
      Final velocity, v1 =0
      Mass of the bullet, m2 = 0.01 kg
      Initial velocity of the bullet, u2 = ?
      Final velocity, v2 =0
      According to the law of conservation of momentum,
                 m1u1 + m2u2 = m1v1 + m2v2

      0.4 x 10 + 0.01 x u2 = 0 + 0
      rightwards double arrow space space straight u subscript 2 space equals space minus fraction numerator 0.4 space cross times space 10 over denominator 0.01 end fraction space equals space minus 400 space straight m divided by straight s squared. space space
      Negative sign shows that the bullet is moving in a direction opposite to that of the motion of target.

      Question 347
      CBSEENSC9005971

      A girl of mass 40 kg jumps with a horizontal velocity of 5 ms–1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction.

      Solution

      Let v be the velocity of the girl on the cart as the cart starts moving.
      The total momenta of the girl and cart before the interaction,
       p = 40 kg x 5 ms-1 + 3 kg x 0 ms-1
          
      = 200 kg ms-1

      The total momenta after the interaction is, 
      p' = (40 + 3) kg x v ms-1 = 43 v kg m s-1

      According to the law of conservation of momentum, the total momentum is conserved during the interaction.
      Therefore,
              43 v = 200 
      rightwards double arrow         v = 200/43
                      = +4.65 ms-1

      The girl on cart would move with a velocity of 4.65 ms–1 in the direction in which the girl jumped onto the cart. 

      Question 348
      CBSEENSC9005972

      Two hockey players of opposite teams, while trying to hit a hockey ball in the ground collide and immediately become entangled. One has a mass of 60 kg and is moving with velocity 5.0 ms–1 while the other has a mass of 55 kg and is moving faster with velocity 6.0 ms–1 towards the first player. In which direction and what velocity will they move after they become entangled ? Assume that the frictional force acting between the feet of the two players and ground is negligible.

      Solution

      Consider the first hockey player be moving from left to right.
      Using sign convention the left to right is taken as positive direction and thus right to left is negative direction.
      Let 'm' and 'u' represent the mass and initial velocity of the two players, respectively.  
      Then
      Mass of the 1st palyer, m1 = 60 kg
      Velocity, u1 = 5 ms-1
      Mass of 2nd player, m2 = 55 kg
      Velocity, u2 =-6 ms-1.
      The total momentum of the two players before the collision,
       = 60 kg x (5 ms-1) + 55 kg x (-6 ms-1)
       = -30 kg ms-1

      If v is the velocity of the two entangled players after the collision then the total momentum is, 
      = (m1 + m2) x v
      = (60 + 55) kg x v ms-1
      =
      115 x v ms-1 

      Acc, to the law of conservation of momentum, we have,
             115 x v = -30
      rightwards double arrow             v = -30/115
                          = -0.26 ms-1

      Thus the two entangled players would move with a velocity 0.26 ms–1 from right to left, that is in the direction the second player was moving before the collision. 

      Question 349
      CBSEENSC9005973

      Define the term force.

      Solution

      Force is a push or pull which tends to change the state of rest or direction of uniform motion uniform motion along a straight line. 


      Question 350
      CBSEENSC9005974

      Is force a scalar or a vector quantity?

      Solution

      Force acts in a particular direction. So, it is a vector quantity.

      Question 351
      CBSEENSC9005975

      What does balanced force cause?

      Solution

      Balanced forces are the forces which are equal in size and opposite in direction. Balanced forces do not cause any change in motion. 

      Question 352
      CBSEENSC9005976

      What is the cause of unbalanced forces on a body ?

      Solution

      Unbalanced forces change or tend to change the state of rest or of uniform motion of a body. Unbalnced forces vary in size and may not be in opposite direction. 

      Question 353
      CBSEENSC9005977

      Can the velocity of a body change when no unbalanced force is acting on it?

      Solution

      No. The velocity would remain constant.

      Question 355
      CBSEENSC9005979

      State Galileo’s law of inertia.

      Solution

      Galileo's law of inertia states that the body moving in a straight line with a certain speed will continue moving in the same straight line with the same speed in the absence of external force. 

      Question 356
      CBSEENSC9005980

      What do you mean by inertia?

      Solution

      The inherent property or tendency of a body to resist or oppose any change in its state of rest or of uniform motion in a straight line is called inertia. 

      Question 357
      CBSEENSC9005981

      What is the measure of inertia of a body?

      Solution

      Mass is a measure of inertia of a body.

      Question 358
      CBSEENSC9005982

      Why do bodies of small mass require small initial effort to bring them into motion?

      Solution

      The mass of a body and inertia are directly related to each other. Smaller the mass, lesser is the opposition to the change of state of motion.  

      Question 359
      CBSEENSC9005983
      Question 360
      CBSEENSC9005984

      Define the term inertia with respect to motion.

      Solution

      Inertia of motion is the tendency of a body to remain in its state of uniform motion.

      Question 361
      CBSEENSC9005985
      Question 362
      CBSEENSC9005986

      State Newton’s first law of motion.

      Solution

      An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.

      Question 363
      CBSEENSC9005987

      Which law of motion is also called ‘law of inertia’?

      Solution

      Newton’s first law of motion is also called 'law of inertia'.

      Question 364
      CBSEENSC9005988

      Can a body in linear motion be in equilibrium?

      Solution

      Yes, a body in linear motion can be in equilibrium provided the vector sum of the forces acting upon the body is zero. 

      Question 365
      CBSEENSC9005989

      A body is acted upon by a number of external forces. Can it remain at rest?

      Solution

      If the vector sum of all the external forces is zero, then the body will remain at rest. 

      Question 366
      CBSEENSC9005990

      If a body is not at rest, the net external force acting on it cannot be zero. Is it true or false?

      Solution

      When a body is moving with a constant velocity, the net external force acting on the body is zero. 



      Question 367
      CBSEENSC9005991

      A kite is skilfully held stationary in the sky. Give the magnitude and the direction of the force acting on it?

      Solution

      It is given that a kite is held stationary on the sky. That is, no net force is acting on it. The force exerted by air on the kite is balanced by tension produced in the string. 

      Question 368
      CBSEENSC9005992

      Give the magnitude and direction of force acting on a small water drop falling down with constant velocity? 

      Solution

      The water drop is falling with constant velocity. Therefore, no net force acts on it. 

      Question 369
      CBSEENSC9005993

      If you jerk a piece of a paper placed under a book very quickly, the book will not move. Why?

      Solution

      When a piece of paper placed under a book is moved quickly, the books will not move. This is because the book continues to be in it's state of rest due to inertia.

      Question 370
      CBSEENSC9005994

      What happens to a stone tied to the end of a string and whirled in a circle if the string suddenly breaks?

      Solution

      A stone tied to the end of a string is whirled and suddenly the string breaks. The stone flies off tangentially along a straight line due to inertia of direction. 
       


       


      Question 371
      CBSEENSC9005995

      Define momentum of body.

      Solution

      Momentum of a body is the quantity of motion possessed by a body and is equal to the product of the mass and velocity of the body. 

      Question 372
      CBSEENSC9005996

      Is momentum a scalar or a vector quantity?

      Solution

      Momentum is a vector quantity and has the same direction as that of velocity. 

      Question 373
      CBSEENSC9005997

      Write the SI unit of momentum.

      Solution

      SI unit of momentum is kg m/s. 

      Question 374
      CBSEENSC9005998

      What is the name given to the product of mass and velocity of a body?

      Solution

      Momentum gives us the product of mass and velocity.

      Momentum, p = mass (m) × velocity (v) 

      Question 375
      CBSEENSC9005999

      Name the scientist who introduced the concept of momentum.

      Solution

      Issac Newton. 

      Question 376
      CBSEENSC9006000

      Which law of motion gives the measure of force?

      Solution

      Second law of motion.

      Question 377
      CBSEENSC9006001

      State Newton’s second law of motion.

      Solution

      Newton's second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. 

      Question 378
      CBSEENSC9006002

      Which physical quantity corresponds to the rate of change of momentum?

      Solution

      Force gives us the measure of rate of change of momentum. 

      Force, F = dpdt
      where, p is the mometum. 

      Question 379
      CBSEENSC9006003

      Define one newton force.

      Solution

      One newton force is that force which produces an acceleration of 1 m/s2 in a body of mass 1 kg. 

      Question 381
      CBSEENSC9006005

      What is the relationship between force and acceleration?

      Solution

      Acceleration produced in a body by an unbalanced force is directly proportional to the force.

      According to the first law of motion, 
                           F = ma 

                      a  = Fm



      Question 382
      CBSEENSC9006006

      What is the relationship between force and mass?

      Solution

      The force acting on a body is directly related to the mass of the body.

                                     F  

      That is, massive the object of a body, more force is to be applied on the body. 

      Question 383
      CBSEENSC9006007

      State the relation between force and momentum.

      Solution

      Force is defined as the rate of change of momentum.

      That is, 

                                F = dpdt 

      where, dpdt is the change of momentum with respect to time. 

      Question 384
      CBSEENSC9006008

      State Newton’s third law of motion. 

      Solution

      Newton’s third law of motion states that action and reaction are equal and opposite and they act on different bodies.

      Question 385
      CBSEENSC9006009

      Do action and reaction act on the same body or on different bodies?

      Solution

      Action and reaction, both act on different bodies. 

      Question 386
      CBSEENSC9006010

      State the law of conservation of momentum.

      Solution

      The law of conservation of momentum states that  the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. 

      Question 387
      CBSEENSC9006011
      Question 388
      CBSEENSC9006012

      If two bodies collide in the absence of any external force, what will be the total change in the momentum of the system?

      Solution

      Collision of two bodies in the absence of any external force is zero. the total momentum of the system has to bec onserved. 

      Question 389
      CBSEENSC9006013

      What is the total momentum of the cannon and shell

      (i) before firing; and (ii) after firing?

      Solution

      (i) The total momentum of both canon and shell before firing is zero, because both are at rest.

      (ii) According to the law of conservation of momentum, total momentum of the cannon and shell after firing is also zero.

      Question 390
      CBSEENSC9006014

      Since action and reaction forces are always equal in magnitude and opposite in direction, how can anything ever be accelerated?

      Solution

      Action and reaction forces are always equal in magnitude and opposite in direction. But, the action and reaction forces always act on different bodies.

      Question 391
      CBSEENSC9006015

      Name the principle on which a rocket works.

      Solution

      The working of rocket is based on the principle of third laws of motion.
      For the rocket to move forward, the exhaust gases fire backwards, against which the the atmosphere gives an equal push back, which thrusts the rocket forward.

      Question 392
      CBSEENSC9006016

      Why are heavy loads moved by placing them on rollers?

      Solution

      Rollers are used for moving heavy loads because rolling friction is less than sliding friction. 

      Question 393
      CBSEENSC9006017

      Why are tyres corrugated and made rough?

      Solution

      Tyres are corrugated and made rough so as to increase the friction and provide better grip on the road. This avoids circumstances of slipping on the road. 

      Question 394
      CBSEENSC9006018

      A heavy trunk is lying on the floor. A boy applies a force of 100 N to pull it, but fails. Then a man pulls it by a force of 300 N, but fails. Then an athlete applies a force of 600 N and is able to pull it. What force does it prevent from moving in the first two cases ? What conclusions can you draw about the force from these three observations?

      Solution

      Given, that the trunk is very heavy. So, the force of friction between the trunk and the floor prevents  the motion in the first two cases. 
      So, the force of friction is greater than 300 N but less than 600 N. 

      Question 395
      CBSEENSC9006019

      Why does Newton’s first law of motion appear to be contradicted in our day-to-day life?

      Solution

      In our daily practical life, we have observed that when we apply external force on a moving body, it covers some distance before it stops. This is in contradiction to Newton’s first law.
      However, the motion of the body is being opposed by force of friction between the body and the ground and also by air resistance. In the absence of such forces, the body would continue to move on its own. 

      Question 396
      CBSEENSC9006020

      A passenger sitting loosely in a bus falls backward when the bus suddenly starts moving, why?

      Solution

      The bus which was at rest, when suddenly starts moving, the  lower part of the passenger’s body begins to move along with the bus. But, the upper part tends to remain at rest due to inertia of rest. This is the reason, a passenger standing or sitting loosely in a bus falls backward when the bus suddenly starts moving. 

      Question 397
      CBSEENSC9006021

      A man jumping out of a moving train falls with his head forward. Why?

      Solution

      When a man jumps out of a moving train, his feet suddenly comes to rest on touching the ground. But, the upper part of the body continues to be in motion and has a tendancy to move forward. That is why the person falls with his head forward.
      In order to save himself from falling headlong, the person should run through some distance in the forward direction.

      Question 398
      CBSEENSC9006022

      Why is it that an athlete always runs some distance before taking a jump?

      Solution

      An athlete always runs for some distance before taking a jump so that inertia of motion may help him in his muscular efforts to take a longer jump.

      Question 399
      CBSEENSC9006023

      If a ball is thrown up in a moving train, it comes back to the person’s hands, why?

      Solution

      When a ball is thrown up in a moving train, there is no horizontal force acting on it. So, the ball comes back to the person as it moves with the same horizontal speed. 

      Question 400
      CBSEENSC9006024

      A bullet fired against a window glass pane makes a hole in it and the glass pane is not cracked. On the other hand, a stone striking the same glass pane will smash it. Explain why?

      Solution

      When a bullet is fired and hits the glass, the part of the glass which comes in contact with the bullet immediately has a tendency to share  the large velocity of bullet. This way it makes a hole while the remaining part of the glass due to inertia  of rest remains at rest. Hence, the glass pane is not smashed.
      When a stone with low velocity strikes the glass, the surrounding part of the glass takes time to share the velocity of the stone. Thus the glass is smashed.

      Question 401
      CBSEENSC9006025

      Why do the blades of an electric fan continue to rotate for some time after the current is switched off ?

      Solution
      The blades of an electric fan continue to rotate for some time because of inertia of motion. The blades of the fan were in motion. Hence, they will take time to come to rest after being switched off. 
      Question 402
      CBSEENSC9006026

      Why are the passengers thrown outwards when a car in which they are travelling suddenly takes a circular turn?

      Solution

      The passemgers are thrown outwards when a car in which they are travelling suddenly takes a circular turn because the passengers tend to maintain their direction of motion (inertia of direction) while the direction of car changes when it takes the turn. 

      Question 403
      CBSEENSC9006027

      What happens when you shake a wet piece of cloth? Explain your observation.

      Solution

      When we shake a wet piece of cloth, water molecules come out of it. On shaking, the cloth comes into motion while water molecules tend to remain at rest due the the property of inertia of rest and get detached.

      Question 404
      CBSEENSC9006028

      A javelin throw is marked foul if the athlete crosses over the line marked for the throw.

      Explain, why athletes often fail to stop themselves before that line?

      Solution

      The athletes who throw the javelin fails to stop themselves before the line due to inertia of motion. So, an athlete may cross the marked line before throwing the javelin.

      Question 405
      CBSEENSC9006029

      Two similar trucks are moving with the same velocity on a road. One of them is loaded while the other is empty. Which of the two will require a larger force to stop it?

      Solution

      Loaded truck has more inertia of motion or greater momentum than the empty truck. So more force is required to stop the loaded truck than the empty truck.

      Question 406
      CBSEENSC9006030

      Why are the wheels of vehicles are provided with mudguards?

      Solution

      The mud sticking to the wheel from the road flies off tangentially, when the wheel rotates at a high speed because of inertia of direction. In order that the flying mud does not spoil the clothes of passerby, the wheels are provided with mudguard.

      Question 407
      CBSEENSC9006031

      Explain the meaning of the following equation in mechanics:

      F =  ma, where symbols have their usual meanings.

      Solution

      The equation, F =  ma means that the force F on an object is equal to the product of its mass m and acceleration a.
      The mathematical expression depicts Newton’s second law of motion.

       

      Question 408
      CBSEENSC9006032

      Explain how Newton’s first law of motion is a special case of second law?

      Solution

      According to Newton’s second law,

                               F = ma

      Since, mass of a body cannot be zero therefore, when when force is equal to zero, acceleration is also zero. So, there wont be any  change in the velocity of the body.  
      So, in the absence of any external force, a body will continue to remain in its state of rest or of uniform motion in a straight line. This is Newton’s first law of motion.

      So, Newton’s first law of motion is contained in the second law.                                                      


      Question 409
      CBSEENSC9006033

      A cricket player lowers his hand while catching a ball. Explain?

      Solution

      The player increases the time of catching the ball, by lowering his hand. Therefore, the rate of change of momentum is small and the force exerted on his hand is less. So, the person is less likely to get hurt. 


      Question 410
      CBSEENSC9006034

      In oil tankers some space is left at the top while filling them. Explain, why?

      Solution

      During transportation, the oil may rise up due to sudden jerks. If some space is left, the oil will take some time to reach the top of the tanker and hence it will exert less force on the tanker. This will prevent the damage caused by overflowing of the oil.

      Question 411
      CBSEENSC9006035

      Chinawares are wrapped in straw paper before packing?

      Solution

      The straw paper between the chinawares increases the time of experiencing the jerk during transportation. Hence, they strike against each other with less force and are less likely to broken.

      Question 412
      CBSEENSC9006036

      Why we are hurt less when we jump on a muddy floor in comparison to a hard floor?

      Solution

      A sandbed is motioned in the direction of the athlete when he jumps and the time interval is increased. This decreases rate of change of momentum. Hence, the force of reaction is decreased. So, the person who is jumping do not get hurt.


      Question 413
      CBSEENSC9006037

      Why are shockers used in cars, scooters and motorcycles?

      Solution

      Shockers are used in cars, scooters and motorcycles so that the time interval of the jerk increases. so, the rate of momentum decreases. Hence, comparitively a lesser force is exerted on the passengers during the jerk. 


      Question 414
      CBSEENSC9006038

      Why are buffers provided between the bogies of a railway train?

      Solution

      Boggies are the carriages which are connected with each other. Use of buffers increase the time interval of jerks during shunting. This reduces the force with which bogies push or pull each other. Hence, this way severe jerks are avoided. 

      Question 415
      CBSEENSC9006039

      A brick can be pushed gently on the smooth floor by applying force with our foot. But if we kick the brick, the foot is hurt. Why?

      Solution

      When a brick is kicked by a person, time for which the force imparted is short. Therefore, rate of change of momentum of the brick is large. So, a large reaction force is applied back by the brick on the foot.
      So, force may hurt our foot. 


      Question 416
      CBSEENSC9006040

      Why is it necessary to bend knees while jumping from greater height?

      Solution
      When jumping from a great height, our feet at once come to rest. For the small time when our legs comes to rest, a large value of force is exerted on our feet. 
      When the knees are bent slowly,  the value of time of impact increases and less force acts on feet.  
      Therefore, the probability of getting injured is less. 
      Question 417
      CBSEENSC9006041

      How does a karate player break a pile of tiles or slab of ice with a single blow?

      Solution

      When the karate player strikes the pile or slab fast, the entire momentum of the hands reduces to zero in a very short span of time. Therefore, the force exerted on the ice slab or pile of tiles is large enough to break it. 

      Question 418
      CBSEENSC9006042

      A boxer moves his head backward to protect himself from the attack of his rival. Why?

      Solution

      The boxer gets more time before being attacked by his opponent, on moving his head backward. Thus, this reduces the effect of the punch. 

      Question 419
      CBSEENSC9006043

      Why are seat belts provided in automobiles and aeroplanes?

      Solution

      When the vehicle suddenly moves, it results in a sudden state of motion of the vehicle. Hence, our feet which is in contact with the vehicle comes to motion. but, the upper part of our body which is at rest, tends to remain at rest due to the property of inertia of rest. 
      Therefore, Seat belts are provided to protect the passengers from falling backward or forward during such situations. 

      Question 420
      CBSEENSC9006044

      If someone jumps to the shore from a boat, the boat moves in opposite direction. Explain, why.

      Solution

      When a man jumps out from a boat, he presses the boat with his feet (action), and the boat also exerts an equal force on him in the opposite direction (reaction). As a result, the man jumps to the bank and the boat moves in the backward direction. 

      Question 421
      CBSEENSC9006045

      A book rests on a table:

      (i) Show the forces on the table and the corresponding reaction forces.

      (ii) Why do not the forces acting on the table cause it to move?

      Solution

      (i) The action and reaction forces have been shown in Fig. below


      (ii) The sum of all the forces acting on the table is zero. So there is no net force acting on it. Thus, the table will not move. 


      Question 422
      CBSEENSC9006046

      Indicate the forces of action and reaction in the following cases:

      (i) a man standing on the ground;

      (ii) a stone suspended by a thread from the ceiling;

      (iii) a book lying on the table;

      (iv) motion of moon around the earth; and

      (v) when a magnet attracts a piece of iron.

      Solution

      (i) A man standing on the ground:

      The action force is when the man exerts a downward force equal to his weight on the ground. And, as a reaction, the ground exerts an equal upward force on the man. 

      (ii) A stone suspended by a thread from the ceiling: 

      The weight of the stone acts vertically downwards (action). A tension equal to the weight of the stone acts vertically upwards in the thread.

      (iii) A book lying on the table: 

      The book exerts a downward force equal to its weight on the table (action). The table exerts an equal force on the book in the upward direction (reaction).

      (iv) Motion of moon around the earth: 

      The moon attracts the earth with some force (action). The earth also attracts the moon with an equal and opposite force (reaction). Hence, it kees orbiting around the Earh. 

      (v) When a magnet attracts a piece of iron:

      When a magnet and a piece of iron are brought closer to each other, the magnet attracts the iron piece with a certain force (action). The iron piece also attracts the magnet with an equal and the opposite force (reaction).

      Question 423
      CBSEENSC9006047

      Do action and reaction act on the same body or on different bodies? How are they related in magnitude and direction? Are they simultaneous or not?

      Solution

      Action and reaction always acts on different bodies. Both the forces are equal in magnitude but act in opposite directions. Action and reaction forces are both simultaneous and reaction appear as long as there is action.  

      Question 424
      CBSEENSC9006048

      Why we can’t walk with our legs kept perpendicular to the ground?

      Solution
      When a person is walking with legs kept perpendicular to the ground, our body exerts a force on the ground. The reaction of the ground  is such that it exerts a force vertically upwards. So, the person become static and motion is not possible.  
      Therefore, we always have to push the ground by exerting a force with our legs slanting backwards. The reaction of the ground makes us move forward.


      Question 425
      CBSEENSC9006049

      A horse cannot pull a cart and run in empty space?

      Solution
      The horse pushes the earth with its feet and reaction of the earth makes it move in the forward direction when pulling a cart or running. In empty space there is no reaction force. Hence, the horse cannot run. 
      Question 426
      CBSEENSC9006050

      A horse is pulling a cart.

      (i) What is the force that causes the horse to move forward?

      (ii) What is the force that causes the cart to move forward?

      Solution

      (i) The reaction force of the ground on the feet of the horse, causes the horse to move forward.

      (ii) The force which the horse exerts on the cart, causes the cart to move forward. 

      Question 427
      CBSEENSC9006051

      A gun recoils on firing. Why?

      Solution

      When a bullet is fired from the gun with some force in the forward direction (action), the bullet in turn exerts an equal and opposite force on the gun. The reaction force by the bullet causes the gun to recoil.

      Question 428
      CBSEENSC9006052

      When a boatman wants to go away from the bank, he pushes the bank with a pole. Why? 

      Solution

      When a boatman wants to go away from the bank, he pushes the bank with a pole (action). The bank in turn, pushes away the boat with an equal and opposite force (reaction).
      This helps the boatman to move. 

      Question 429
      CBSEENSC9006053

      When a balloon filled with air with its untied mouth, directing downwards, is released, it moves upwards. Why?

      Solution
      The air rushes out vertically downwards (action), when a balloon filled with air is held with its untied mouth directing downwards. The vertically downwards rush of air downwards let the balloon moves vertically upwards (reaction).
      Question 430
      CBSEENSC9006054

      While swimming, a person pushes the water backward. Why?

      Solution

      A person pushes the water with his hands in the backward direction while swimming. In turn, water in the pool, pushes him forward due to the reaction force.

      Question 431
      CBSEENSC9006055

      Why is it difficult to drive a nail into a wooden block without supporting it?

      Solution

      The end of the nail is pointed and it needs to be supported. If not given a proper support, the nail will fall and will be unable to offer suitable reaction force.


      Question 432
      CBSEENSC9006056

      Why it is difficult to climb up a greasy pole?

      Solution

      When a person climbs up a pole, he presses the pole downward with his feet. The pole, in turn, pushes the person upwards with an equal force.
      If the pole is greasy, its surface becomes slippery and the person is not able to press it. As there is no action, there will be no reaction.
      Hence it becomes difficult for the person to climb up. 

      Question 433
      CBSEENSC9006057

      A man is at rest in the middle of a pond on perfectly frictionless ice. How can he get himself to the shore?

      Solution

      Given, a man is at rest in the middle of a pond on perfectly frictionless ice. If the man wants to move himself to the shore, the man must throw away his shirt in a direction opposite to the desired direction of motion. 

      Question 434
      CBSEENSC9006058

      Can a sailboat be propelled by air blown at the sails from a fan attached to the boat?

      Solution

      From Newton’s second law, a body can be accelerated when some external force is applied over the body. A fan, blowing air, attached to the boat is a part of the boat and not an external agent. Therefore, the motion of the boat cannot be propelled.


      Question 435
      CBSEENSC9006059

      When a body falls to the earth, the earth also moves up to meet it. But the earth’s motion is not noticeable. Why?

      Solution

      Acceleration is the ratio of applied force to the mass of the body. Since, the earth is a massive body, therefore its acceleration is very small.
      Therefore, the movement of the Earth is negligibe and hardly noticeable. 

      Question 436
      CBSEENSC9006060

      Vehicles stop on applying brakes. Does this phenomenon violate the principle of conservation of momentum?

      Solution

      The stopping of vehicles when brakes are applied does not violate the law of conservation of momentum because an external force is acting on the system. 

      Question 437
      CBSEENSC9006487

      What is sound and how is it produced?

      Solution

      Sound is a form of energy. Sound is produced due to vibration of different objects thus, produces a sensation of hearing in our ears. 

      Question 438
      CBSEENSC9006488

      Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

      Solution
      Propagation of sound through air:
      i) Sound travels through air in the form of longitudinal waves.
      ii) Consider a vibrating tuning fork producing sound waves as shown in Fig. 

      Fig. Tuning fork producing sound waves in air.

      iii) The prongs of the tuning fork first move inward and then outwards and so on.
      iv) Now, focusing our attention on prong B only, we will see that this prong moves from B to B1  and it compresses the layer of air in front of it.
      v) As this compressed layer moves forward, it compresses the next layer and so on. So a wave of compression moves forward. When the prong moves backward from B1 to B2, the pressure of air in the adjoining layer decreases.
      vi) The next layer, being at higher pressure tends to move it towards right and so on. So a wave of rarefaction moves forward.
      vii) The vibrating tuning fork continues to send a series of compressions and rarefactions.
      viii) These rarefactions and compressions will finally reach the ear and force the tympanic membrane to vibrate with their frequency and thus cause the sensation of hearing.
      Compression is the region of high pressure and rarefaction is the region of low pressure. Higher the pressure in a region, higher is the number of particles per unit volume and hence higher is the density of the medium. So, a sound wave propagates through a medium as the variation in its pressure or density.
      Question 439
      CBSEENSC9006489

      Cite an experiment to show that sound needs a material medium for its propagation.

      Solution

      Sound is a mechanical wave which needs a material medium to travel.
      It can travel through air, water, steel, etc. but cannot travel through vacuum. This can be demonstrated by the following simple experiment.

      Experiment:
      i) Suspend an electric bell inside a glass bell jar by passing the connecting wires through an airtight cork fitted at the mouth of the jar.
      ii) Place the jar over a disc which has a pipe connected to a vacuum pump. 
      iii) When we turn on the switch, we hear the sound of the bell.
      iv) Now, we pump out the air from the jar gradually with the help of the vacuum pump. The sound becomes fainter.
      v) When most of the air has been removed, we hear a very feeble sound.

      vi) When there is air inside the jar, sound travels through it to the wall of the jar. This makes the wall to vibrate which in turn, sends sound to us.
      vii) When air is removed, sound from the bell cannot travel to the wall of the jar.
      Hence, the sound needs a material medium for it's propagation. 

      Question 440
      CBSEENSC9006490

      Why is sound wave called a longitudinal wave?

      Solution

      Longitudnal waves travel in the form of compression and rarefactions. In the waves the individual particles of the medium move in a direction parallel to the direction of the propagation of the distirbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. This is exactly how a sound wave propagates. Hence, sound waves are longitudnal waves. 

      Question 441
      CBSEENSC9006491

      Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

      Solution

      The characteristic of the sound that will help us to identify our friend's voice is timbre. The quality or timber of sound is that characteristic which enables us to distinguish one sound from another having the same pitch and loudness.

      Tips: -

      The quality or timber of sound is the characteristic which helps us to identify our friend’s voice.

      Question 442
      CBSEENSC9006492

      Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

      Solution

      The speed of sound which is equal to 344 ms–1 is much smaller than the speed of light, equal to 3 x 108 ms–1. Therefore, thunder is heard a few seconds after the lightning flash is seen. 

      Question 443
      CBSEENSC9006493

      Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

      Solution

      The speed of sound (344 ms–1) is much smaller than the speed of light (3 x 108 ms–1).
      Therefore, thunder is heard a few seconds after the flash is seen.

      Question 444
      CBSEENSC9006494

      Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

      Solution

      Let, length of the aluminium rod = 'd' metre

      Speed of sound in air = 344 ms-1

      Time taken by sound in air to reach the child, t(air) = d346 sec 

      Speed of sound in aluminium = 6420  ms-1 

      Time taken by sound  in aluminium rod to reach the child, t (aluminium) = d6420 s

      Ratio of times taken by sound wave in air to aluminium is given by,

      t (air)t (aluminium) = d346d/6420=d346×6420d = 6420346 = 18.55

      Question 445
      CBSEENSC9006495

      The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

      Solution

      Frequency of the source = 100  Hz = 100 s-1

      That is,

      Number of vibrations in 1 s = 100

      So,

      Number of  vibrations in 1 minute or 60 s = 100 x 60 = 6000

      Question 446
      CBSEENSC9006496

      Does sound follow the same laws of reflection as light does? Explain.

      Solution

      The laws of reflection followed by sound is same as that of light. 
      i) The directions in which sound is incident and reflected make equal angles with the normal to the reflecting surface.
      ii) Angle of incidence, Angle of reflection and normal all the three lie in the same plane of incidence. 

      Question 447
      CBSEENSC9006497

      When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

      Solution

      The speed of sound is proportional to temperature. As the temperature increases, the speed of sound also increases. So, the reflected sound returns to source earlier than 0.1 s. Hence, a distinct echo cannot be heard. 

      Question 448
      CBSEENSC9006498

      Give two practical applications of reflection of sound waves.

      Solution

      Two practical applications of reflection of sound waves are: 

      (i) In a stethoscope, the sound of a patient’s heartbeat is guided along the tube of the stethoscope to the doctor’s ears by multiple reflections of sound.

      (ii) Reflection of sound plays an important role in the designing of concert halls.

      Question 449
      CBSEENSC9006499

      A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms–2 and speed of sound = 340 ms–1.

      Solution

      First, we find the time taken by the stone to reach the base of the tower.
      We use the relation,
      space space space space space space space space space space space space space straight s space equals space ut space plus space 1 half gt squared space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
H e r e comma space
D i s tan c e comma space straight s space equals space 500 space straight m

Initial space velocity comma space space space space straight u space equals space 0

P u t t i n g space t h e s e space v a l u e s space i n space f o r m u l a space left parenthesis 1 right parenthesis comma space

A c c e l e r a t i o n space d u e space t o space g r a v i t y comma space straight g space equals space 10 space ms to the power of negative 2 end exponent
space therefore space space space space space space space 500 space equals space 0 space plus space 1 half space cross times space 10 space cross times space straight t squared
rightwards double arrow space space space space space space space space straight t squared space equals space 500 over 5 space equals space 100
rightwards double arrow space space space space space space space straight t space equals space 10 space straight s
 
      Time taken by sound to travel from base of the tower to its top,
      straight t apostrophe space equals space fraction numerator Distance over denominator Speed space of space sound end fraction space equals space fraction numerator 500 space straight m over denominator 340 space ms to the power of negative 1 end exponent end fraction space equals space 1.47 space straight s
      Total space time space after space which space the space splash space is space heard
space space space space space space space space equals straight t space plus space straight t apostrophe space equals space 10 plus 1.47 space equals space 11.47 space straight s. space

      Question 450
      CBSEENSC9006500
      Question 451
      CBSEENSC9006501

      What is reverberation? How can it be reduced?

      Solution

      The repeated reflection that results in this persistence of sound is called reverberation. A sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible.

      Reverberation can be reduced by following methods:

      (i) By covering the walls and roof of the halls with sound absorbing materials.

      (ii) By providing heavy curtains with folds, so that reflection of the sound is avoided. 

      Question 452
      CBSEENSC9006502

      What is loudness of sound ? What factors does it depend on?

      Solution

      Loudness is a physiological response of the ear to the intensity of sound. Even if the intensity of the two sound waves are same, the loudness of the two would differ.
      Loudness depends on two factors:

      (i)Intensity of sound, which is directly proportional to the square of amplitude of the sound wave.

      (ii)Sensitivity of the ear.

      Question 453
      CBSEENSC9006503

      Explain how bats use ultrasound to catch a prey.

      Solution

      i) Bats can produce and receive ultrasonic waves.
      ii) During its flight, a bat emits ultrasonic waves. The bat receives back these waves after being reflected by the obstacle in its path.
      iii) The bat gets information where the obstacle or prey is and what is its size, because of the nature of these waves. 

      Fig. Ultrasound is emitted by a bat and it is reflected back by the prey.
      Porpoises also use ultrasound for navigation and location of food even in total darkness.

      Question 454
      CBSEENSC9006504

      Explain how bats use ultrasound to catch a prey.

      Solution

      Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves. The high-pitched ultrasonic squeaks of the bat are reflected from the obstacles or prey and returned to bat’s ear. The nature of reflections tells the bat where the obstacle or prey is and what it is like. 

      Question 455
      CBSEENSC9006505

      Explain the working and application of a sonar.

      Solution

      The acronym SONAR stands for Sound Navigation and Ranging.

      Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.

      Principle: It uses the phenomenon of echoes in determining the sea-depth and locating the presence of under-water objects.

      Working: 

      i) Sonar consists of a transmitter and a detector and is installed in a boat or a ship.

      ii) The transmitter produces and transmits strong ultrasonic waves.

      iii) These waves travel through the water and after striking the object the beam is reflected from the seabed and is received by an under-water detector which is also mounted on the ship.

      iv) The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted.

      v) The time interval between transmission and reception of the ultrasonic signal is noted. 

      The SONAR method is also called echo ranging. This technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc. 

      Question 456
      CBSEENSC9006506

      A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the device from the submarine is 3625 m.

      Solution

      Given,

      Time between transmission and echo, t = 5 s

      Distance of the sonar from the submarine, d = 3625 m

      Total distance covered by sound = 2d = 2 x 3625 = 7250 m

      Therefore,

      Speed of sound , v = 2dt = 72505 = 1450 m/s 
                                           

      Question 457
      CBSEENSC9006507

      Explain how defects in a metal block can be detected using ultrasound.

      Solution

      Ultrasound is generally used to clean parts located in hard-to-reach places, for example, spiral tube, odd shaped parts, electronic components etc. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned. 

      Question 458
      CBSEENSC9006508

      Explain how the human ear works.

      Solution

      i) Human ear is a highly sensitive part of the human body which enables us to hear a sound.
      ii) It converts the pressure variations in air with audible frequencies into electric signals which travel to the brain via the auditory nerve.

      iii) The human ear has three main parts. Their auditory functions are as follows:

      Fig. 12.27. Auditory parts of the human ear
      1. Outer ear: The outer ear is called ‘pinna’. It collects the sound from the surrounding. The collected sound passes through the auditory canal. At the end of the auditory canal there is a thin membrane called the ear drum or tympanic membrane.
      When compression of the medium produced due to vibration of the object reaches the ear drum, the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches. In this way the ear drum vibrates.

      2. Middle ear: The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear which act as levers. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear.

      3. Inner ear: In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.

      Question 467
      CBSEENSC9006517
      Question 469
      CBSEENSC9006519
      Question 478
      CBSEENSC9006528

      Define periodic motion. Give examples.

      Solution

      When the motion of a body repeats itself regularly after a fixed interval of time, the motion is called a periodic motion. 

      Some examples of periodic motion are:

      (i) Motion of a planet around the sun.

      (ii) Motion of the hands of a clock.

      (iii) Motion of a ball bering rotated in a circle at the end of a string. 

      Question 479
      CBSEENSC9006529

      Define oscillatory motion. Give examples.

      Solution

      When a body moves to and fro repeatedly about a fixed position, called it's mean position, the motion is said to be oscillatory or vibratory motion. 

      Examples of oscillatory motion are:

      (i) Motion of the pendulum of a wall clock.

      (ii) Motion of a mass attached to a spring.

      (iii) Motion of a swing.

      Question 480
      CBSEENSC9006530

      Can a motion be periodic but not oscillatory?

      Solution

      Yes, a motion can be periodic and not oscillatory. 

      For example, uniform circular motion is periodic but not oscillatory.

      Question 481
      CBSEENSC9006531

      Motion of the earth about its axis is periodic but not oscillatory. Justify.

      Solution

      Motion of the earth about its axis is periodic because earth takes 24 hours to complete its rotation about its axis. But the motion is not a to and fro motion, hence it is not oscillatory.

      Question 482
      CBSEENSC9006532

      Define the term oscillation or vibration.

      Solution

      The back and forth motion of a vibrating body around it's mean position is called an oscillation or vibration.

      Example: We say that the pendulum of a clock is oscillating. 



      Question 483
      CBSEENSC9006533

      Define the terms time period and frequency of an oscillating body. Give their units and write the relation between them. 

      Solution

      Time period: The time taken by an oscillating body to complete one oscillation is called its time period.

      It is denoted by T.

      SI unit of time period is second (s).

      Frequency: The number of oscillations or vibrations completed by an oscillating body in one second is called its frequency.

      Frequency is symbolically denoted by v (Greek letter nu).

      SI unit of frequency = per  socond (s-1)

                                     = cycles per second (cps)

                                     = hertz |(Hz.)


      Relation between time period and frequency:

      Let T be the time period of an oscillating body.

      Then number of oscillations completed in T second = 1

      So, number of oscillations completed in 1 second =1T

      But frequecny(ν) is the number of oscillations completed in 1 second.

      Therefore, ν = 1T 

      Hence, number of oscillations is equal to the recoprocal of time period. 

      Question 484
      CBSEENSC9006534

      Define the terms wave and wave motion. Explain the formation of waves with the help of an example.

      Solution

      Wave: The repeated vibrations of the particles of the medium, creates a disturbance which travels through a medium is called a wave. The disturbance moves from one particle to the next particle.

      Wave motion: The motion of the disturbance is called wave motion.

      Let us look at the wave formation using an example. 
      When we drop a pebble into a pond of still water, a circular pattern of alternate crests and troughs spreads out from the point where the pebble strikes the water surface. The particle oscillates as a result of kinetic energy of the pebble. The oscillating particles, in turn, transfer energy to the particles of next layer which also begin to oscillate. Energy is further transferred to the next layer of particle. This way energy is transferred from one particle to another particle. 


      Question 485
      CBSEENSC9006535

      State some important characteristics of wave motion.

      Solution

      Characteristics of wave motion are: 

      (i) Wave is the propagation of disturbance in the forward direction through the medium. The particles of the medium, merely vibrate about their mean positions.

      (ii) Each particle receives vibrations a little later than its preceding particle. 

      (iii) Velocity of the wave is different from the velocity of the particles with which they vibrate about their mean positions. 

      (iv) The wave velocity remains constant in a given medium while the particle velocity changes continuously during its vibration about mean position. 

      Question 486
      CBSEENSC9006536

      What are mechanical waves? Give examples. 

      Solution

      The waves which require a material medium for their propagation are called mechanical waves. The propagation of mechanical waves depend upon the elastic propertie so the medium. Hence, they are also called elastic waves. 

      Examples of mechanical waves :

      (i) Sound waves in air.

      (ii) Waves over water surface.

      (iii) Waves produced during earthquake also, known as seismic waves. 

      Question 487
      CBSEENSC9006537

      What are electromagnetic waves? Give examples.

      Solution

      The waves which do not require a material medium for their propagation are called electromagnetic waves. Such waves travel through vacuum with a speed of 3 x 108 m/s.

      Examples of electromagnetic waves :

      (i) Light waves 

      (ii) X-rays

      (iii) Radiowaves 

      (iv) Microwaves.

      Question 488
      CBSEENSC9006538

      Mention the important properties which a medium must possess for the propagation of mechanical waves through it?

      Solution

      Essential properties of a medium for the propagation of mechanical waves:

      (i) Elasticity: For the particles to return to their mean positions after being disturbed, the medium must possess elasticity.

      (ii) Inertia: The medium must possess inertia or mass so that particles can store kinetic energy.

      (iii) Minimum friction: The medium should have minimum frictional force between its particles so that they can continue oscillating for a sufficiently long time.

      Question 489
      CBSEENSC9006539

      What are transverse waves? Explain with an example.

      Solution

      Transverse waves are the waves in which particles of the medium vibrate in a direction perpendicular to the direction of wave motion.

      Fig. 12.2 (a). Pulse in a slinky.

      i) As shown in the fig. above, take a long slinky and attach it to a wall at its one end.
      ii) Give the other end an upward jerk. An upward hump is created in the slinky which travels along the slinky towards the fixed end. Such a sudden disturbance that lasts for short duration is called a pulse. 
      iii) As shown in Fig. 12.2, if we continuously give up and down jerks to the free end of the slinky, a number of waves begin to travel along the slinky forming a wave train. 

      iv) Each part of the slinky vibrates up and down while the waves travel along the rope. So the waves in the rope are transverse in nature.

      Fig. 12.2 (b). Formation of transverse waves in a slinky.

      Question 490
      CBSEENSC9006540

      What are longitudinal waves? Explain with an example.

      Solution

      Longitudinal waves:
      The waves in which particles of the medium vibrate along the direction of wave motion are called longitudnal motion.

      As shown in Fig. below
      Fig. Formation of longitudinal waves in a slinky.
      i) Consider a slinky held horizontally and its one end tied to a rigid support.
      ii) If we continuously give motion to its end slightly, the turns of the slinky begin to oscillate back and forth about their original positions along the length of the spring.
      iii) These turns alternately come closer together and move away from the one another so that there are alternate regions of increased and decreased pressure. These regions are called ‘compressions’ and ‘rarefactions’ respectively.
      iv) Thus, the waves in the slinky are longitudinal as the vibrations of the particles are along the direction of wave motion.

      Question 491
      CBSEENSC9006541

      Through what type of media, can

      (i) the transverse waves and
      (ii) the longitudinal waves be transmitted?

      Give reason.

      Solution

      Transverse waves travel in the form of crests and troughs. During the course of motion, they involve changes in the shape of the medium. So they can be transmitted through a medium which has elasticity of shape. As solids have elasticity of shape, so transverse waves can be transmitted through solids. 

      Longitudinal waves travel in the form of compressions and rarefactions. They involve changes in the volume and density of the medium. All media—solid, liquids and gases have elasticity of volume. Hence longitudinal waves can be transmitted through all the three types of media.

      Question 492
      CBSEENSC9006542

      Give four differences between transverse and longitudinal waves. 

      Solution

      Difference between transverse and longitudinal waves are as given below:

      Transverse waves

      Longitudinal waves


      1. In transverse waves, the particles of the medium vibrate perpendicular to the direction of wave motion. 

      1. In longitudinal waves, the particles of the medium vibrate along the direction of wave motion.

      2. These waves travel in the form of alternate crest and troughs.

      2. These waves travel in the form of alternate compressions and rarefactions.

      3. These waves can be transmitted through solids or over liquid surfaces. 

      3. These waves can be transmitted through all the three types of media, viz., solids, liquids and gases.

      4. They do not cause pressure changes in the medium through which they pass. 

      4. They cause changes in the pressure of the different parts of the medium through which they pass. 

      5. Examples:
      (i) Waves formed over water surface,
      (ii) Waves in stretched strings.

      5. Examples:
      (i) Sound waves in air,
      (ii) Waves formed along a compressed spring. 

      Question 493
      CBSEENSC9006543

      Distinguish between mechanical and electromagnetic waves.

      Solution

      Differences between mechanical and electromagnetic waves are:

      Mechanical waves

      Electromagnetic waves

      1. The waves require a material medium for their propagation.

      1. These waves do not require a material medium for their propagation.

      2. These are caused due to vibrations of the particles of the medium. 

      2. These are caused due to varying electric and magnetic fields.

      3. These waves have low speeds, e.g. speed of sound in air is 332 m/s at 0°C.

      3. These waves travel with a very high speed of 3 x 108 m/s through vacuum.

      4. These waves have usually low frequency and large wavelength.

      4. These waves have usually high frequency and low wavelength.

      5. These can be transverse or longitudinal.

      5. These are only trnasverse waves.

      Question 494
      CBSEENSC9006544

      What is sound and how is it produced?

      Solution

      Sound is a form of energy which produces a sensation of hearing in our ears. Vibrating any object produces sound.

      For example, when a bell is rung, we hear a sound. We can feel the vibrations by touching the ringing bell.
      Plucking a string of guitar or sitar produces sound. The vibrations of the string can be observed.

      The sound of human voice is produced due to the vibrations in the vocal cord.

      When a bird flaps its wings, again a sound is produced.

      Question 495
      CBSEENSC9006545

      Sound can make a light spot dance. Describe a simple experiment to illustrate this fact.

      Solution
      Experiment to show that sound can make a light spot dance. 

      Fig. 12.7. A beam of light from a light source is made to fall on a mirror. The reflected light is falling on the wall.
      i) Take a tin can. Remove both ends to make it a hollow cylinder.
      ii) Take a balloon and stretch it over the can, then wrap a rubber band around the balloon.
      iii) Now, take a small piece of mirror. Stick the piece of mirror to balloon using a glue stick. 
      iv) Allow the light to fall on the mirror through the slit. After reflection, we can see the light spot on the wall as seen in the fig. 
      v) Shout directly into the open end of the can and observe the dancing light spot on the wall.
      vi) The sound wave incident on the back of the mirror makes set it into vibration. As a result, the reflected spot of light appears to dance on the wall.
      Question 496
      CBSEENSC9006546

      How does the sound produced by a vibrating object in a medium reach your ear?

      Solution

      Sound moves through a medium from the point of generation to the listener. When an object vibrates, it sets the particles of the medium around it vibrating. The particles do not travel all the way from the vibrating object to the ear. A particle of the medium in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle. As a result of which the adjacent particle gets displaced from its position of rest. After displacing the adjacent particle the first particle comes back to its original position. This process continues in the medium till the sound reaches your ear. 

      Question 497
      CBSEENSC9006547

      Give an experiment to show that sound needs a material medium for its propagation.

      Solution

      Sound needs a material medium for propagation.
      Sound is a mechanical wave which needs a material medium to travel. It can travel through air, water, steel, etc. but cannot travel through vacuum.
      This can be demonstrated by the following simple experiment.

      i) Suspend an electric bell inside a glass bell jar by passing the connecting wires through an airtight cork fitted at the mouth of the jar.
      ii) Place the jar over a disc which has a pipe connected to a vacuum pump, as shown in Fig. 12.9.

      Fig. 12.9. Experiment showing sound needs a medium to travel.
      iii) When we turn on the switch, we hear the sound of the bell.
      iv) Now, with the help of a vacuum pump, we pump out the air from the jar. The sound becomes fainter.
      v) When most of the air has been removed, we hear a very feeble sound. When there is air inside the jar, sound travels through it to the wall of the jar. This makes the wall to vibrate which in turn, sends sound to us.
      vi) When air is removed, sound from the bell cannot travel to the wall of the jar.
      Therefore, we can see that the sound waves need material medium for propagation. 

      Question 498
      CBSEENSC9006548

      Explain how is sound produced by your school bell.

      Solution

      When the hammer is beaten on a school bell, it begins to vibrate and produces sound. We are able to feel the vibrations, if we gently touch the bell. Therefore, we can say that sound is produced by a vibrating body.

      Question 499
      CBSEENSC9006549

      Why are sound waves called mechanical waves?

      Solution

      Mechanical waves is characterised by the motion of the particles of the medium. Sound needs a material medium like air, water, steel, etc. for its propagation. Hence, sound waves are called mechanical waves.

      Question 500
      CBSEENSC9006550

      Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend ?

      Solution

      The moon has no atmosphere or material medium through which sound can travel. Hence, the sound produced by our body cannot be propagated while we are on the surface of the moon. 

      Question 501
      CBSEENSC9006551

      What is the nature of the sound waves?

      Solution

      Sound waves are longitudinal waves.
      When a sound wave passes through a medium, such as air, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles of the medium thus move to and fro along the direction of propagation of the sound waves.

      Question 502
      CBSEENSC9006552

      How can a transverse wave be represented graphically?

      Solution

      Graphical representation of a transverse wave.
      In the fig. given below, the curve represents displacement-distance graph for a transverse wave at any given instant of time.


      Fig. 12.11. Displacement-distance graph of a transverse wave.

      X-axis represents the direction of propagation of the wave.
      Y-axis represents the displacements of the particles from their mean positions.
      In a transverse wave, the displacements of the particles are perpendicular to the direction of wave motion.
      So, the displacement-distance graph is essentially the picture of the wave itself. 
      The distance between two successive crests or troughs is called wavelength.
      It is usually represented by λ (lambda). 

      Question 503
      CBSEENSC9006553

      How can a longitudinal wave be represented graphically?

      Solution

      Graphical representation of a longitudinal wave:

      Fig. Sound propagates as density or pressure variations as shown in (a), (b) and (c) represent graphically the density and pressure variations.
      i) When a longitudinal wave passes through a medium, the particles of the medium alternately come closer together and move away from one another. Thus, alternate regions of increased and decreased density is created. These regions are called compressions and rarefactions respectively.
      ii) In the figure below, (a) and (b) represent the density and pressure variations, respectively as the sound wave propagates through a medium. Fig. (c) represents the variations of density.
      iii) The variation of density increases or decreases as the pressure of the medium at a given time increases or decreases with distance, above and below the average value of density and pressure.

      iv) The distance between two successive compressions (C) or two rarefactions (R) is called wavelength. It is usually represented by λ (lambda).

      Question 504
      CBSEENSC9006554

      Define the terms amplitude, intensity, time period, frequency, wavelength and velocity of a wave. Give their SI units.

      Solution

      Sound waves are characterised by the following physical quantities:

      (i) Amplitude: The maximum displacement suffered by the particles of the medium from their mean positions during the wave propagation is called ‘amplitude’ of the wave.
      It is denoted by A.
      Its SI unit is metre (m).
      Crests and troughs have amplitude A.

      (ii) Intensity: The amount of sound energy passing each second through unit area is called the intensity of the sound wave.
      Larger the amplitude of a wave, larger is its intensity. In fact.
      Intensity space proportional to space space space left parenthesis Amplitude right parenthesis squared
Intensity space of space straight a space wave space equals space fraction numerator Energy over denominator Time space cross times space Area end fraction
therefore space space space SI space unit space of space intensity space space equals space fraction numerator straight J over denominator straight s space straight m squared end fraction space equals space Js to the power of negative 1 end exponent straight m to the power of negative 2 end exponent.

      Fig. 12.13. Characteristics of a wave.

      (iii) Time period: The time period of a wave is defined as the time in which a particle of the medium completes one vibration to and fro about its mean position.
      Time period is equal to the time taken by two successive compressions or rarefactions of a sound wave to cross a fixed point.
      It is represented by the symbol T.
      Its SI unit is second (s).

      (iv) Frequency: The number of waves produced per second is called frequency of the wave.
      It is equal to the number of compressions or rarefactions crossing a point per unit time. It is denoted by v (nu).
      It is equal to the reciprocal of time period.
      Thus,
      Frequency space equals fraction numerator 1 over denominator Time space period end fraction space
therefore space space space SI space unit space of space frequency space equals space 1 over second space equals straight s to the power of negative 1 end exponent space equals hertz space left parenthesis Hz right parenthesis

      Wavelength: Wavelength is the distance travelled by the wave during the time a particle of the medium completes one vibration.
      It is equal to the distance between two successive crests or troughs or between two successive compressions and rarefactions and is denoted by λ (lambda).
      Its SI unit is metre (m).

      Wave velocity: Wave velocity is the distance travelled by a wave per unit time.
      It is the speed with which a disturbance (crest, trough, compression or rarefaction) propagates through a medium.
      Its SI unit is ms–1.

      Question 505
      CBSEENSC9006555

      Q. 24. Define wave velocity, frequency and wavelength. Give SI unit of each term.

      Solution

      Wave velocity: Wave velocity is the distance travelled by a wave per unit time. It is the speed with which a disturbance of the particle like crest, trough, compression or rarefaction propagates through a medium.

      Its SI unit is ms–1.

      (iv) Frequency: The number of waves produced per second is called frequency of the wave. It is equal to the number of compressions or rarefactions crossing a point per unit time.

      Symbolically, it is denoted by v (nu).

      It is  equal to the reciprocal of time period.

      SI unit of frequency = hertz (Hz). 

      Wavelength: The distance between two successive crests or troughs or between two successive compressions and rarefactions is called the wavelength.

      It is denoted by λ (lambda).

      SI unit of wavelength is metre (m).


      Question 506
      CBSEENSC9006556

      Derive a relation between wavelength, frequency and velocity of a wave. 

      Solution

      Frequency, wavelength and wave velocity are related as follows: 
      Wavelength is the distance travelled by the wave during the time a particle of the medium completes one vibration.
      Therefore, if λ be the wavelength and T the time-period, then the wave travels a distance λ in time T.
      Hence, 
      Wave space velocity space equals space Distance over Time space space space
rightwards double arrow space space space space space space space space space space space space space space space space space space straight v space equals straight lambda over straight T
rightwards double arrow space space space space space space space space space space space space space space space space space straight u space equals space straight v space straight lambda space space space space space space space space space space space space space space space space space open square brackets because space space 1 over straight T space equals space frequency space left parenthesis straight v right parenthesis close square brackets

therefore space space Wave space velocity space equals space Frequency space cross times space space Wavelength
      The wave velocity in a medium remains constant under the same physical conditions. 

      Question 507
      CBSEENSC9006557

      Does the speed of sound through a medium depend on frequency?

      Solution

      No, the speed of sound remains almost same for all frequencies in a given medium under the same physical conditions. That is, speed of sound is independent of frequency. 

      Question 508
      CBSEENSC9006558

      Distinguish between particle velocity and wave velocity.

      Solution

      The distance travelled by a wave per unit time is known as the wave velocity. Wave velocity remains constant in a given medium. Wave velocity is given by,
                                        υ = vλ

      Particle velocity is the velocity of a particle in a medium as it transmits a wave. Particle velocity continuously changes with time. It is maximum at the mean position and zero at the extreme position.

      Question 509
      CBSEENSC9006559

      On what factor does the frequency of a wave depend

      Solution

      Frequency of a wave is given by, 

                                 ν = vλ 

      where, 

      v is the speed with which the wave propagates, and
      λ is the wavelength of the wave. 

      Different sources produce  oscillations of different frequencies depending on the wavelength of the sources. Frequency changes such that the speed remains constant. 

      Question 510
      CBSEENSC9006560

      What are the wavelength, frequency, time period and amplitude of a sound wave?

      Solution

      Wavelength(λ): Wavelength is defined as the distance between two consecutive compressions or rarefactions of a sound wave. 

      Frequency(v): Frequency is the number of complete oscillations per unit time.

      Time period(T): The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the sound wave.

      Amplitude(A): The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave.

      Question 512
      CBSEENSC9006562

      Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 ms–1 in a given medium.

      Solution

      Given, 
      Frequency of the wave, ν = 220 Hz 

      Speed of the wave, v = 440 ms-1

      So, wavelength is given by, λ =speed frequency= 440220= 2 m 

      Question 513
      CBSEENSC9006563

      A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source?

      Solution

      Given, 

      Frequency of the sound, ν = 500 Hz 
      distance of the person from the source = 450 m

      Therefore,

      Time interval between two successive compressions is given by, 

                      T = 1ν = 1500 =0.002 s 

      Question 515
      CBSEENSC9006565

      25 waves are produced per second, what is the frequency in Hz? 

      Solution
      Number of waves produced = 25 

      So, frequency = Number of wavestime taken= 251 = 25 Hz 
      Question 516
      CBSEENSC9006566

      What is the frequency of a wave time-period is 0.05 s?

      Solution

      Time period of the wave, T = 0.05 s


      So,
      Frequency of the wave, ν = 1T = 120 = 0.05 s

      Question 517
      CBSEENSC9006567

      A human heart, on an average, is found to beat 75 times a minute. Calculate its frequency.

      Solution

      Number of beats in 1 min or 60 s = 75

      Number of beats in 1 second = 7560=1.25

      Number of heartbeats in 1 second is the frequency of the wave. 

      Thererefore,

      Frequency of the heartbeat = 1.25  Hz

      Question 519
      CBSEENSC9006569

      Sound waves travel with a speed of about 330 m/s. What is the wavelength of sound waves whose frequency is 550 hertz?

      Solution

      Speed of sound, u = 330 m/s
      Frequency of sound waves = 550 Hz

      Wavelength comma space straight lambda space equals space straight u over straight v space equals space 330 over 550 space equals space 0.6 space straight m. space space

      Question 521
      CBSEENSC9006571

      A wave pulse on a string moves a distance of 8 m in 0.05 s.

      (a) Find the velocity of the pulse.

      (b) What would be the wavelength of the wave of the same string if its frequency is 200 Hz?

      Solution

      a) Given, 
      Distance travelled by the string, s  = 8 m
      Time taken for the pulse to move through the string, t = 0.05 s
      Velocity of the pulse, v = s/t 
      v=  fraction numerator 8 over denominator 0.05 end fraction space equals space 160 space straight m divided by straight s
      left parenthesis straight b right parenthesis space So comma space

Velocity space of space the space pulse comma space straight v space equals space 160 space straight m divided by straight s

Frequemcy space of space the space pulse comma space straight v space equals space 200 space Hz

therefore space Wavelength comma space straight lambda space equals space straight v over straight v space equals space fraction numerator 160 space straight m divided by straight s over denominator 200 space Hz end fraction space equals space 0.8 space straight m

      Question 524
      CBSEENSC9006574

      A source of wave produces 40 crests and 40 troughs in 0.4 second. Find the frequency of the wave.

      Solution
      Number of crests and trough produced = 40

      Time taken, t = 0.4 second

      One wave corresponds to 1 crest and 1 trough.

      Hence, 40 waves is equal to 40 crests and 4 troughs. 

      Therefore, 

      Frequency of the wave, 

      ν = Number of wavestime taken = 400.4 = 100 Hz
      Question 525
      CBSEENSC9006575

      A sound wave has a frequency of 2 kHz and wavelength 35 cm. How long will it take to travel 1.5 km?

      Solution
      Frequency comma space straight v space equals space 2 space kHz space equals space 2000 space Hz

Wavelength comma space straight lambda space equals space 35 space cm space equals space 0.35 space straight m

We space know space that space speed comma v space of space the space wave space equals space wavelength space cross times space frequency

space i. e. 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 space space space space space space space v space equals space λv

space space space space space space space space space space space space space space space space space space space space space space space space space 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 0.35 space straight m space cross times space 2000 space Hz space equals space 700 space straight m divided by straight s space space space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space
      The time taken by the wave to travel a distance, d of 1.5 km is, 
      straight t space equals space straight d over v space equals fraction numerator 1.5 space cross times space 1000 space straight m over denominator 700 space ms to the power of negative 1 end exponent end fraction space equals space 15 over 7 space straight s space equals space 2.1 space straight s.
      Thus sound will take 2.1 s to travel a distance of 1.5 km.
      Question 529
      CBSEENSC9006579

      The velocity of sound in a medium is 1400 m/s, and its wavelength is 1 km. (i) What is its frequency 1 (ii) Can you hear this sound?

      Solution
      left parenthesis straight i right parenthesis
V e l o c i t y space o f space s o u n d space w a v e comma space v space equals space 1400 space straight m divided by straight s

Wavelength comma space straight lambda space equals space 1 space km space equals space 1000 space straight m
therefore space space Frequency comma space space space space straight v space equals space v over straight lambda space equals 1400 over 1000 space equals space 1.4 space Hz.
      (ii) We cannot hear this sound because sound waves having frequency less than 20 Hz are not audible. 
      Question 530
      CBSEENSC9006580

      During a thunder storm, sound is heard 10 s after the lightning is seen. What is the distance of the cloud from the observer ? Take velocity of sound as 330 m/s.

      Solution

      Given,

      time lapse between thinder and lightning, t = 10 s

      Velocity of sound, v = 330 m/s

      So, distance travelled, s = vt 

                                          = 330 x 10

                                          =  3300 m 




      Question 531
      CBSEENSC9006581

      Find the velocity of the wave shown in the figure below.




      Solution
      Distance between two successive crests is λ, therefore
      λ = 2.5 m
      Now a wave travels a distance equal to wavelength in one time period.
      Therefore, time-period is 
      straight T space equals space 4.5 space minus space 2.5 space equals space 2 space ms space equals space 2 space cross times space 10 to the power of negative 3 end exponent straight s
Velocity space of space wave comma space v space equals space straight lambda over straight T space equals space fraction numerator 2.5 over denominator 2 space cross times space 10 to the power of negative 3 end exponent end fraction space equals space 1250 space straight m divided by straight s.
      Question 533
      CBSEENSC9006583

      Distinguish between the terms music and noise.

      Solution

      Music: Music is a pleasing sensation to the ears and is of rich quality. It is produced by regular and periodic vibrations, without any sudden change in loudness.

      Example: The sound produced by plucking the string of a sitar, sound from a tabla, etc.

      Noise: The sound which has a non-pleasing or jarrying effect on the ears is called noise.
      Noise is produced at irregular intervals and there is a sudden change in its loudness.

      Example: The sound produced by an explosion, sound from a market are examples of noise. 

      Question 534
      CBSEENSC9006584

      Distinguish between the terms tone and note.

      Solution

      A sound of single frequeny is called a tone.

      The sound which is a mixture of several frequencies is called a note.

      Question 535
      CBSEENSC9006585

      Explain the characteristics of sound. On what factors do they depend?

      Solution

      The three characteristics of sound are,
      (i) Loudness, (ii) Pitch and (iii) Quality or timbre.
      (i) Loudness: The physiological response of the ear to the intensity of sound is called as loudness.
      It distinguishes between a loud sound and low sound.
      Loudness depends on two main factors:
      (a) Intensity of sound which is directly proportional to the square of the amplitude of the sound wave.
      (b) Sensitivity of the ear.

      Fig. Soft sound has small amplitude and louder sound has large amplitude. The graph shows the wave shapes of a loud sound and a soft sound of the same frequency. 

      (ii) Pitch: Pitch is the sensation which helps a listener to distinguish betwen a high and a grave note. Pitch depends on frequency.
      The faster the vibration of the source of sound, the higher is the frequency and higher is the pitch.
      From the fig. below we can see that a high pitch sound corresponds to more number of compressions and rarefactions passing a given point per unit time.

      Fig. 12.17. Low pitch sound has low frequency and high pitch sound has high frequency.

      (iii) Quality or timbre: The characteristic which enables us, to distinguish one sound from another having the same pitch and loudness is called as the quality or timbre. 

      Question 536
      CBSEENSC9006586

      Which wave property determines

      (a) loudness, (b) pitch?

      Solution

      (a) Intensity of the wave determines the loudness.

      (b) Pitch is determined by the frequency of the wave. 

      Question 537
      CBSEENSC9006587

      Guess which sound has a higher pitch guitar or car horn?

      Solution

      The sound of a guitar is more shill than that of a car horn. Therefore, the sound of a guitar has a higher pitch. 

      Question 538
      CBSEENSC9006588

      Distinguish between loudness and intensity of sound.

      Solution

      Intensity: Intensity of sound wave is the amount of sound energy passing each second through a unit area.

      Loudness: It is a physiological response of the ear. Loudness depends both on the intensity of sound and the response of the ear towards that sound.

      Question 539
      CBSEENSC9006589

      On what factors does the speed of sound through a medium depend?

      Solution

      The speed of sound through a medium primarily depends on two factors:

      (i) Nature of the medium.

      (ii) Temperature of the medium.

      Question 540
      CBSEENSC9006590

      In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?

      Solution

      In different media, the sound of speed is different. The speed of sound depends on the density and volume of the medium. Elasticity of solids is much greater than that of liquids which in turn, is greater than that of gases. Therefore, sound travels fastest through solids and slowest through gases. 

      The speeds of sound in various media at 25°C are listed in the below table. 

      Table: Speed of sound in different media at 25°C

      State

      Substance

      Speed in ms–1

      Solids

      Aluminium

      6420

       

      Nickel

      6040

      Steel

      5960

      Iron

      5950

      Brass

      4700

      Glass (Flint)

      3980

      Liquids

      Water (Sea)

      1531

       

      Water (distilled)

      1498

      Ethanol

      1207

      Methanol

      1103

      Gases

      Hydrogen

      1284

       

      Helium

      965

      Air

      346

      Oxygen

      316

      Sulphur dioxide

      213

      Question 541
      CBSEENSC9006591

      In which of the three media : air, water or iron, does the sound travel the fastest at a particular temperature? 

      Solution

      The propagation of sound need variation in density and volume of the mediium. Solid (Iron) is the most elastic out of three. therefore, sound travels fastest through iron with a speed of 5950 ms–1

      Question 542
      CBSEENSC9006592

      State the factors on which the speed of sound in a gas depends.

      Solution

      The speed of sound in a gas depends on the following factors:

      (i) Effect of pressure: Pressure has no effect on the speed of sound in a gas.

      (ii) Effect of density: At constant pressure, the speed of sound in a gas is inversely proportional to the square root of its density.
      Speed space of space sound space proportional to space space fraction numerator 1 over denominator square root of Density space of space gas end root end fraction

      (iii) Effect of humidity: Sound travels faster in moist air than in dry air.

      (iv) Effect of temperature: The speed of sound in a gas is directly proportional to the square root of its absolute temperature.
      Speed space of space sound space proportional to space space fraction numerator 1 over denominator square root of Absolute space temperature space of space gas end root end fraction

      So, the speed of sound increases with the increase in temperature of the gas.
      For example, the speed of sound in air is 331 ms–1 at 0°C and 344 ms–1 at 22°C.

      (v) Effect of wind: Sound is carried by air, so the speed of sound increases when the wind blows in the direction of sound and speed of sound decreases when the wind blows in the opposite direction of sound.

      Question 543
      CBSEENSC9006593

      Why do we hear the sound of a horn of approaching car before the car reaches us?

      Solution

      We hear the sound of a horn of the approaching car first because the velocity of sound is much greater than that of the fastest car.

      Question 544
      CBSEENSC9006594

      Why is the flash of lightning from clouds seen much before the thunder, although both occur simultaneously?

      Solution

      The velocity of light which is equal to 3 x 108 ms-1 is much greater than the velocity of sound equal to 344 ms–1. Hence, we see the flash of lightning much before the sound of thunder reaches us.

      Question 545
      CBSEENSC9006595

      What do you understand by the terms supersonic speed, shock waves and sonic boom? 

      Solution

      Supersonic sound is when a body moves in air with a speed greater than the speed of sound.
      Example: Bullets, jet aircrafts etc. travel at supersonic speed. 

      When a source of sound travels in air at a supersonic speed, it leaves behind itself a conical region of disturbance which spreads continuously. A region consisting of a very high pressure layer followed by a lower pressure layer travels through the space together with the cone. This is called a shock wave. Shock wave carries a large amount of energy. 

      Shock waves have large variation in the air pressure. So, when a shock wave reaches a person, it produce a sharp and loud sound in his ears because of the differences in air pressure. This is called sonic boom.

      The sonic booms from a high speed jet aircraft can break glass dishes, windowpanes and can even cause serious damage to a building.

      Question 546
      CBSEENSC9006596

      What is reflection of sound? Does sound follow the same laws of reflection as light does?

      Solution

      When a sound wave travelling in a medium strikes the surface separating two media, a part of the incident wave is reflected back into the first medium. This is called reflection of sound waves.

      Like light, sound gets reflected at the surface of a solid or liquid. Same laws of reflection of sound is followed by reflection of sound also. 

      When sound is reflected, the directions in which the sound is incident and reflected make equal angles with the normal at the reflecting surface and the three are in the same plane. 

      Question 547
      CBSEENSC9006597

      From which types of surfaces the reflection of sound is better?

      Solution

      Reflection of sound will be heard better from the smooth surface. The reason is that the rough or porous surface allows for many internal reflections, resulting in more absorption and less reflection. 

      For example, sound reflected from a plastic tray is louder than that reflected from a cardboard. 

      Question 548
      CBSEENSC9006598

      What types of surfaces are needed for reflection of sound waves?

      Solution

      The types of surfaces needed for reflection of sound waves are: 

      i) obstacle of large size
      ii) A surface which is well polished or rough. 

      Question 549
      CBSEENSC9006599

      What is an echo? Name the conditions for the production of an echo? 

      Solution

      The phenomenon of repetition of a sound due to its reflection from the surface of a large obstacle is termed as echo.

      For example, when a loud sound is produced in an empty hall, the same sound is heard again after some time.  

      To produce an echo, the conditions required a are:  

      (i) There should be sufficient time gap between original and reflected sounds. So, if we want to hear distinct echo, the reflected sound must reach the ear 0.1 s after the direct original sound.

      (ii) The distance between the source of sound and the obstacle should be sufficient. An echo will be heard distinctly if the minimum distance between the source of sound and the obstacle is 34.4/2 = 17.2 m.

      (iii) Nature of the obstacle plays an important role. For the formation of echo, the reflecting surface must be rigid such as a building, hill or a cliff.

      (iv) Size of the obstacle should be quite large for the echoes to be heard. 

      Question 550
      CBSEENSC9006600

      What are multiple echoes? Give examples.

      Solution

      The successive reflection of a sound wave from a number of obstacles, causes us to hear echo of the sound transmitted one after other. When a sound is repeatedly reflected between two parallel distant buildings or cliffs, multiple echos are produced. 

      For example, rolling of thunder is due to successive reflections between clouds and land surfaces. 

      Question 551
      CBSEENSC9006601

      What is reverberation? How can it be reduced?

      Solution

      The waves suffer multiple reflections from the walls, ceilings and other materials present, when a sound is produced in a big hall or auditorium. As a result of this, sound persists for some time even after the source has stopped producing sound.
      Revertebration is the process of persistence of sound due to repeated reflection. 

      Methods of reducing reverberation.

      In a big hall or auditorium, excessive reverberation is highly undesirable. The following below methods can be used to reduce them.

      (i) By covering the walls and roof of the auditorium with sound absorbent materials like compressed fireboard, rough plaster, etc.

      (ii) Providing open windows in the space.

      (iii) Providing heavy curtains with folds, so that the sound gets absorbed. 

      (iv) By using good sound absorbing materials for the seats.

      Question 552
      CBSEENSC9006602

      An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms–1?

      Solution
      Given,

      Speed of sound, v = 342 ms-1

      Time taken for hearing the echo, t = 3 s 

      Distance traversed by the sound wave = vt

                                                                = 342 x 3 

                                                                =  1026 m

      In 3 s, sound has to travel twice the distance between the reflecting surface and the source.

      So, Distance of the reflecting surface from the source , d = 10262 = 513 m 
      Question 553
      CBSEENSC9006603

      A person clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the person if the speed of the sound, v is taken as 342 ms–1?

      Solution
      Speed of sound in air, v = 342 ms-1

      Time taken for hearing the echo, t = 3 s 

      Distance  travelled by the sound = vt

                                                       = 342 x 3

                                                       =  1026 m

      In 3 s, sound has to travel twice the distance between the the cliff and the person.

      Hence the distance between the cliff and the person is given by, d=  1730 m/2 = 865 m
      Question 554
      CBSEENSC9006604

      A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced. How far away is the cliff from the child?

      Solution
      Time taken for the echo to reach the child's ear = 4 seconds

      Therefore time taken by the sound to reach the cliff = 42 = 2 sec

      Speed of sound in air, v = 340 m/s

      Distance of cliff from the child = vt

                                                   =340  x 2

                                                   = 680 m.
      Question 555
      CBSEENSC9006605

      A stone is dropped into a well, 44.1 metres deep. The sound of the splash is heard 3.13 seconds after the stone is dropped. Find the velocity of sound in air.

      Solution

      The time t taken by the stone to reach the water level in the well by using the relation,
      straight s space equals space ut space plus space 1 half at squared space
We space have comma space

straight s space equals space 500 space straight m space comma space straight u space equals space 0 comma space straight g space equals space 10 space straight m divided by straight s squared space

That space is comma space

space space space space space space space space 500 space equals space 0 plus space 1 half cross times 10 cross times straight t squared space
rightwards double arrow space space space space space straight t squared space space equals space 500 over 5 space equals space 100 space

That space is comma space straight t space equals space 10 space sec space

Time space taken space by space sound space to space travelfrom space base space to space the space tower space
to space it apostrophe straight s space top

straight t apostrophe space equals space fraction numerator Distance over denominator Speed space of space sound end fraction space equals space 500 over 340 equals space 1.47 space straight s

Total space time space after space which space the space splash space is space heard comma space

straight t subscript 1 space equals straight t space plus space straight t to the power of apostrophe
space space space space equals space 10 space plus space 1.47 space
space space space space equals space 11.47 space sec

      Question 556
      CBSEENSC9006606

      Describe some practical applications from daily life based on multiple reflection of sound.

      Solution

      Uses of multiple reflection of sound:

      1. Megaphones or loudhailers, horns, musical instruments such as trumpets and shehanais, are all designed to send sound in a particular direction without spreading it in all directions.
      In these instruments, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.
      The fig. below shows us some musical instruments. 

      Fig. Multiple reflection of sound in a megaphone or horn.

      2. Stethoscope: In a stethoscope, the sound of a patient’s heartbeat is guided along the tube of the stethoscope to the doctor’s ears by multiple reflections of sound.

      Fig. Mutliple reflections of sound in the tube of a stethoscope.

      3. Curved ceilings concert halls The ceilings of concert halls, conference halls and cinema halls are made curved so that sound after reflection reaches all corners of the hall. 

      Fig. 12.21. Curved ceiling of a conference hall

      4. Sound boards: In large halls or auditorium, large concave wooden boards are placed behind the speaker. The speaker stands at the focus of this concave reflecting surface. After reflection, the sound is spread evenly towards the audience. This makes the speech readily audible even at a distance.

      Fig. 12.22. Sound board used in a big hall.

      Question 557
      CBSEENSC9006607

      Why are the ceilings of concerts halls curved?

      Solution

      The ceilings of the concert halls are made curved. This is done so that the sound after reflecting from the ceiling reaches all corners of the hall, as shown in Fig. below. 

       
      Fig.  Curved ceiling of a concert hall.

      Question 558
      CBSEENSC9006608

      What is the audible range of the average human ear?

      Solution

      In human beings, frequency range of hearing from 20 Hz to 20,000 Hz is called audible range. Human ear is sensitive to frequencies between 20 Hz to 20,000 Hz.

      Human ears cannot hear sounds of frequencies less than 20 Hz and above 20,000 Hz. 

      Question 559
      CBSEENSC9006609

      What are infrasonic and ultrasonic waves? Give examples. 

      Solution

      Infrasonic waves: If the frequency of the waves is less than 20 Hz then the waves are called infrasonic waves. Infrasonic waves are produced by large vibrating bodies. 

      The waves produced before the actual shockwaves during an earthquake is infrasonic waves. Animals ike whales and elephants produce infrasonic waves. Human beings cannot hear these waves. 

      Ultrasonic waves: When the frequency of the waves is greater than 20,000 Hz they are called ultrasonuic waves. 

      Bats, dogs, rats, cats, birds and insects can produce and also hear these waves. 

      Question 560
      CBSEENSC9006610

      What types of waves are used by rhinoceroses during communication?

      Solution

      Rhinoceroses communicate using infrasonic waves of frequency as low as 5 Hz. 

      Question 561
      CBSEENSC9006611

      Why do the animals behave in a peculiar manner before an earthquake?

      Solution

      When the earth quakes, they produce low frequency infrasound before the main shock waves. These infrasonic waves are not audible to the human ears. But animals are able to detect these waves and hence some animals get disturbed before earthquakes and start behaving in a different manner.


      Question 562
      CBSEENSC9006612

      What is a hearing aid ? Briefly describe its working principle.

      Solution

      A battery operated electronic device used by persons who cannot hear properly is called as hearing aid. 

      Working of the hearing aid is as follows:

      Hearing aid receives sound through a microphone. The microphone converts the sound waves to electrical signals. These electrical signals are amplified by an amplifier. The amplified electrical signals are given to a speaker of the hearing aid. The speaker converts the amplified electrical signal to sound and sends to the ear for clear hearing. 

      Question 563
      CBSEENSC9006613

      What is the audible range of the average human ear?

      Solution

      The audible range of frequency extends from 20 Hz to 20,000 Hz, for an average human ear. 

      Question 564
      CBSEENSC9006614

      What is the range of frequencies associated with,

      (a) Infrasound?

      (b) Ultrasound? 

      Solution

      (a) Sound of frequency less than 20 Hz is called infrasound.

      (b) Sound of frequency higher than 20 kHz is called ultrasound.

      Question 565
      CBSEENSC9006615

      What is ultrasound? State the special properties of ultrasound that make it useful to us.

      Solution

      Ultrasound or ultrasonic wave is a wave in which the compressions and rarefactions in a material medium propagates with a frequency above 20 kHz.

      Ultrasonic wave travels freely in solids and liquids but its intensity decreases rapidly in gases. 

      Ultrasonic waves have the following two important properties, because of their high frequencies:

      1. They have high power.

      2. They are able to travel along well-defined straight paths, even in the presence of obstacles. 

      The above two properties make the ultrasonic waves highly useful for industrial and medical purposes.

      Question 566
      CBSEENSC9006616

      Explain some important applications of ultrasound in industry and medicine. 

      Solution

      Industrial applications of ultrasound in industry and medicine are: 

      Industrial application

      1. Ultrasonic detection of defects in metals: The construction of big structures like buildings, bridges, machines, scientific equipments, etc use metallic components. The cracks or holes inside the block reduce the strength of the structure. Such defects are not visible from outside. Ultrasonic waves can be used to detect such defects. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back and does not reach the detector. This is how the presence of a flaw or defect is detected. 

      Medical application:

      1. Echocardiography: In this technique, ultrasonic waves are made to reflect from various parts of the heart to form the image of the heart.

      2. Ultrasonography: Ultrasonic waves can be used to develop three dimensional photographs of different parts of the human body. This technique is called ultrasonography.

      This technique is also used to monitor the growth of a foetus inside its mother’s womb. In this technique the ultrasonic waves travel through the tissues of the body and get reflected from a region where there is a change of tissue density. These waves are then converted into electrical signals that are used to generate images of the organ. These images are then displayed on a monitor or printed on a film.

      3. In surgery: Ultrasonic waves are commonly used in cataract removal. Ultrasonic waves are used to grind small stones formed in the kidneys. These grinded grains are flushed with urine. 

      Question 567
      CBSEENSC9006617

      What is a SONAR? Briefly explain its working principle.

      Solution

      SONAR- The acronym SONAR stands for Sound Navigation and Ranging.

      Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.

      Principle: It uses the phenomenon of echoes in determining the sea-depth and locating the presence of under-water objects.

      Working:
      i) A strong beam of ultrasonic waves is sent from a transmitter mounted on the ship.
      ii) The beam is reflected from the seabed and is received by an under-water detector which is also mounted on the ship.
      iii) The time interval between transmission and reception of the ultrasonic signal is noted. 

      Question 568
      CBSEENSC9006618

      What is a SONAR? Briefly explain its working principle.

      Solution

      SONAR- The acronym SONAR stands for Sound Navigation and Ranging.

      Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. SONAR is also called as echo ranging.

      Principle: It uses the phenomenon of echoes in determining the sea-depth and locating the presence of under-water objects.

      Working:
      i) A strong beam of ultrasonic waves is sent from a transmitter mounted on the ship.
      ii) This beam is then reflected from the seabed and is received by an under-water detector which is also mounted on the ship.
      iii) The time interval between transmission and reception of the ultrasonic signal is noted.

      Fig. Ultrasound sent by a transmitter and received by a detector.

      iv) If ultrasonic waves travel with speed v in sea-water and time t is elapsed between the transmission and reception of the ultrasound signal, then the sea depth will be
                       space straight d space equals space fraction numerator straight v cross times straight t over denominator 2 end fraction

      Uses:
      SONAR technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc. 

      Question 569
      CBSEENSC9006619

      A submarine emits a SONAR pulse, which returns from an obstacle underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff? 

      Solution

      Given, 

      Time between transmission and detection of sonar pulse, = 1.02 s  

      Speed of sound in salt water, v=  1531 ms-2

      Let, distance of the cliff be equal to d (say).

      Then the distance travelecd by sound = 2d 

      Now, using the distance-time formula, we have

                  2d = speed of sound x time

                       = vt 

                       = 1531 x 1.02 m 



      So, distance of the cliff, d = 1531×1.022 = 780.81 m

      Question 570
      CBSEENSC9006620

      A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the distance of the seabed from the ship?

      Solution

      Time between transmission and direction, t = 3.42 s

      Speed of ultrasound in sea water, v = 1531 m/s 

      Let, the depth of the sea be 'd'. 

      Distance travelled by the ultasound = 2 x depth of the sea = 2d 

      Therefore,
                      2d  = speed of sound X time

                            = 1531 m/s x 3.42 s

                            = 5236 m  

      Therefore,

      Distance of the seabed from the ship, d =52362 = 2618 m = 2.62 km

      Thus, the distance of the seabed from the ship is 2.62 km. 

      Question 571
      CBSEENSC9006621

      Using the SONAR, sound pulses are emitted at the surface of water. These pulses after being reflected from bottom of water, are detected. If the time interval from the emission to the detection of the sound pulses is 2 seconds, find the depth of the water.

      Solution

      Time between the transmission and detection of the pulses = 2 seconds

      Speed  of sound in water, v = 1450 m/s

      Time taken by the pulse to travel to the bottom of water = 2/2 sec = 1 sec 

      Therefore, using the distance-time formula, we have

      Depth of water = v×t = 1450 x 1 = 1450 m.

      Question 572
      CBSEENSC9006622

      Define a vibration.

      Solution

      Vibration or oscillation is defined as one complete back and forth motion of an oscillating body about its mean position. 

      Question 574
      CBSEENSC9006624

      How does the motion of the particles of the medium differ from that of a wave through it?

      Solution

      Particles of the medium simply vibrate around their mean position whereas a wave travels through different parts of the medium with a definite velocity. 

      Question 575
      CBSEENSC9006625

      What are the two types of wave motions? How do the vibrations of the particles differ during the propagation of the two types? 

      Solution

      The wave motion is of two types namely: 

      (i) Transverse wave motion, and
      (ii) Longitudinal wave motion

      In transverse wave motion, the particles of the medium vibrate in a direction perpendicular to the direction of wave motion

      In longitudinal wave motion, the particles of the medium vibrate along the direction of wave motion. 

      Question 576
      CBSEENSC9006626

      What is a transverse wave?

      Solution

      Transverse waves are the waves in which the particles of the medium vibrate in directions perpendicular to the direction of wave motion.

      For example, when a stone is thrown in a pond of water, the waves formed over water surface are transverse waves. 

      Question 577
      CBSEENSC9006627

      What are longitudinal waves? Give two examples.

      Solution

      Longitudnal waves are the waves in which the particles of the medium vibrate along the direction of wave motion.

      Sound waves in air, the waves produced along the length of a compressed spring are all examples of longitudnal waves. 

      Question 578
      CBSEENSC9006628

      Define the terms crests and troughs of wave.

      Solution

      The point or the maximum displacement in the positive y-direction, to which the particle of the medium rises to are called crests.

      The particles of the medium has maximum displacement in the negative y-direction. These valleys are called troughs. 


      Question 579
      CBSEENSC9006629

      Define the terms compressions and rarefactions of a wave.

      Solution

      A longitudnal wave passes through a medium, and causes pressure variations in different parts of the medium. Compressions are the regions of increased pressure and the regions of decreased pressure are called rarefactions. 

      Question 582
      CBSEENSC9006632

      Are sound waves in air longitudinal or transverse?

      Solution

      Longitudinal waves.

      Question 583
      CBSEENSC9006633

      When the wire of a sitar is plucked, what type of waves are produced in

      (i) the wire, and

      (ii) air?

      Solution

      (i) Transverse waves in the wire.

      (ii) Longitudinal waves in the air.

      Question 584
      CBSEENSC9006634

      Define wavelength.

      Solution

      Wavelength is the distance between successive crests of a wave, in a time the particle of the medium completes one vibration.

      Question 585
      CBSEENSC9006635

      What is the SI unit of the wavelength of a wave?

      Solution

      SI unit of wavelength is metre. 

      Question 586
      CBSEENSC9006636

      Define frequency.

      Solution

      Frequecy of wave is the number of waves produced per second. 

      Question 587
      CBSEENSC9006637

      Tuning fork is marked with number 512. What does this number signify?

      Solution

      The number 512 indicates the frequency of the tuning fork. When struck against a rubber pad, the tuning fork will make 512 vibrations in one second.   

      Question 588
      CBSEENSC9006638

      Define velocity of a wave.

      Solution

      Velocity of a wave is the distance travelled by the wave in one second. 

      Question 589
      CBSEENSC9006639

      What is the relationship between time-period and frequency?

      Solution

      Relationship between time period and frequency is given by: 

                                         T = 1ν

      Time-period (T) is equal to reciprocal of frequency (v). 

      Question 590
      CBSEENSC9006640

      What is called the reciprocal of time-period?

      Solution

      Frequency. 

      Question 591
      CBSEENSC9006641

      What is the SI unit of frequency?

      Solution

      Hertz (Hz).

      Question 592
      CBSEENSC9006642

      What is the relationship between frequency, wavelength and wave velocity?

      Solution

      Relationship between frequency, wavelength and wave velocity is given as, 

             Wave velocity = Frequency x Wavelength. 

      Question 593
      CBSEENSC9006643
      Question 595
      CBSEENSC9006645

      What is the distance between a crest and the next trough?

      Solution

      The distance between a crest and the next trough is  h alf space the space wavelength space open parentheses straight lambda over 2 close parentheses. 

      Question 596
      CBSEENSC9006646

      Is the velocity of a wave same as the velocity of the particles of the medium?

      Solution

      No, particle velocity and wave velocity are both different. Velocity of a wave is constant in a given medium while particle velocity changes continuously during its vibration about mean position. 

      Question 597
      CBSEENSC9006647

      What is mechanical elastic wave?

      Solution

      Mechanical elastic wave is a wave which requires an elastic medium for its propagation. 

      Solid has a good elasticity strength. 

      Question 598
      CBSEENSC9006648

      Can sound travel through vacuum?

      Solution

      No, sound cannot travel in vacuum. Sound requires a medium for its propagation. Large obstacles are required for the reflection of sound waves. 

      Question 599
      CBSEENSC9006649

      Why a sound cannot be heard on the moon?

      Solution

      The surface of the moon has no atmosphere and hence the sound waves cannot be reflected in vacuum. Sound needs a medium for propagation.  

      Question 600
      CBSEENSC9006650

      What is the nature of sound waves in air?

      Solution

      Sound travels through air in the form of longitudinal waves.

      Question 601
      CBSEENSC9006651

      Mention two points of distinction between elastic waves and electromagnetic waves.

      Solution

      Difference between elastic waves and electromagnetic waves are as follows: 

      i) Elastic waves require a medium for their propagation while electromagnetic waves do not require any medium for their propagation.

      (ii) Elastic waves may be longitudinal or transverse while electromagnetic waves are only transverse in nature. 

      Question 602
      CBSEENSC9006652

      What is the range of human audibility?

      Solution

      Human ear can hear Sounds waves having frequencies between 20 Hz and 20,000 Hz. This range of frequency is called human audible range.  

      Question 603
      CBSEENSC9006653
      Question 604
      CBSEENSC9006654

      What is a wave-pulse?

      Solution

      Wave pulse is a wave of short duration. It is a non-sinusoidal waveform, which is similar to a square waveform. 

      Question 605
      CBSEENSC9006655

      Define wave number.

      Solution

      The number of waves contained in unit length of the medium is called ‘wave number’. 

      Question 606
      CBSEENSC9006656

      What is an echo?

      Solution

      The phenomenon of repetition of sound due to its reflection from the surface of a large obstacle is called Echo.

      Question 607
      CBSEENSC9006657

      Suppose you and your friend land on the moon. Would you and your friend be able to talk each other? Give reason.

      Solution

      No, the friends won't be able to talk to each other because the moon has no atmosphere or material medium through which sound can travel. Sound waves cannot travel in vacuum. 

      Question 608
      CBSEENSC9006658

      Why do we hear sound of a horn of an approaching car before the car reaches us?

      Solution

      We can hear the sound of a horn before actually seeing the car because the velocity of sound is much greater than that of the fastest car. 

      Question 609
      CBSEENSC9006659

      State the characteristics of longitudinal waves.

      Solution

      The characteristics of longitudinal waves are as follows: 

      (i) The vibrations of the particles of the medium are along the direction of wave motion.

      (ii) They travel in the form of alternate compressions and rarefactions causing pressure variations in different parts of the medium.

      (iii) Longitudnal waves can be transmitted through all the three types of media, viz., solids, liquids and gases. 

      Question 610
      CBSEENSC9006660

      Can transverse waves be produced in air? Give reason.

      Solution

      No, air has no elasticity of shape. So, transverse waves cannot be produced in air. 

      Question 611
      CBSEENSC9006661

      Why we cannot hear an echo in a small room?

      Solution

      The length of a room is generally less than 17 m. And, inorder to produce an echo, minimum distance between a source of sound and the obstacle is 17 m. Hence, we do not hear an echo in a small room. 

      Question 612
      CBSEENSC9006662

      What does cause the rolling sound of thunder?

      Solution

      The rolling sound of thunder is caused by the multiple reflections of the sound of lightning between the clouds and the earth’s surface. 

      Question 613
      CBSEENSC9006663

      Why are the longitudinal waves also called pressure waves?

      Solution

      Longitudinal waves travel in a medium as series of alternate compressions and rarefactions. That is as they travel, there is variations in pressure. Therefore, they are called pressure waves. 

      Question 614
      CBSEENSC9006664

      Why is it possible to detect the approaching of a distant train by placing the ear very close to the railway line?

      Solution

      We can detect a train approaching by placing the ear very close to the railway line because sound travels sixteen times faster in iron than in air.

      Question 615
      CBSEENSC9006665

      When a stone is thrown on the surface of water, a wave spreads out. From where does the energy come? 

      Solution

      The kinetic energy of the stone imparts energy on the water molecules, on which it falls. So, the energy of the waves spreading on the water surface is seen as ripples. 

      Question 616
      CBSEENSC9006666

      Why explosions on other planets cannot be heard on the earth?

      Solution

      Explosions cannot be heard on planets because there is no material medium in the space between the planets. Sound needs a material medium for propagation. Hence, we do not hear explosions happeing on a different planet. 

      Question 617
      CBSEENSC9006667

      List any two differences between transverse and longitudinal waves. What is the nature of sound waves in air.

      Solution

      (i) In transverse waves, the particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave.

      In longitudinal waves, the particles of the medium vibrate along the direction of propagation of the wave. 

      (ii) Transverse waves travel in the form of crests and troughs.

      Longitudinal waves travel in the form of compressions and rarefactions. 

      Sound waves are longitudnal waves in nature.


      Question 618
      CBSEENSC9006668

      Define waves. Classify waves on the basis of material medium. Give one example of each. 

      Solution

      The disturbance arising due to repeated vibrations of the particles of the medium, moving from one particle to another is known as a wave. 

      Waves can be classified as follows, on the basis of the medium in which they travel. 

      1. Mechanical/elastic waves: The waves which require a material medium for their propagation are known as mechanical waves.

      e.g., sound waves.

      2. Electromagnetic waves: Waves which do not require any material medium for their propagation.

      e.g., light waves. 



      Question 619
      CBSEENSC9006669

      Give some examples of waves which are combination of both transverse and longitudinal wave motion. 

      Solution

      A wave has both transverse and longitudnal waves, when it propagates along the surface of a medium. The particles move either along a circle or an ellipse. 

      Examples of such waves are: 

      (i) Seismic waves travelling along the surface of the earth during an earthquake or explosion. 

      (ii) Ocean waves in deep water.

      (iii) Wave pulses produced by beetle’s motion along the sand’s surface. 

      Question 620
      CBSEENSC9006670

      How can bats ascertain distances, directions, nature and size of the obstacles without eyes? 

      Solution

      Ultrasonic waves are produced and received by bats during it's flight. After being reflected by the obstacle in it's path, the bast receives these waves. From the time interval between transmission and reception of ultrasonic waves, the bat gets information about the distance, nature of obstacle and its direction of location. Hence bats can move about freely even in total darkness. 

      Question 622
      CBSEENSC9006672

      Distinguish between a wave pulse and a periodic wave. 

      Solution

      Pulse

      Periodic wave

      1. A pulse is a wave produced by a sudden disturbance of short duration. 

      1. A periodic wave produced by continuous and regular vibrations of the particles of the medium.

      2. The medium oscillates for a short time and then returns to its undisturbed position. 

      2. The medium vibrates for a long time after being disturbed, in a periodic wave.

      3. The wave forms are not repetitive.

      3. The wave form has a basic pattern which repeats itself after a fixed interval. 

      4. It is formed in a small portion of the medium. 

      4. It is stretched over the entire length of the medium.

      Question 623
      CBSEENSC9006866

      What is meant by a substance?

      Solution

      A substance which consists of a single type of particle and it cannot be separated into other kinds of matter by any physical process is called a substance.

      Question 624
      CBSEENSC9006867

      List the points of differences between homogeneous and heterogeneous mixtures.

      Solution

      Homogeneous mixture

      Heterogeneous mixture

      It is the mixture in which different constituents are mixed uniformly.

      It is the mixture in which different constitutes are not mixed uniformly

      They cannot have physically distinct parts.

      They have physically distinct parts.

      The constituents cannot be easily seen.

      The constituents can be easily seen.

      The constituents cannot be easily separated.

      The constituents can be easily separated.

      Example: sugar solution, soda water, soft drinks, vinegar, air etc.

      Example: sugar and sand mixture, milk, ink, paint, etc.

      Question 625
      CBSEENSC9006868

      Differentiate between homogenous and heterogeneous mixtures with examples.

      Solution

      s.no

      Homogenous mixture

      Heterogeneous mixture

      1.

      It has a uniform composition throughout its mass

      It has no uniform composition.

       

      2.

      It has no visible boundary or boundaries of separation between its constituents. For example solution of sugar and salt.

      It has visible boundary or boundaries of separation. For example mixture of sugar and sand.


      Question 626
      CBSEENSC9006869

      To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature. 

      Solution

      Mass of sodium chloride = 36g
      Mass of solution =100g
      Mass of the solution = 100+36g =136g
      Concentration of the solution = 36/136 x 100 =26.47%

      Question 627
      CBSEENSC9006870

      How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than  25 degree straight C), which are miscible with each other?

      Solution

      We can separate a mixture containing kerosene and petrol by simple distillation because kerosene and petrol do not decompose on heating and their boiling points are sufficiently apart. The apparatus used for the simple distillation is given below.

      Question 628
      CBSEENSC9006871

      Name the technique to separate:
      (i) butter from curd,

      (ii) salt from sea-water

      (iii) camphor from salt.

      Solution

      (i) Butter from curd = centrifugation
      (ii) Salt from sea-water = crystallization
      (iii) Camphor from salt = Sublimation

      Question 629
      CBSEENSC9006872

      What type of mixtures are separated by the technique of cyrstallisation?

      Solution

      Crystallization is used to purify salt that we get from sea water and to obtain pure substance from their impure samples.

      Question 630
      CBSEENSC9006873

      Classify the following as chemical or physical changes:
      (i) cutting of trees.

      (ii) melting of butter in a pan.

      (iii) rusting of almirah.

      (iv) boiling of water to form steam.

      (v) passing of electric current, through water and the water breaking down into hydrogen and oxygen gases.

      (vi) dissolving common salt in water.

      (vii) making a fruit salad with raw fruits, and

      (viii) burning of paper and wood.


      Solution

      (i) Cutting of tree is a physical change.
       
      (ii) Melting of butter in a pan is a physical change.
      (iii) Rusting of almirah is a chemical change. 
      (iv) Boiling of water to form steam is a physical change.
      (v) Passing of electric current through water and the water breaking down into hydrogen and oxygen gases is a chemical change. 
      (vi) Dissolving common salt in water is a physical change. 
      (vii) Making a fruit salad with raw fruits is a physical change.
      (viii) Burning of paper and wood is a chemical change.

      Question 631
      CBSEENSC9006895

      A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s–1.

      Solution

      For a sound wave,
      Speed  = wavelength x frequency
      i.e.,                space straight v space equals space λv space
      Given that, the speed of sound in air = 344 m/s
      (i) For, v = 20 Hz,
                     straight lambda subscript 1 space space equals space straight v over straight v subscript 1 space equals space 344 over 20 space equals space 17.2 space straight m 
      (ii) For v = 20,000 Hz
                                     straight lambda subscript 2 space equals space straight v over straight v subscript 2 space equals space 344 over 20000 space equals space 0.0172 space straight m 
      Hence, for the humans the wavelength range for hearing is 0.0172 m to 17.2 m. 

      Question 632
      CBSEENSC9006896

      2 ml of dettol is added to a beaker containing 500 ml of water and stirred. State the observations that you can make.

      Solution

      1)The solution becomes colloidal in nature.
      2) The 2 ml dettol mixed with 500 ml of water and formed homogenous solution.
      3) A white turbid colour appear when dettol is mixed with water.

      Mock Test Series

      Sponsor Area

      Sponsor Area

      NCERT Book Store

      NCERT Sample Papers

      Entrance Exams Preparation

      3