Sponsor Area
Adhesive force is the attractive force between two unlike molecules. They are caused by forces acting between two substances like electrostatic force (attraction due to opposing charges), mechanical force etc.
E.g., Force of attraction between gum and paper.
Inter-atomic or inter-molecular forces are electrostatic in nature.
The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity.
When a body is stretched by two equal forces applied normal to its cross-sectional area, the dimension of the body is increased.The restoring force per unit area in this case is called tensile stress.
If a body is compressed under the action of applied forces resulting in decrease in it's dimension. The restoring force per unit area is known as compressive stress.
Sponsor Area
When two equal and opposite deforming forces are applied parallel to the cross-sectional area of an object, there is relative displacement between the opposite faces of the body. The restoring force per unit area developed due to the applied tangential force is known as tangential stress.
Which of the followings are dimensionless?
(i) Longitudinal strain
(ii) Poisson’s ratio
(iii) Shear strain
(iv) Volumetric strain
Sponsor Area
Tips: -
Whenever a deforming force is applied on a body, the shape of the body changes. Strain is defined as the ratio of change in configuration to the original configuration.
Different types of strain are:
(i) Longitudinal strain: The deformation produced in the length of the body as a result of the deforming force is longitudnal strain.
So, longitudinal strain is defined as the ratio of change in length (∆L) of body to its original length L.
(ii) Volumetric strain: When the deforming force produces a change in volume only then the strain produced in the body is called volumetric strain. The volumetric strain is defined as the ratio of change in volume (∆V) of the body to its original volume V.
(iii) Shearing strain: When only the shape of the body deforms without any change in it's volume, it is shear strain.
So, shearing strain is defined as the ratio of the relative displacement of the opposite faces of the body to the length of the cylinder L.
All types of strains are dimensionless.
The elasticity of a material is affected by the following factors:
(i) Effect of temperature: On heating, mostly the elasticity of materials decreases.
(ii) Effect of impurities: Depending upon the nature of impurity, the elasticity of materials may increase or decrease.
(iii) Annealing: Annealing decreases the elasticity of materials.
(iv) Hammering and rolling: This process increases the elasticity of materials.
Note: Annealing is a heat process where a metal is heated to a specific temperature /colour and then allowed to cool slowly. This softens the metal which means it can be cut and shaped more easily. Iron rod is heated till it turns red-hot.
Plot the graph between intermolecular forces. What is the range of these forces? What is the range of these forces?
Sponsor Area
Let a wire of length l, diameter d be loaded by force F.
Let dl be the change in length and dV the change in volume, and dD be the change in the diameter.
Volume of wire before the force is applied is,
Volume of wire after the force is applied is,
Change in volume is,
dV = V' - V
(a) Upto what point on the curve is the Hooke’s law obeyed? (This point is sometimes called ‘Proportional limit’).
(b) Which point on the curve corresponds to ‘elastic limit’ or yield point of the wire?
(c) Indicate the elastic and plastic regions of the stress-strain graph.
(d) Describe what happens when the wire is loaded upto a stress corresponding to the point A on the graph, and then unloaded gradually. In particular, explain the dotted curve.
(e) What is peculiar(odd) about the portion of the stress-strain graph from C to B? Upto what stress can the wire be subjected without causing fracture?
(a) Hooke's law states that that stress is directly proportional to strain. From the graph, we can say that Hooke’s law is obeyed from point O to P.
(b) E point on the curve corresponds to elastic limit or the yield point.
(c) Elastic region is the region in which the body regains its state after removal of stress and is from O to E. Plastic region is the region in which there is residue strain after removing the stress and that region is E to B.
(d) From O to P, the strain is directly proportional to stress, beyond P increase in strain is more for same increase in stress. Beyond E, when stress is increased, it does not retrace its path on decreasing the stress, but returns to point O' along path AO'. Therefore there is residue strain of 00' when stress is increased corresponding to the point A.
(e) From C to B, strain increases even if the wire is being unloaded and at B, it fractures. Till point C, stress can be applied without causing the fracture.
The stress-strain graphs for materials A and B are as shown in figure. Which of the material:
(i) has greater value of Young's modulus,
(ii) is more ductile or more brittle?
(i) For graph A, slope of OE is more than that of graph B. Therefore, Young's modulus of material A is greater than that of B.
(ii) Material A shows wider plastic region than that of B. Thus, A is ductile than B and B is more brittle than A.
At point A, the beam undergoes tensile strain, therefore hole at A will increase in size along the length of the beam and becomes oblate.
At point C, the beam undergoes compression strain, therefore hole at C will decrease in size along the length of beam and becomes prolate.
The point B is in neutral zone, therefore the hole at B will be circular.
Radius of the wire is given by r and 2r.
The breaking force of wire is given by,
F = σA
where,
σ is the breaking stress (constant of material) and A is the area of cross section.
Let F1 and F2 be the breaking force for wires respectively.
Therefore,
Hence the breaking force required to break the thick wire is four times the thin wire.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
a) False, because the modulus of elasticity is inversely proportional to the strain for a given stress. If steel and rubber are under the same stress, then strain in steel is less than rubber. Hence steel is more elastic than rubber.
(b) True, because the change takes place in the shape of the coil spring. Therefore, stretching is determined by its shear modulus.
When a rod of metal is heated, compressive stress will be set up in the rod. The wire is not allowed to bend and the force is applied normal to it's cross sectional area.
Let 'L' be the length of wire, A be the area of cross section, Y be Young's modulus and '' be the coefficient of linear expansion, heated to a temperature ''.
If the rod was free then increase in length of wire due to rise in the temperature will be Lθ.
Here, since rod is not allowed to expand, therefore, compression in rod is equal to increase in the length of rod i.e. Lθ.
If S is the compression stress in rod then,
Hence stress is independent of the length of the wire.
Given, an ideal gas of bulk modulus 'K' and is heated to a temperature .
Increase of temperature of gas at constant volume is equivalent if the gas is first heated at constant pressure and then compressed isothermally to original volume.
Let V be the volume of gas. When it is heated through a temperature θ at constant pressure, then the increase in volume of gas is,
∆V = Vγθ
Now compress the gas isothermally so that the volume decreases by ∆V = Vγθ.
If ∆P be the stress set up in the gas, then
Given,
Wire is increased of it's length, i.e., strain = 10-3
Stress = 108N/m2
Given the radii of two wires is of the ratio 2:1.
Using the formula as per Hooke's law,
i) When both the wires are under the same stress, strain produced will be the same.
ii) When both the wires are loaded by the same weight, then
A wire loaded by weight of density 7800 kg/m3 is stretched to length of 1m. On immersing the weight in liquid of density 1300kg/m3, the length shortens by 6mm. Find the original length of the wire.
Given,
Density of the load, ρ = 7800 kg/m3
Length of the wire, l = 1 m
Density of liquid, = 1300 kg/m3
Let, V be the volume of weight suspended from the wire.
Therefore, tension in the wire when load is in the air is equal to,
Given,
Length of one end of wire, L = 1m
Angular velocity,=12 rad/s
Mass of the load, M=2kg
Therefore tension in the string is,
Increase in the length of the wire = (5.53 - 1) = 4.53 m
Length of the wire, l = 4m
Radius of the wire, r = 4 mm = 0.004 m
Force on the wire, F = 80
Strain = 0.01% = 0.0001
Now, using the formula of stress,
=
Young's modulus of the wire is given by,
Y =
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Length of steel wire, Ls = 4.7m
Length of copper wire, Lc=3.5m
Area of cross-section of steel wire, As=3.0 x 10-5 m2
Area of copper wire, Ac = 4.0 x 10-5 m2
To find = the ratio of Young's modulus of steel to that of copper?
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Given,
Diameter of the wires = 0.25 cm
That is, ds = dB = 0.25 cm
Therefore,
Radius of the wires will be given by,
Unloaded length of the steel wire, Ls=1.5 m
Unloaded length of the brass wire, LB = 1 m
Young's modulus of steel, Ys = 2 Pa
Young's modulus of brass, YB =
m1 = 4 kg and m2 = 6 kg
Brass wire is under a tension of load 6 kg.
Therefore, increase in length is given by
Since, brass wire is under a tension of load 6 kg.
"
The given question is illustrated in the figure. below.
From right angled ,
The different forces acting on mass are:
(i) Weight mg in vertically downward direction.
(ii) Tension T in the string along AS.
Now resolving the components of T as shown in fig. above.
The component Tcosθ balances the weight mg of the mass and component Tsinθ provides the necessary centripetal force to whirl the mass.
Given,
Weight of the load = 200 N
Elongation caused in the wire is given by,
To find - Elastic potential energy stored in the wire = ?
Therefore, energy stored in the wire is given by,
The volume of a spherical body is decreased by10–3 % when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 Atmosphere pressure =1.01x105 N/m2).
Volume of a spherical body decreased = 10-3 %
Pressure, P = 40 atm
We know,
Bulk modulus,
Percentage change in volume is given by 10-3 %
is the bulk modulus of the body.
Sponsor Area
We have two barometers. One barometer contains mercury and second water. Which one will register more pressure? (Neglect the surface tension effect).
The laws of floatation are:
(i) The weight of body is equal to the weight of liquid displaced.
(ii) Centre of gravity of body and that of liquid displaced are in same vertical line.
The buoyant force on body is,
U = Vρg
where,
V is the volume of body immersed in liquid,and
ρ is the density of liquid.
The distance between meta centre and centre of buoyancy is called meta-centre height.
Thrust: The force exerted by liquid normal to any surface in contact with it, is called thrust.
Pressure: The thrust exerted by a liquid per unit area of surface in contact with liquid is called pressure.
The SI unit of pressure is N/m2 or Pascal (Pa).
Application of pressure:
i) Hydraulic machines
ii) Blood pressure - Pressure exerted by the circulating blood upon the walls of the blood vessels.
iii) Variation of pressure with depth - submarines are designed to withstand such enormous pressure.
So, the force must be normal to surface.
Thus, the free surface of liquid in equilibrium adjusts normal to the direction of force acting on it.
PAscal's law states that, if some pressure is applied at any point of incompressible liquid then the same pressure is transmitted to all the points of liquid and on the walls of the container.
Let us imagine an arbitrary right angled prismatic triangle in the liquid of density ρ. This prismatic element is very small so, every part is considered at the same depth from the liquid surface. Therefore, effect of gravity is the same at all these points. That, the small element is in equilibrium.
The area of faces ABFE, ABDC and CDFE are ad, bd and cd respectively. Let the pressure of liquid on faces ABFE, ABDC and CDFE be P1, P2 and P3respectively.
The pressure of liquid exerts the force normal to the surface. Let us assume pressure P1exerts the force F1 on the surface ABFE, pressure P2 exerts force F2 on the surface ABDC and pressure P3 exerts force on the surface CDFE.
So, Force F1 is given by,
Since the prism is in equilibrium, so net force on the prism is zero.
Thus,
Hydraulic lift is a machine which is based on Pascal's law. In these devices fluids are used for transmitting presure.
Hydraulic lift consists of two cylinders of different cross-sectional areas connected with a pipe. The cylinders are filled with incompressible liquid and frictionless pistons are fitted in both the cylinders as shown in figure below.
Let 'a' and 'A' be the area of cross-section of smaller piston and bigger piston respectively. The load to be lifted is placed on bigger cross-section and effort is applied on smaller piston.
Let a force 'f' be applied on the smaller piston.
The pressure exerted by f on piston is,
... (1)
Now, according to Pascal's law, the pressure transmitted to bigger piston is also P.
Thus, force on bigger piston is, given by,
That is, the transmitted force gets multiplied by a factor of applied force.
Consider a liquid of density ρ in a vessel as shown in figure.
To find: Pressure difference between two points A and B separated by vertical height h.
Consider an imaginary cuboid of area of cross-section a of liquid with upper and lower cap passing through A and B respectively in order to evaluate the pressure difference between points A and B.
Volume of the imaginary cylinder is, V = ah
Mass of liquid of imaginary cylinder, m = ρah
Let, P1 and P2 be the pressure on the upper and lower face of cylinder.
Forces acting on the imaginary cylinder are:
(i) Weight, mg = ρahg in vertically downward direction.
(ii) Downward thrust of F1 =P1a on upper cap.
(iii) Upward thrust of F2 = P2a on lower face.
As the imaginary cylinder in the liquid is in equilibrium, therefore the net force on the cylinder is zero.
Thus, the pressure difference between two points separated vertically by height h in the presence of gravity is ρgh.
Note: In the absence of gravity this pressure difference becomes zero.
Atmosphere is a gaseous envelope surrounding the earth and the pressure exerted by the atmosphere is called atmospheric pressure.
The cause of atmospheric pressure is motion of air molecules which are in continuous mtion.
The molecules strike the surface of body placed in it and exert a huge force.
To measure the atmospheric pressure,
i) Torricelli took a meter long graduated tube and filled it with clean and dry mercury.
ii) By closing the tube with thumb, he inverted the tube in a cistern ( a tub filled with mercury) as shown in the figure.
iii) He observed that the level of mercury first fell down and finally stayed with a column of height 76cm in the tube above the free surface of mercury in the cistern leaving behind vacuum.
iv) At point C, vacuum is formed. Therefore, pressure at this point is zero. Point B in the tube is 76cm below C. So, pressure at B is given by,
PB =PC + ρgh = 0 + ρgh = ρgh
where ρ is density of mercury and h be the height column of mercury in the tube above point B.
In the given figure, point A is at the interface of air and mercury, therefore it is both in air and as well as in mercury.
Thus pressure at point A is equal to atmospheric pressure, i.e.
PA = Atmospheric pressure as vertical height between A and B is zero, therefore pressure at A and B is same, i.e.
PA=PB
Atmospheric pressure = pgh
= 13600 x 9.8 x 0.76
= 1.013 x 105 N/m2
Absolute pressure is the pressure above that of vacuum.
Absolute pressure = Gauge pressure + One atmospheric pressure.
Gauge pressure is the pressure above that of one atmospheric pressure. Since the pressure of vacuum is zero, it is referenced against an ambient air pressure.
Note: Atmospheric pressure is the pressure at any point equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere.
The pressure due to height column, P = ρgh
If the pressure due to height column of different liquids is same, then the relation between the height coulumn is given by,
,
is the required height of the water column which will produce the same pressure as 50 cm of Hg column.
Given,
Weight of the girl = 50 kg = 490 N
Circular diameter of the heel = 1.0 cm = 10-2 m
Area of the heel is given by,
Threfore, the pressure exerted by the heel on the ground is given by,
The atmospheric pressure at sea level is 760mm of mercury. At what height it will be 740mm of mercury? Take the density of air 1.293kg/m3 constant upto height about 1 km.
Unit of surface tension:
SI system: N/m
CGS system: dyne/cm
Dimensional formula of surface tension is [M1L0T–2].
Which part of liquid is responsible for phenomenon of surface tension: bulk or surface film?
Does the surface tension of a liquid depend on the area of the free surface of liquid?
Pressure due to 25 cm of water = Pressure due to x cm of mercury + Pressure due to 27.5 cm of spirit
A manometer reads the pressure of a gas in an enclosure as shown in the figure. The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
Give the absolute and gauge pressure of the gas in the enclosure.
Archimedes principle: When a body is immersed partially or wholly in a liquid, its weight appears to be reduced and loss of weight is equal to the weight of liquid displaced by the body.
Proof: Consider a body of height h and area of cross-section A placed in liquid of density σ at a depth x below the free surface.
Let ρ be density of the body.
Mass of liquid = 100 gm
Density of liquid = 2 gm/cc
Volume of ballon containing liquid = 50cc
When it is fully immeresed in water, it will displace 50cc of water.
Therefore upward thrust on ballon is given by,
U = 50 x 1
= 50 gmf
Tension in string is,
T=100 - 50
= 50 gmf
Fractional volume submerged,
(a) Ice floats in water with about nine-tenth of its volume submerged. What is the fractional volume submerged for an iceberg floating on a fresh water lake of (hypothetical) a planet whose gravity is ten times that of the earth?
(b) What is the fractional volume submerged of an ice-cube in a pail of water placed in an enclosure which is freely falling under gravity?
(a) Let a be the density of ice and ρ be the density of water.
The ice floats in water.
Therefore, weight of ice is equal to the weight of liquid displaced.
Let the volume of ice be V and volume of water displaced be V'.
i.e. fractional volume displaced by ice is independent of g.
Therefore, on the new hypothetical planet, fractional volume displaced by ice is also same as that on the earth i.e. nine-tenth.
(b) In free falling state, ice-cube is in weightlessness state and also the thrust exerted by water is zero. Therefore, it can float with any volume submerged.
Surface tension of the liquid decreases with increase in temperature, reaching a value of 0 at the critical temperature.
This dependence of surface tension on temperature is given by the equation,
Tt = T0(1–αat)
i) Take a wired rectangular frame whose three sides are rigid and one side AB is free to slide.
ii) Dip the frame in a soap solution to form the film.
iii) The surface tension force acts on all the sides of the frame.
iv) Side AB is free to slide and the other three sides being fixed cannot displace. Due to surface tension force slides in left direction and free surface tries to contract.
Let surface tension of soap solution be T and length of movable rod AB be l.
There are two free surfaces of liquid which touch the movable wire AB.
The surface tension of total inward force on the wire = 2Tl.
If frame AB is displaced by distance x in right direction, then work done against surface tension to increase the surface area is,
Increase in surface area of film is
Let R be the radius of big drop.
Given that 'n' drops coalesce to form a single drop.
During the process, energy is released which is converted into heat.
Energy released is given by,
This energy is converted to heat.
Therefore,
where, θ is the increase in temperature drop.
The surface area of liquid increases, when the sprayed in small droplets. Work has to be done against the cohesive force to increase the surface area. This work is done at the cost of internal energy and hence the temperature of drop falls.
When a vertical film of soap solution is formed, soap particles being heavy shift to lower side of film and hence surface tension on upper part is greater than the lower part which provides the necessary force to balance the weight of film. Hence, the liquid film is stable.
But in vertical film of pure water surface tension forces are same at all the points i.e. on A' B' and C'D'. Therefore, a film of pure water will break in the absence of extra force required to balance the weight.
Applications of Surface tension in daily life are:
(i) The surface tension of antiseptic ointments is low which make them to spread over the wound.
(ii) Adding detergent in water decreases the surface tension of water and enhances the cleaning action of water and detergent both.
(i) For the molecules which are perpendicular to the surface of the tube, adhesive force acts.
(ii) For liquid molecules at an angle of 45o, cohesive force will act on the molecules.
So,
,
Resultant force or liquid molecule is outside the tube and meniscus is concave, and
,
then, meniscus is convex.
Therefore there is more pressure inside than outside.
Let pi be the pressure inside the liquid drop and po be the pressure outside the drop.
Therefore excess of pressure inside the liquid drop is,
p = p1– p0
Due to excess of pressure inside the liquid drop the free surface of the drop will experience the net force in outward direction due to which the drop will expand.
Let the free surface displace by dR under isothermal conditions.
Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.
The work done by excess of pressure in displacing the surface is,
dW = Force x displacement
= (Excess of pressure x surface area) x displacement of surface
Increase in the potential energy is,
dU = surface tension x increase in area of the free surface
The above expression gives us the pressure inside a liquid drop.
Consider a soap bubble of radius R and surface tension T.
There are two free surfaces of soap bubble. Due to surface tension the molecules on the surface film experience the net force in inward direction normal to the surface.
Therefore there is more pressure inside than outside.
Let pi be pressure inside the liquid drop and po the pressure outside the drop.
Therefore excess of pressure inside the liquid drop is,
p =p1–Po
Due to excess of pressure inside the liquid drop, the free surface of the drop will experience the net force in outward direction due to which the drop expands.
Let the free surface be displaced by dR under isothermal conditions.
Therefore excess of pressure does the work in displacing the surface and that work will be stored in the form of potential energy.
The work done by excess of pressure in displacing the surface is,
dW= Force x displacement
= (excess of pressure x Surface area) x displacement of surface
Increase in the potential energy is given by,
dU = surface tension x increase in area of the free surface
From (1) and (2)
p is the excess of pressure inside a soap bubble.
Why oil spreads on cold water and may remain as a drop on hot water?
Given,
Radius of the circular plate, r = 5 cm
Mass, m = 2 gm
Surface tension of water, T = 72 dyne/cm
Weight of plate is, W = mg=2 x 980 = 1960 dyne
The circumference of circular plate is given by,
The down pull on plate due to surface tension is given by,
Therefore, total force required to take plate away from water surface is,
F=W+f
=1960+2260.8
=4220.8 dyne
Let,
Radius of small droplet be 'r',
Radius of big drop be 'R'.
Volume of water will remain constant on splitting the drop into droplets.
Therefore,
R3 = nr3
Here n = 1000,
Therefpre, R = 10 r
Now, the surface of the potential energy of drop is,
Let r be the radius of small droplet.
The volume of mercury will remain constant on splitting the drop into droplets.
Two beakers containing water and honey are stirred rapidly and kept on a table. Which liquid will come to rest earlier?
The coefficient of viscosity of honey is very large as compared to that of water. Therefore, honey kept in the beaker after stirring will come to rest earlier than water.
A soap bubble of radius 10mm is blown from soap solution of surface tension 0.06 N/m. Find the work done in blowing the bubble. What addition work will be done in further blowing to double the radius?
The different forces acting on the plate are:
(i) Weight W vertically downward,
(ii) Upward thrust U vertically upward,
(iii) Surface tension force T in downward direction .
These forces are given by,
Weight, W=8.6 x 980 = 8428 dyne
Upward Thrust, U=weight of water displaced
=Volume of water displaced x density of water x g
Let,
T = Total length of the water in touch with plate x surface tension
=2(1.5 + 0.2) x 70 = 238 dyne
Now the apparent weight,
Wa + W + T - U = 8428 + 238 - 1470
= 7196 dyne
= 7.434 gf
What will be the pressure inside a small air bubble of 0.1 mm radius situated just below the free surface of water?
(S.T. of water = 72dyne/cm; Atmospheric pressure 1.013 x 106 dyne/cm2 )
Excess of pressure inside the air bubble is given by,
If an air bubble of 5.0 mm is formed at depth of 40.0 cm inside a container containing soap solution (of relative density 1.20), what would be the pressure inside the bubble?(1AP = 1.013 x105 Pa and surface tension of soap solution is 0-025 N/m)
So, total length of free surface that touches with slider is 2L = 60cm = 0.60 m.
Here,
Angle of contact,
Radius of the narrow tube, r = 1.0mm=10-3m
Surface tension, T = 0.465 N/m
Density of mercury, = 13600 kg/m3
We know the height by which liquid rises in the tube is,
Negative sign means that mercury will dip down.
Stoke’s law states that the viscous force acting on a spherical body of radius 'r' moving with velocity 'v' in viscous media of viscosity 'η' is given by,
F= – 6πηrv.
Consider a tube of varying cross-section through which an ideal liquid is in streamline flow.
Let,
P1 = Pressure applied on the liquid at A,
P2 = Pressure at end B, against which the liquid is to move out
a1, a2 = Area of cross section at the tube at A and B respectively.
h1, h2 = mean height of section A and B from ground or a reference level.
v1, v2 = normal velocity of the liquid flow at section A and B, = density of ideal liquid flowing though the tube.
Liquid flows from A to B.
That is,
P1 > P2
Mass m of the liquid crossing per second through any section of the tube is in accordance with the equation of continuity, given by
a1v1p = a2 v2 p = m
Given, depth of orifice, h = 4.9 m
Therefore, the velocity of effulx is given by,
Define coefficient of viscosity. What are its units and dimensions?
SI unit of viscosity is deca-poise and CGS unit of viscosity is poise.
Dimensional formula of viscosity is [M1L–1T–1].
Let during fall, v be the velocity of body at any instant.
The different forces acting on the body at that instant are:
(i) Weight W, acting downward,
(ii) Upward thrust U, acting upward,
(iii) Viscous force F, acting upward.
Net force on body is,
We can see that the body falls under gravity.
Therefore, its velocity will increase due to gravity.
But as the velocity increases, the viscous force (factor II) acting on the body also increase.
Since factor I is constant, therefore the net force on the body decreases with time.
At some velocity v = vo, the resultant of all the forces acting on body reduces to zero and body acquires the constant velocity known as terminal velocity.
Therefore,
The above expression represents the terminal velocity acquired by a spherical body falling in viscous media under gravity.
The internal energy of a fluid due to pressure exerted on it's container is called pressure head. is called the pressure head.
The static pressure required to produce the velocity of fluid is known as velocity head.
The factor is called velocity head.
What are the differences between solid friction and viscosity?
Tiny droplets of water are contained in clouds. We kniow that terminal velocity acquired by a spherical body falling through a viscous media is directly proportional to the square of the radius of a body.
So, very small drops falling in air under gravity have very small terminal velocity and remain in sky. Hence, clouds appear to be floating in the sky.
The formula for viscous force is given by,
Here given that,
Horizontal force, F = 30720 dyne
Area of the plate, A = 200 cm2
Velocity with which the plate is moved = 3 cm/sec
Thickness of the layer of honey between the plate and table = 2.5 mm
Putting the values in equation (1), we get
is the required coefficient of viscosity of honey.
Given that,
Coefficient of viscosity, =0.01 poise
p = 1 gm/cc]
Coefficient of kinematic viscosity =0.01 stokes
Difference between streamline motion and turbulent motion.
i) It is an orderly type of motion in which the liquid flows in parallel layers while, turbulent motion is disorderly type of motion.
ii) Every particle of the liquid follows the path of it's preceeding particle and travel with the same velocity in magnitude and direction whereas, the motion of particles of the liquids becomes different at different points in turbulent flow.
iii) The velocity of streamline flow is less than critical velocity but, in turbulent motion the liquid moves with a velocity greater than the critical velocity of the liquid.
iv) Streamline flow is laminar whereas, turbulent flow is non-laminar.
It is the distance between two nearestive particles vibrating in same phase.
or
It is distance travelled by a wave in the direction of propagation in the duration equal to time period of vibrating body that produces the wave.
The harmonic waves in terms of wavelength and time period is given as,
where,
Harmonic wave in terms of wavelength and velocity is given as,
where,
The velocity of particle of the media located at x in a simple harmonic wve at any instant t is given by,
In a harmonic wave, the ratio of maximum particle velocity to the wave velocity in a harmonic wave is,
One-dimensional wave in the differential form is written as,
The properties of wave motion are:
(i) Wave motion is a disturbance propagated in media due to periodic motion of the particles of media.
(ii) In wave motion, it is the momentum and energy which is propagated, not the mass.
(iii) In wave motion, the particles of media vibrate about mean position.
(iv) The velocity of vibrating particle is different at different position.
(v) The velocity of wave is constant.
(vi) Different particles of media vibrate with different phase and phase difference between two consecutive particles is constant.
Transverse wave:
Transverse wave motion is the disturbance in which particles of medium vibrate in the direction perpendicular to the direction of propagation of wave. Transverse wave is propagated through media in the form of crests or troughs.
Longitudinal wave:
Longitudnal wave motion is the disturbance in which particles of media vibrate in the direction of propagation of wave. These waves propagate in the form of compressions and rarefactions.
Propagation of transverse waves in a media.
Consider nine particles of media on a reference line AB.
i) Let the particles vibrate perpendicular to line AB with amplitude ‘a’ and the wave propagates along AB from left to right.
ii) When the wave propagates, the different particles of media vibrate in different phase because it takes some time to transfer the disturbance (momentum and energy) from one particle to next particle.
iii) Consider that the disturbance takes T/8 seconds to travel from one particle to next.
iv) At t =0, all the particles are at mean position.
v) At t = T/8 sec, particle 1 gets displaced by 0.707a distance in the upward direction, while the disturbance reaches the particle 2.
vi) At t = 2T/8 sec, particle 1 reaches positive extreme position, the particle 2 gets displaced by 0.101a distance in the upward direction and the disturbance reaches the particle 3.
vii) At t = 3T/8 sec, after completing three eighth of vibration, particle 1 comes back to 0.707 a; the particle 2 reaches positive extreme position; particle 3 undergoes the displacement of 0.707a and the disturbance just reaches the particle 4.
viii) In this way the disturbance continues and the position of different particles at 4778, 5778, 6778 and 7778 seconds is as shown in the figure above.
viii) After T seconds, the particle 1 completes one vibration and particle 9 is just at the point to start its first vibration.
Thus, the particle 1 leads the particle 9 in phase by an angle of
Propagation of longitudinal waves in a media:
i) Consider nine particles of media on a reference line AB.
ii) Let the particles vibrate perpendicular to line AB with amplitude ‘a’ and the wave propagates perpendicular to AB from left to right.
iii) When the wave propagates, the different particles of media vibrate in different phase because it takes some time to transfer the disturbance (momentum and energy) from one particle to the next particle.
iv) Let disturbance take 778 seconds to travel from one particle to next.
v) At t = 0, all the particles are at mean position.
vi) At t = T/8 sec, particle 1 gets displaced by 0.707a distance in the right direction, while the disturbance reaches the particle 2.
vii) At t = 2T/8 sec, particle 1 reaches positive extreme position, the particle 2 gets displaced by 0 707a distance in the right direction and the disturbance reaches the particle 3.
viii) At f = 3T/8 sec, particle 1 after completing three eighth of vibration, comes back to 0.707a, the particle 2 reaches positive extreme position, particle 3 undergoes the displacement of 0.707a and the disturbance reaches the particle 4.
The disturbance of the particles is as shown below:
viii) In this way, the disturbance continues and the position of different particles at 4778, 5778, 6778 and 7778 seconds is as shown in fig. above.
ix) After T seconds, the particle 1 completes one vibration and particle 9 is just at the point to start its first vibration. Thus the particle 1 leads the particle 9 in phase by angle
The longitudinal wave propagates in the form of compression and rarefaction. Let us see how:
i) Draw the instantaneous position of different particles and their relative displacement from their mean positions after the particle 1 has completed on vibration.
The illustration of the disturbance is as shown below:
ii) After one complete vibration, the particles 1,5,9 are at the mean position; particles 2,3 and 4 move close towards particle 1 and particles 6,7 and 8 move towards particle 9. That is the particles crowd nearby 1 and 9. Thus the positions of particles 1 and 9 are the positions of compression.
iii) On the other hand, the particles 2,3 and 4 move away towards left from particle 5 and particles 6,7 and 8 move away towards right from particle 5. Thus the position of particle 5 is the position of rarefaction.
iv) Thus, there is an alternate formation of compression and rarefaction in media and hence the wave propagates in the form of compression and rarefaction.
v) In the longitudinal wave, the distance between the two consecutive position of maximum compression or rarefaction is equal to the wavelength of wave.
For one mole of gas, ideal gas equation is given by,
Now the velocity of sound in gas at temperature T is,
Since,
∴
Hence the velocity of sound in gases varies directly to the square root of temperature in Kelvin.
Velocity of the sound wave is given by,
Humid air contains large proportion of water vapours.
Therefore, the density of humid air is less than dry air.
Since velocity of sound is inversely proportional to the square root of density, therefore, sound travels faster in moist air than dry air.
Wavelength of the wave is given by,
Wavelength corresponding to frequency 20 Hz is,
and
Wavelength corresponding to frequency 20KHz is,
Wave length range of audible sound is 0.017m to 17m.
Given,
Velocity of longitudnal wave, v = 1450 m/s
And the formula for velocity of a longitudanal wave is,
∴
The given values are,
∴
, is the adiabatic compressibiltiy of water.
Velocity of wave in stretched string is given by,
where,
T is the tension, and
m is the mass per unit length.
If is length of string then time taken by wave to reach the other end of string is,
We have,
Therefore,
Time taken to reach the other end of the string is,
(a) Frequency of wave,
Velocity of reflected wave
∴ Wavelength,
(b) Velocity of wave in water = 1486 m/s
Since the frequency of wave does not change on refraction, therefore wavelength of transmitted wave is,
Let ρ1 be the density of hydrogen and ρ2 be the density of oxygen at 0°C.
The molecular weight of oxygen is 16 times as that of hydrogen.
Therefore,
ρ2= 16 ρ1
Density of the mixture consisting of two parts by volume of hydrogen and one part by volume of oxygen is.
As, velocity of longitudnal wave is given by,
∴
Given,
Frequency of the wave, v = 400HZ
Velocity of the wave, v = 340m/s
∴ Distance between two nearest points which are out by phase 120o is given by the wavelength of the wave,
Velocity of a wave is given by,
∴ ...(1)
Also,
∴ ...(2)
From (1) and (2),
Increase in wavelength is,
Given,
Therefore velocity of sound in sea water,
Let d be the depth of the sea.
As the echo is heard after 1.8s, therefore the wave travels 2d distance in water in time 1.8s.
i.e.
The displacement of particle O at any instant is given by y(0,t) = A sin ωt
The disturbance is handed over from one particle to next and will reach point P at a distance x from O, a bit later.
Therefore, phase of particle at P lags behind the phase of particle at O by,
where,
Therefore the displacement of particle at P is,
Therefore simple harmonic progressive wave propagating towards positive direction of x-axis is,
y(x,t) = A sin (ωt – kx).
If wave propagates towards left direction then replacing x by -x, we get
y(x,t) = A sin (ωt + kx).
Wave velocity:
Wave velocity is the velocity with which progressive wave front travels forward. It is constant throughout the media if the media is homogenous and given by,
v = v λ.
Particle velocity:
Particle velocity is the velocity with which particles vibrate during the wave propagation. The particle velocity is changing and given by,
v = rω cos(ωt – kx).
The given travelling wave is,
On comparing with general wave equation
We get,
Now,
Maximum particle velocity of media is,
...(1)
The velocity of wave is,
...(2)
Maximum deformation in media,
Now,
...(3)
From (1) and (3),
Given that,
Frequency of the sound wave = 512 Hz
Amplitude of the sound wave = 2.4 10-3 m
Density of air = 1.293 kg/m3
Intensity of the wave is,
Therefore,
Intensity, I =
I =
When the two waves superpose, then what is the resultant wave?
According to the principle of superposition of waves,
The fundamental frequency of the vibration of string is,
Frequency of vibrations of a string between two points vibrating in the mode,
The frequency of the vibration of the string is given by,
The frequency of the vibration of string is given by,
The overtone of vibration is second.
The fundamental frequency of stationary waves produced in one end closed organ pipe is given by,
The fundamental frequency of stationary waves produced in both end open organic pi[e is given by,
Frequency of stationary transverse waves produced in the sonometer wire is given by,
The frequency of vibration of air column in resonance pipe is,
Why have tuning fork two prongs?
Tuning fork B is loaded. Therefore, B has higher frequency.
Amplitude of two waves = a
Phase difference,
Therefore,
Amplitude of resultant wave =
Resultant Intensity of the wave, I =
where, is the phase differnce.
Resultant Intensity of the waves which are in phase is given by,
I =
Phase differnce between two waves for constructive interference is.
The relation between path difference and wavelength for the destructive interference between two waves is,
where,
x is the path differnce, and is the wavelength.
The phase difference between two waves for destructive interference is given by,
Amplitude of the crest = a
Amplitude of trough = b
Amplitude of the resultant wave = a + b
The range of variation of intensity on the screen is,
Ratio of maximum to minimum intensity in interference pattern is given by,
Travelling waves |
Stationary waves |
|
1. Each particle of media transfers disturbance to next particle 2. Amplitude of vibration of each particle is constant.
5. None particle is permanently at rest. 6. There is a flow of energy along the length of propagation of wave. |
1. Disturbance is not 2. Amplitude of vibration 3. Every particle in a loop passing through mean position has different 4. All the particles between two consecutive nodes vibrate in same phase. 5. The particles at nodes are permanently at rest. 6. There is no flow of energy along the length of propagation of wave |
|
Characteristics of stationary waves:
(i) The disturbance is confined to a particular region between the starting point and the reflecting point of the wave.
(ii) There is no onward motion of the disturbance from one particle to the adjoining particle and so on beyond this particular region.
(iii) The total energy assosciated with a stationary wave is twice the energy of each of the incident and the reflected wave. There is no flow of energy along the stationary waves.
(iv) In a stationary wave, the medium splits up into a numvber of segments. Each segment is vibrating up and down as a whole. All the particles in one particular segment vibrate in the same phase.
(v) There are certain points in the medium of standing wave, which is permanently at rest. These are called nodes. And certain other points in the medium where the amplitude is maximumwhich are called antinodes.
i) Sonometer consists of a hollow wooden box.
The arrangement for producing sonometer waves is shown in the figure.
ii) A string is stretched over the sonometer box with its one end fixed rigidly at one end of the sonometer box.
iii) The other end is connected to a hanger. The string is passed over the frictionless pulley fixed at the other end of rigid support.
iv) Two wedges are placed below the string. The two wedges behave as two rigid supports for the string.
v) Transverse stationary waves are produced in the string between two rigid supports.
According to Poiseuille’s equation, the rate of flow of fluid through capillary tube is directly proportional to fourth power of diameter.
i.e., V ∝ d4
If d is changed to d/2, then rate of flow will decrease to V/16.
Yes, it does matter if one uses gauge pressure instead of absolute pressure if atmospheric pressure at the two points. Bernoulli's equation applied is different at different points.
If the atmospheric pressure is same at all the points then it does not matter whether we use absolute or gauge pressure.
Limitations of Bernoulli's equation are:
(i) The Bernoulli's equation is derived by assuming that there is no loss of mechanical energy i.e. sum of pressure energy, potential energy and kinetic energy is constant. But when fluid is in motion, a part of mechanical energy is converted into heat energy.
(ii) The Bernoulli's equation is derived by assuming that liquid is non-viscous. But actually none of the liquids is non-viscous.
(iii)The Bernoulli's equation is derived by assuming that velocity of every fluid particle across any cross-section of pipe is same. But the particles of central layer have maximum velocity and velocity decreases towards the walls of pipe.
Let the liquid enter at A with velocity 'v1' whose area of cross-section is 'a1' and exit from end B with velocity 'v2' whose area of cross-section is 'a2'.
Let 'ρ1' and 'ρ2' be the densities of liquid at ends A and B respectively.
Now the volume of liquid that enters in one second at end A is given by,
V1 = a1 v1
Mass of liquid entering per second at end A is,
m1 = a1v1ρ1
Similarly the mass of liquid leaving per second at end B is,
m2 = a2v2ρ2
If there is no source or sink of liquid, then mass of liquid that enters at end A in one second is equal to the mass of liquid that leave in one second.
i.e. a1 v1 ρ1 = a2 v2 ρ2 ... (1)
If the liquid is incompressible, then
ρ1 = ρ2.
Therefore equation (1) reduces to
a1v1= a2v2 ....(2)
Equation (2) is called the equation of continuity.
According to the equation of continuity,
According to Poiseuille’s equation, the rate of flow of fluid is,
(a) The equation represents a stationary wave.
(b) The equation does not represent any wave.
(c) We have,
y = 3sin(5x–0.5t) + 4cos(5x – 0.5t) = 5 sin [5x – 0.5t + tan-1 (4/3)].
This is an equation for a travelling wave.
(d) y = cos x sin t + cos 2xsin 2t, is the equation for superposition of two stationary waves.
Length of one end closed pipe is, L = 20 cm = 0.20m
Frequency of source is, v = 430 Hz
We know that the frequency of nth mode is given by,
Let nth mode of pipe be in resonance with the source.
Therefore,
Therefore, the fundamental mode of pipe is in resonance with the source.
If both the ends of pipe are open then frequency of nth mode is,
Substituting the values, we get
n = 0.5, which is not possible. Therefore, if both the ends are open, then pipe cannot be in resonance with the source.
If y is the resultant displacement due to superimposition, then
where,
A = 2r cos (v1 – v2)t is the amplitude of the resultant wave.
As intensity is directly proportional to the square of amplitude, therefore, the intensity is maximum when amplitude is maximum.
i.e.
That is at times,
intensity will be maximum.
Therefore, the interval between two successive loud sounds is,
Therefore, Beat frequency, b = v1 – v2
i.e. Beat frequency is equal to the difference of frequencies of two interfering harmonic waves.
Pressure energy is the energy possessed by liquid by virtue of pressure.
To derive the expression for pressure energy, take a wide tank connected to a narrow pipe fitted with frictionless piston at the base as shown in figure.
Let A and a be the area of cross-sections of tank and pipe respectively.
By holding the piston, fill the tank to a height h above the pipe with a liquid of density ρ.
The pressure of liquid on the piston is,
P = ρgh
Force on the piston in outward direction is,
F= ρgha
Force to be exerted on the piston to keep it in equilibrium is,
F = ρgha
As A >>a, therefore if the piston is pushed inward by a small distance x, the height of liquid in the tank remains constant and hence the force on the piston is constant.
The work done to push the piston inward by distance x is, W = Fx = ρghax
This work done against pressure is stored in the form of pressure energy.
This energy is stored in the volume of liquid pushed inside i.e. ax.
Therefore, pressure energy stored per unit volume of liquid is,
Let v be the frequency of the note.
The note produces 7 beats with tuning fork A of frequency 256Hz.
Therefore, the possible frequencies of note are, v = 256 ± 7
i.e. 249 or 263 Hz.
When the note is sounded with tuning fork B of frequency 266 Hz, it produces 3 beats.
Therefore the possible frequencies of note are, v = 266 ± 3.
i.e. 263 or 269 Hz.
The common frequency in both the cases is 263 Hz.
Thus, the frequency of the note is 263 Hz.
The Bernoulli's equation is given by, ,
if liquid is non-viscous.
But if liquid is viscous, then the quantity on the left hand side will decrease along the direction of the flow of fluid.
Statement: For the streamline flow of non-viscous and incompressible liquid, the sum of potential energy, kinetic energy and pressure energy is constant.
Interference: The pattern is produced when two waves of same frequency travelling in same direction superimpose.
Beats are formed when frequencies are slightly different.
In the interference, the position of maximum and minimum are permanent i.e. intensity at a given point does not change with time.
In beats the intensity of wave changes at given point in time.
The conditions to obtain interference of sound waves is:
(i) The two sources of sound must emit the waves of same frequency and the same amplitude.
(ii) The initial phase difference between the waves emitting from two sources should be constant.
(iii) The two wave trains must travel in the same direction.
Venturimeter is based on Bernoulli's theorem. It consists of a two truncated tubes connected by a pipe at narrow ends. The pipe connecting the two tubes is called throat.
i) Venturimeter tube is positioned horizontally and the liquid is made to enter in it from end A and after passing through throat BC, it leaves tube at end D.
ii) Let at A,
Area of cross-section of tube = a1,
Pressure of liquid = P1
Velocity = v1
At the throat,
Area of cross-section of tube = a2
Pressure of liquid = P2
Velocity be v2
According to the equation of continuity,
Let the two coherent waves be,
Using principle of superposition, the resultant wave is,
...(3)
where,
r is the amplitude of resultant wave.
Amplitude of the resultant wave is given by,
...(4)
If both the waves have same amplitude, then
The intensity of a wave is directly proportional to the square of the amplitude.
Therefore intensities of waves are
where,
p is constant of proportionality.
Now from equation (4),
If then
The variation of intensity of the resultant wave with phase difference is as shown in the figure below.
A garden sprinkler has 150 small holes, each of 2mm2 in area. If water is supplied at the rate of 0.3 liter/s, then find the average velocity of spray.
What is the condition for the two waves to interference constructively and destructively:
(a) in terms of path difference.
(b) in terms of the phase difference.
The condition for the two waves for constructive interference and destructive interference are:
(a)
(b) (constructive interference)
(destructive interference)
Area of cross-section of cylindrical tube, A = 8 cm2
Velocity of flow in the tube, v = 1.5 m/min = 2.5 cm/sec
Diameter of the hole, d = 1 mm = 0.1 cm
Area of one hole, ao =
Total area of 40 holes, a = 40 ao =
= 0.1
Let v' be the velocity of the ejection of the liquid through the holes.
Using equation of continuity,
AV = av
Putting the values, we get
Let P1, a1 and v1 be the pressure, cross-sectional area and velocity at end A respectively.
P2, a2 and v2 be the same quantities at end B.
Apparent frequency of wave is given by,
Bernoulli's equation.
Change in frequency of wave received when source is moving away from stationary observer is given by,
The apparent frequency of wave received by stationary observer when source is moving towards observer is given by,
Change in frequency of wave received when source is moving towards stationary observer is given by,
The apparent frequency of wave received by an observer when it approaches towards stationary source is given by,
The change in frequency of wave received when observer is moving given by,
(i) ; when observer is moving towards stationary source.
(ii) ; when observer is moving away from the stationary source.
Apparent frequency of wave received by an observer when it moves away from stationary source is given by,
When an observer moves with velocity u or the source of light moves towards stationary observer with same speed is,
Change in frequency,
Change in the frequency, when an observer moves away with velocity 'u' or the source of light moves away from stationary observer is given by,
Frequency of the wave =
Speed of the wave with which the wave is approaching = u
Number of beats observed along with with the
source of the wave is,
Speed with which the train is approaching and leaving the platform = u
Frequency of waves =
The number of beats heard by stationary observer on the platform =
The formula of doppler effect when both source and observer are in motion is given by,
where,
uo and us are positive if they move towards each other.
The values are taken as negative if they move away from each other.
The characteristics of musical sound are:
(i) Loudness
(ii) Pitch, and
(iii) Quality.
The music intervals are:
(i) The major tone has interval 9/8.
(ii) The minor tone has interval 10/9.
Given,
Height of pipe from ground, h1 = 2m = 200 cm
Height of the pipe, h2 = 6m = 600 cm
Velocity, v1 = 80 cm/s
P1 = 2 x 106 dyne/cm2
From equation of continuity,
a1v1=a2v2
v2 is the velocity at end B.
The bernoulli's theorem is,
Substituting the respective values, we have
P2 is pressure at end B.
What are different intervals in diatonic scale?
Simple harmonic motion is produced by the projection of uniform circular motion onto one of the axes in the x-y plane.
For a particle exhibiting Simple Harmonic motion, acceleration is maximum and speed is zero at extreme position.
Number of heart beat = 75
Time, t = 1 minute = 60 seconds
The beat frequency of heart =
Given here,
Therefore velocity of sound in sea water is,
Let d be the depth of the sea.
As the echo is heard after 1.8s, therefore the wave travels 2d distance in water in time 1.8s,
i.e.
Mass of the load = 1 kg
Distance by which spring extends, x = 1 cm
We have,
kx = mg
Therefore,
The spring constant of the spring is,
The time period of persistence of ear for simple sound = 0.1s.
Therefore, to hear the distinct echo of simple sound, the minimum time taken by the sound to reach the listener after reflection should be 0.1s.
Let d be the distance between source and reflector.
The time taken by sound to reach the listener after reflection = t
Sound will travel the total distance of 2d.
∴ 2d = vt d = vt/2
Now, putting the values,
v = 340 m/s, t = 0.1s
We get,
d = 17 m
Therefore minimum distance of reflector from the source should be 17m.
When pendulum is taken on jupiter, there will be no change in the time period of oscillation of spring pendulum.
The value of 'g' is negligible on jupiter.
A free oscillator of natural frequency ω0 oscillates under the influence of external
periodic force f = f0 cos ωt. What is the frequency of oscillation?
Doppler's effect is the change in the pitch when there is a relative motion between source and observer.
Let a source which is at rest be emitting waves of frequency f and wavelength λ.
Let V be the velocity of the wave emitted by source,
V=fλ.
Let the distance between source and observer be V.
Therefore, the wave will take one second to reach the observer.
Now let the source start moving with velocity u towards the observer.
The wave, which was just emitted by the source, starts moving will take one second to reach the observer and in a second source, will travel 'u' distance.
The source will be at a distance V–u just after 1 second and there will be f waves in V–u distance.
Therefore wavelength of wave when source starts moving is,
The apparent frequency is,
...(1)
When source starts moving away from observer then replace u by –u.
Therefore, the apparent frequency is, ...(2)
From equation (1), the apparent frequency increases when the source moves towards the observer.
From equation (2) the apparent frequency decreases when the source moves away from the observer.
Periodic motion is a motion which repeats itself after a regular interval of time.
Examples of periodic motion are:
(i) Revolution of planets around the sun.
(ii) Motion of the simple pendulum.
(iii) Motion of the moon around the earth.
(iv) Motion of the wings of a bee.
To and fro motion about its mean position is called oscillatory motion.
Examples of oscillatory motion are:
(i) Motion of a simple pendulum.
(ii) Motion of a body floating in liquid.
(iii) Motion of liquid in a U-tube.
Some application of Doppler's effect are:
(i) It is used to estimate the speed of distant stars, the speed of aeroplane or submarines.
(ii) Observing red shift or blue shift, we can find the direction of motion of the object.
(iii) Using Doppler effect, we can find the magnitude and direction of revolution of the sun.
Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?
(a) It is not periodic motion because motion does not repeat itself after a regular interval of time.
(b) the given graph illustrates a periodic motion, which is repeating itself after every 4 seconds.
(c) The given graph does not exhibit a periodic motion because the motion is repeated in unequal intervals of time.
(d) It is periodic motion which is repeating itself in 5 seconds.
Let the particle P start from S.
The angular position of the particle at any instant t is given by,
θ = ωt + ϕ
Since the motion of the particle in the circle is not to and fro, it is not a simple harmonic motion.
Let a parallel beam of light be incident on the particle P and take the shadow on Y1Y2 axis.
As the particle revolves, its shadow vibrates between K and L.
If another particle Q is allowed to move along with the shadow, it will move to and fro along the shadow.
So, position of the particle Q along Y1Y2 axis w.r.t. O is given by,
Equivalent length of the simple pendulum is,
Amplitude = A
Displacement from the mean position, where the energy is half kinetic energy and half potential =
Here, we have
Wavelength,
Velocity of the sound in air,
(i) When the observer is on the east of train, the train moving due east will approach the observer.
The apparent wavelength of wave received by observer is,
(ii) When the observer is on the west of train, the train moving due east will move away from the observer.
The apparent wavelength of wave received by observer is,
Where will the particle executing simple harmonic motion have acceleration maximum and speed zero?
The apparent frequency of siren from approaching engine is,
and that from receding engine is,
(i) The apparent frequency of sirens from approaching engines are,
and
The difference in the frequencies of wave received from two engines is,
(ii) The apparent frequency of sirens from receding engine are given by,
and
The difference in the frequency of two sirens as heard by the man is given by,
Displacement of particle executing S.H.M is given by,
y=
Velocity of the particle is,
This implies,
A source of sound of frequency v is moving towards a wall with speed u. Find the number of beats heard by a stationary observer on the ground when:
(i) observer is between source and wall.
(ii) source is between observer and wall.
Let the particle be vibrating between points L and M.
In one complete vibration the particle moves from O to L, then from L to O and then from O to M and finally from M to O as shown in figure.
Therefore the total distance travelled by particle in one complete oscillation is,
S = SOL + SLO + SOM + SMO
= A + A + A + A = 4A
As the particle comes back to same position after one complete oscillation, therefore the net displacement of particle is zero.
The acceleration of particle executing simple harmonic motion, a = – ω2y
We know that the acceleration is second derivative of displacement.
Therefore,
The above equation represent the differential equation of simple harmonic motion.
Musical sound is pleasant to the ears, and it is produced by a series of similar pulses following each other regularly and rapidly at equal intervals without any sudden change in amplitude.
Noise is the sound which produces a displeasing effect to our ears, and it is produced by a series of different pulses following each other after unequal intervals of time.
A tone is a sound of a single frequency.
For example, tuning fork, when struck gently, produces a pure tone.
Note is formed by combination of many tones. That is note has a combination of different frequencies.
Intensity of sound depends on the following factors:
(i) Amplitude of vibration (I ∝ A2)
(ii) The frequency of vibration (I ∝ v2).
(iii) The density of media (I ∝ ρ).
(iv) Surface area of a vibrating body.
(v) The distance of listener from the source.
Position of particle is given by,
y = A = cos + B sin
...(1)
The velocity of particle is given by,
Acceleration of particle is given by,
Since the acceleration of particle is directly proportional to displacement and directed towards mean position, therefore the motion is simple harmonic motion.
Now,
Squaring (2) and (3) and adding, we have
Dividing (2) (3), we have
Epoch =
A particle is executing simple harmonic motion. What is phase relationship between:
(i) displacement and velocity
(ii) velocity and acceleration?
Kinetic energy:
Kinetic energy of particle executing simple harmonic oscillation at any instant is given by,
The velocity of particle at a distance y from the mean position is,
Potential energy:
Potential energy stored in the particle is equal to the work done in displacing the particle from mean position to y.
Let the particle be displaced through a distance x from mean position.
The restoring force F acting on particle is,
Therefore, when the particle is moved through a distance x, work done against the restoring force is given by,
Therefore, total work done by restoring force in displacing the particle from mean position to P is,
The potential energy stored in the body is equal in magnitude and opposite in sign of the work done by restoring force. Thus
Total energy is,
E = K + U
From the above equation, total energy is independent of the position of particle during its motion.
Thus, total energy is constant.
Hence proved.
Kinetic energy of a particle executing S.H.M is given by,
Potential energy is given by,
Therefore the graph for kinetic energy versus displacement is a parabola, which is shown below.
The graph for potential energy versus displacement is also parabola.
Total energy is given by,
Graph is straight line parallel to displacement axis.
The kinetic energy of the particle executing simple harmonic motion is given by,
K.E =
Potential energy is given by,
P.E, U =
Total energy of the particle is given by,
T.E = K.E + P.E
=
=
Given that,
Amplitude of the motion, r= 0.6cm
Velocity at mean position, v0 = 12m/s
Therefore, '
The acceleration of particle at extreme position is,
The basic postulates of kinetic theory of gases are:
(i) All gases consist of atoms or molecules. The atoms or molecules of one gas are all similar to one another and different from the molecules of the other gas.
(ii) Molecules of a gas are in random motion.
(iii)The volume occupied by gas molecules is negligibly small as compared to volume of the container.
(iv) Molecules collide with each other. The collisions are elastic and instantaneous.
(v) Between the two consecutive collisions, the path followed by molecules is straight line.
(vi) Gas molecules do not exert any force of attraction or repulsion upon each other.
The postulates of Kinetic Theory which do not fit exactly for ideal gas are:
(i) Gas molecules do not exert any force of attraction or repulsion upon each other.
(ii) Volume occupied by gas molecules is negligible.
State two postulates of kinetic theory of gases which are obeyed by all the gases.
The two postulates of kinetic theory of gases which are obeyed by all the gases are:
(i) All the gases consist of molecules that are similar to one another for one gas and different from the molecules of other gas.
(ii) Molecules of all the gases are in random motion with speed ranging from zero to infinity.
A body oscillates simple harmonically according to equation y=5 cos( What is displacement, velocity and acceleration at t= 3s?
Simple Harmonic Equation is given by,
y = 5cos(3
Therefore,
Velocity, v
Now, y (t=3) = 5 cos (9
=
Therefore,
v|t=3= -15
Acceleration, a|t=3 =
=
Using the formula of acceleration,
We have
a =
So, time period of the oscillation is given by,
Now, putting the values of acceleration and displacement give, we get, is the time period of the oscillation.
Let A be the amplitude and be the frequency of oscillation.
Therefore,
The position of the particle executing simple harmonic motion is given by
Therefore the time taken by particle to move between two points 8cm on either side of mean position is,
The minimum time is 0.4s.
Avogadro law states that under similar physical conditions of temperature and pressure, equal volume of all the gases contains equal number of molecules.
Let P be the pressure, V be the volume and T be the absolute temperature.
Let us consider two gases having same P, V and T. Let one gas contains n1 molecules, each of mass m1 and second gas contains n2 molecules each of mass m2. Let v1, and v2 be the r.m.s. velocities of two gases.
Now, According to kinetic theory of gases, pressure of an ideal gas is given by
Now, from equations (1) and (2), we have
Now, since temperature of both the gases is the same, the average K.E of both the gases will be same.
That is, ... (4)
So, from equation (3) and (4), we have
n1 = n2
That, is the number of molecules for both gases is the same.
Hence, Avogadro's law is proved.
The velocity of air molecules is of the order of 500 m/s. But, the fragrance spreads at a much slower rate because the vapour molecules do not travel straight with this velocity. Due to collisions the molecules follow zig-zag path and the effective displacement of molecules per unit time i.e drift motion is very less.
Maximum acceleration is,
The vertical motion of a huge piston having a small block of mass m placed over it in simple harmonic motion with a frequency is ω . What is the maximum amplitude of piston’s simple harmonic motion for block and piston to remain together?
If the acceleration of piston is less or equal to acceleration due to gravity at that point, the block will not leave the piston at any point.
The value equal to 'g' for the block not to leave the piston is the maximum acceleration at the extreme point.
Therefore,
where A is amplitude of vibration.
The displacement of particle executing simple harmonic oscillation is given by,
y = A sin ωt
Let the particle take to time to go from mean position to half of extreme position.
Therefore,
So, at a time of the simple harmonic oscillator is equal to half of amplitude from the mean position.
The velocity of the particle executing simple harmonic motion when at a distance x from mean position is given by,
The average thermal energy of atom/molecule is given by ,
(i) Given,
Room temperature = 27°C = 300 K.
Therefore the average thermal energy of He atom at room temperature is,
(ii) Given,
Temperature at surface of the sun=6000K
Therefore the average thermal energy of He atom at temperature of the sun is,
(iii) Thermal energy at temperature "
So, the average thermal energy of the He atom is,
Let the equation of motion of two particles be,
where,
ϕ is the phase difference between two particles.
Each time the particles cross at a distance 0.707A.
Therefore,
0.707A=A sin
and 0.707A=A sin(wt+)
0.707=sin
and,
0.707= sn
Now,
sin(=sin(
=sin
Since,
We can see that ϕ = 0 is not possible.
Therefore the phase difference between two particles is 90°.
Two containers of same size are at the same temperature. One of them contains 1 kg of H2 gas and other 1 kg of O2 gas.
(a) Which vessel contains more molecules?
(b) Which vessel is under greater pressure?
Kinetic energy of particle at a distance y from the meain position is,
Potential energy of particle at a distance y from the mean position is,
The ratio of potential energy to kinetic energy at a distance y from the mean position is,
Let the particle of mass m vibrate with frequency and A be the amplitude of oscillation.
The kinetic energy of particle at a distance y from mean position is,
A is the amplitude of the oscillation.
Which of the following function represents:
(a) simple harmonic motion,
(b) periodic but not simple harmonic motion,
(c) non-periodic motion: (i) a cosωt + b sin ωt (ii) sin ωt + sin 2 ωt
(iii) sin2 ωt ± cos 2 ωt
(iv) log |cos (ωt) |
(V) e–ωt
(vi) cos2 ωt + cos ωt
(i) Simple harmonic
(ii) periodic but not simple harmonic motion
(iii) non-periodic
(iv) periodic but not simple harmonic motion
(v) non-periodic
(vi) periodic but not simple harmonic motion
Which of following is example of (a) simple harmonic motion (b) periodic but not simple harmonic motion and (c) non-periodic motion:
(i) Planet revolving around the sun.
(ii) A body dropped from a certain height and colliding elastically on the earth.
(iii)Motion of ball inside the smooth concave bowl.
(iv) Rotation of the earth about its own axis.
(v) Motion of body under force F = ky.
(vi) Motion of simple pendulum in free falling lift.
(i) Periodic but not simple harmonic.
(ii) Periodic but not simple harmonic.
(iii) Simple harmonic motion.
(iv) Periodic but not simple harmonic.
(v) Non-periodic.
(vi) Non-periodic.
At temperature t,
Average kinetic energy of gas molecule is given by E = kT
Which of following conditions are not sufficient for simple harmonic motion? Explain:
(a) Acceleration ∝ displacement.
(b) Force acting on particle should be directed towards mean position.
(a) Acceleration is directly proportional to displacement. But for a motion to execute S.H.M, the direction of acceleration must be directed towards mean position is not indicated here.
Therefore thiscondition is not sufficient for the particle to execute simple harmonic motion.
(b) Force is directed towards mean position and hence the acceleration. But this does not give the variations of acceleration w.r.t. displacement.
Therefore this condition is also not sufficient.
But both the conditions jointly are sufficient for simple harmonic motion.
For oscillating simple pendulum:
(i) What is the direction of acceleration of the bob at: (a) mean position (b) end points?
(ii) Is the tension in the string constant throughout the oscillation? If not, when it is,
(a) the least, (b) the greatest?
(i)
(a) At mean position the acceleration of bob is radial towards the point of suspension.
(b) At end position, the acceleration of bob is tangential towards the mean position.
(ii) Tension in the string is different at different positions.
(a) Tension is least at the end points.
(b) Tension is greatest at the mean position.
At the bottom of the lake,
Volume of air bubble, V = 1.0 cm3
Depth of lake = 40 m
Temperature, T = 12o C = 12+273 K = 285 K
At the top of the lake
Volume of air bubble = ?
Changed temperature, T' = 35o C
Pressure at the top of the lake, Pt = 1 atm = 1.01 Pa
Volume at the top, Vt = ?
Now using the gas equation, we have
is the required volume of the air bubble at the surface.
Copper (in g) |
Copper oxide (in g) |
5.015 |
6.284 |
8.286 |
10.357 |
18.443 |
23.169 |
12.011 |
15.051 |
The r.m.s velocity of gas molecule is given by,
Given,
∴
Mean free path of the molecule is given by,
The average wavelength of visible light is 5800 .
Thus the mean free path is less than the average wavelength of visible light.
The two springs shown in the figure are parallel.
Therefore, the spring constant of the combination is k = k1 + k2.
Therefore, time period of oscillation of system is,
Radius of the metallic ring = R
Coefficient of linear expansion =
The ring is heated by a temperature of 1oC.
Therefore, increase in are of the ring =
At the mean position B, tension is given by,
where,
v is the velocity of bob at lowest point B.
Let A be the extreme position of the bob during oscillation and is at a height h from the lowest point B.
From the figure,
As the bob moves from A to B, the potential energy decreases by mgh and is converted into kinetic energy at B.
Yes, the length of two rods of different material is constant at all the temperatures. The length does not vary on heating. It remains the same.
The length of the rods is independent of the temperature.
The relation between the length of the rods is given by,
The quantity of the liquid in the container is given by,
where,
Vliq = Vol. of liquid,
Vcont = Vol. of container = Temp. coefficient of liquid.
= Temp. coefficient of container.
Relation between coefficient of real expansion and that of apparent expansion of liquid is given by,
A spring of force constant 1200 N/m is mounted on a horizontal table as shown. A mass of 30 kg is attached to the free end of the spring, pulled side ways to a distance of 2-0 cm and released.
(a) What is the frequency of oscillation of the mass?
(b) What is the maximum acceleration of mass?
(c) What is the maximum speed of mass?
We have,
Spring constant, k=1200 N/m
Mass of the spring, m=30kg
Amplitude of vibration = 2.0 cm =0.02m
(a) We know frequency of oscillation is,
(b) We know maximum acceleration of mass is,
(c) Maximum speed is,
The temperature coefficient of superficial expansion is defined as the ratio of the increase in area per unit original area per unit rise in temperature.
It is represented by β.
Consider a sheet of area S0 at 0°C.
If we increase the temperature of sheet by 0°C, it will expand.
The increase in the area of sheet is,
(i) proportional to the original area
(ii) proportional to rise in temperature,
i.e.
The area of sheet at temperature is
The ratio of the increase in volume per unit original volume per unit rise in temperature is defined as the temperature coefficient of cubical expansion.
It is represented by γ.
Consider a cube of volume V0 at 0°C.
If the temperature of cube is increased by θ°C, it will expand.
The increase in the volume of cube is,
(i) proportional to the original volume
(ii) proportional to rise in temperature,
i.e.
The volume of cube at temperature is,
The frequency of oscillation is given by,
Given, when m = M ;
Given,
Coefficient of friction, μ = 0.6 and
Amplitude, A = 3 cm
Let m be the mass of block and ω the maximum frequency of oscillation.
The limiting friction between block and table is,
Flimiting = μmg
The maximum acceleration of table is,
amax = ω2A
The maximum pseudo force on the block is,
Fpseudo = mamax = mω2A
For the block not to slip over the table, the maximum pseudo force should not exceed the limiting friction.
Therefore,
is the maximum frequency at which the block does not slip over the table.
A spring of constant 245N/m is loaded by two blocks of mass 4 kg and 5 kg. If 4 kg mass is suddenly removed, then what will be the frequency and amplitude of oscillation of 5kg mass?
We have,
Spring constant, k=245N/m
Mass, m1 = 4 kg
MAss, m2 = 5kg
When 4 kg mass is removed from the loaded spring, the the amplitude of oscillation is equal to extension produced due to 4 kg load.
The spring remains loaded by 5 kg mass. Therefore, the frequency of oscillation is,
Time period of the simple pendulum is given by,
At height h, acceleration due to gravity is,
Define the apparent and absolute coefficient of expansion of liquids. Find the relation between them.
Liquids cannot hold themselves alone but have to be contained in the vessel. Therefore when liquids are heated, naturally the vessel will also expand along with liquids. Thus when liquids are heated, they have also to fill the increased volume of the vessel. We cannot observe the intermediate state. We can only observe the initial and the final levels. This observed expansion of the liquid is known as the apparent expansion of the liquid and is less than its actual expansion.
The apparent coefficient of expansion of liquid is defined as the apparent increase in the volume of liquid in the vessel per unit original volume of liquid in the vessel per unit rise in temperature. It is represented by γa.
The absolute coefficient of expansion of liquid is defined as the actual increase in the volume of liquid (sum of the apparent increase in the volume of liquid in vessel and increase in the volume of the vessel) per unit original volume of liquid per unit rise in temperature. It is represented by γrRelation between γa and γr
Let us consider a vessel of volume V0 at 0°C filled to brim with a liquid. Let γa and γr be the apparent and absolute coefficient of volume expansion of liquid respectively and γv is coefficient of cubical expansion of vessel.
Now heat the vessel to 0°C. Both vessel and liquid will expand. As the container is filled to the brim, therefore on heating, the liquid will overflow. Collect the overflow liquid. The liquid that overflows is the apparent increase in volume. Let it be ∆Va. By definition, the apparent coefficient of expansion of liquid is,
Let ∆Vv be the increase in the volume of the vessel,
Now the actual increase in the volume of liquid is,
...(1)
Also, the actual increase in the volume of liquid is,
...(2)
From (1) and (2),
or
This is the required relation between
Consider a rod of length L, area of cross-section A, clamped between the two rigid supports so that it neither expands nor bends.
Heat the rod so that its temperature increases by θ°C.
As the rod is neither allowed to expand nor bend, therefore a compressive force or stress is set up in the rod. This force or stress is called thermal force or thermal stress.
Let be the coefficient of linear expansion and Y be the Young’s modulus of material.
If the ends of rod were free, then its length would increase by,
The rod being clamped, does not expand but a strain is produced in the rod.
The strain in the rod is given by,
The stress in the wire is,
Thermal stress = Y x strain
=
Thermal force developed in the rod is,
Thermal force = Thermal Stress x Area
The time period of simple pendulum clock at the surface of the earth is,
Let g' be the acceleration due to gravity at the top of mountain.
Then, time period of simple pendulum clock at the top of mountain is,
The acceleration due to gravity at the surface of the earth is,
where,
M is mass, and
R is the radius of the earth.
Let h be the height of the mountain.
Acceleration due to gravity at the top of the mountain is given by,
The loss of time of oscillation of simple pendulum clock in one second is,
Thus loss of time of oscillation of simple pedulum clock in one day is,
Height of the mountain is 54 km.
The amplitude of simple harmonic oscillation is doubled. How would this affect:
(i) Time period,
(ii) Maximum velocity,
(iii) Acceleration at mean position,
iv) Kinetic energy at mean position?
Time period is independent of the amplitude of vibration. Therefore, time period remains unchanged on doubling the amplitude of oscillation.
(ii) For a particle undergoing Simple Harmonic Motion, maximum velocity is given by
v = A ω
where, A is amplitude.
When amplitude is doubled, the maximum velocity becomes twice.
(iii) Acceleration at mean position is zero and is independent of the amplitude of vibration.
Therefore, acceleration at mean position remains unaffected.
(iv) Kinetic energy at mean position is,
If the amplitude of oscillation gets doubled, then kinetic energy at mean position increases to four times.
Place the two glasses in hot water, such that there is no water in the inner glass. The outer glass expands because of hot water and hence a gap is set up between the two glasses and thus we can take the glass out.
If we have cold water, then put the cold water in the inner glass. The cold water makes the inner glass contract leaving a gap between the two glasses and thus we can take the other glass out.
The simple harmonic oscillation is executed when an oscillator is displaced from it's equilibrium position.
The time period of its oscillation depends upon the oscillator's inertia factor and elastic factor.
If the amplitude of oscillations remains constant in the absence of any external source of energy, then the oscillations are called free oscillations and oscillator is called free oscillator.
The frequency with which the oscillator oscillates is called natural frequency.
The differential equation of free oscillator is,
The solution of the above equation differential equation is,
The energy of oscillator is,
As the amplitude of free oscillations is constant, therefore the energy of oscillation is also constant.
A.
Bimetal is used in metal thermometersB.
Bimetal is used in thermostats for regulating the heating or cooling of roomsC.
Bimetal relays are used to open or close the electric circuits.D.
Bimetals are used to generate electricity.Consider a loaded spring oscillating in the vertical direction in a resistive media.
Let m be the mass of load and k be the spring constant.
The load is displaced from the equilibrium position. Let at any instant, y be the displacement from equilibrium position and v is the velocity of the load.
The different forces acting on load are:
(i) Restoring force which is proportional to displacement and directed towards equilibrium position.
i.e., F1= – ky
(ii) Resistive force of medium which is proportional to velocity and opposite to the direction of velocity.
i.e., F2 = –bv
The net force acting on the load is,
F = Fx1 + F2
= – ky –bv
Therefore the equation of motion of load is,
This is the required differential equation of motion of damped oscillations.
The solution of the equation is,
Comparing it with , we get,
.....(1)
Equation (1) gives the amplitude of oscillation, which decreases exponentially with time.
Equation (2) gives the frequency of oscillation which is less than the natural frequency of oscillation ω0.
When an oscillator is made to oscillate by an external periodic force, whose frequency is different from the natural frequency of the free oscillator is defined as forced oscillations.
In forced oscillations, the oscillator is under the influence of two forces simultaneously:
i) Restoring force of its own
ii) Force due to external periodic force.
Both these forces have a tendency to make the oscillator oscillate their own way.
Thus, for the first few, oscillator does not exhibit regular oscillations.
First, the restoring force is more effective and the oscillator oscillates with natural frequency and then external periodic force dominates and makes the oscillator oscillate with a frequency same as that of external periodic force.
Consider a mass m attached to a spring of constant k.
When an external periodic-force is applied on mass, it will oscillate under the influence of following forces:
(i) The restoring force which is proportional to displacement and directed towards equilibrium position.
i.e., F1= – ky
(ii) Resistive force of medium which is proportional to velocity and opposite to the direction of velocity.
i.e., F2 = – bv
(iii) The external periodic force F3 = fo cos ωt.
The net force acting on the load is,
Therefore the equation of motion of load is,
This is the required differential equation of motion of damped oscillations.
The solution of the equation is,
From the above equation, we can deduce that the amplitude of oscillation depends on the difference in the frequency of periodic force and natural frequency.
Smaller is the frequency difference greater is the magnitude of oscillation and vice versa.
Let L and L' be the length of edge at
∴
The edge length of the cube at 60°C is 10.00855 cm.
Therefore area of each face if cube at 60°C is,
And volume of cube at is,
The SI unit of specific heat is J kg-1 K-1.
The CGS unit of specific heat is cal gm-1 K-1.
The dimensional formula for molar specific heat is,
S.I. unit of heat capacity is Joule/Kelvin and the dimensional formula is .
The SI unit of latent heat is J/kg.
The CGS unit of latent heat is cal/gm.
A gas is heated at constant pressure. What fraction of heat is converted into work?
The fraction of heat which is converted into work is given by,
where y is the ratio of the two specific heats.
When an ideal gas is heated at a constant pressure, then fraction of heat is converted into internal energy.
where,
y is the ratio the two specific heats.
For an ideal gas having 2 degree of freedom, the value of specific heat at constant pressure is given by,
(a) What do you understand by degree of freedom? Give an example.
(b) On what factors does the degree of freedom of gas depend?
(a) The number of independent coordinates required to specify the position of particle or system of particles is known as degree of freedom.
For e.g. if a particle moves on the surface of table, it has two degrees of freedom. This is because in order to locate the position of particle two independent coordinates are required.
(b) Degree of freedom depends upon the following factors:
(i) Temperature of gas.
(ii) Atomicity of gas
i.e. whether gas is monoatomic, diatomic, triatomic or polyatomic.
(a) A body constraint to move on the surface of the earth.
(b) A body constraint to move on the surface of the earth only along the equator of the earth.
(c) Bob of a simple pendulum.
(d) Cylinder rotating about its axis only.
(e) Monoatomic gas molecule.
(a) The earth is three-dimensional but the body is constrained to move on the surface of the earth. Therefore, one degree of freedom will decrease and the body has 2 degrees of freedom.
(b) Only one degree of freedom.
(c) One degree of freedom.
(d) One degree of freedom.
(e) Three degrees of freedom.
(a) The degree of freedom is one.
Reason: Diatomic gas molecule has at the maximum six degrees of freedom (2x3 = 6) out of which three are due to translational motion, two are due to rotational motion.
(b) Monoatomic gas molecule has only three degrees of freedom and they are only translational.
Diatomic gas molecule has five degrees of freedom.
Consider a triatomic linear gas molecule like CO2 which has seven degrees of freedom and triatomic angular has six degrees of freedom.
Law of equipartition of energy:
According to this law, the total energy of the dynamical system in thermal equilibrium is equally shared among all the degrees of freedom.
The energy associated with each degree of freedom is 1/2kT,
where,
k is Boltzmann’s constant and T is temperature.
Heat | Temperature |
i) Heat is energy which produces the sensation of warmthness. | i) Temperature is the physical condition that tells us the direction of flow of heat. |
ii) The flow of heat from one body to another does not depend on the amount of heat energy in the body i.e. heat can flow from a body having low heat energy to another body having high heat energy and vice versa. | ii) Heat can flow only from a body at high temperature to another body at low temperature spontaneously. |
iii) When heat flows from one body to another, the amount of heat energy lost by the hot body is equal to the amount of heat energy gained by the cold body. | iii) Fall of the temperature of the hot body need not necessarily to be equal to rise in temperature of the cold body. |
iv) Heat energy of the body can increase without increasing the temperature e.g. during phase transition the temperature of the body remains constant. | iv) Temperature of the body cannot increase without increasing the heat energy of the body |
Some of the heating effects are:
(i) Heat energy brings about the change in dimensions of the body.
(ii) Heat energy changes the temperature of the body.
(iii)Heat energy may also cause chemical changes.
(iv) Heat energy can also change the state of matter.
Thermometer is an instrument used for measuring the temperature.
The different types of thermometers are:
(i) Mercury thermometer: It is based on the principle that substance expands on heating.
(ii) Gas thermometer: It is based on Regnault’s law which states that at constant volume, the pressure of gas varies directly with temperature. Specific heat of gas is constant for all values of the pressure of gas.
(iii) Resistance thermometer: It is based on the fact that resistance of conductor increases with the increase in temperature.
(iv) Thermoelectric thermometer: It is based on Seeback effect.
The Seebeck effect is a phenomenon in which a temperature difference between two dissimilar electrical conductors or semiconductors produces a voltage difference between the two substances.
Advantages of gas thermometers:
(i) Gas thermometers are very sensitive because the expansion of gases is considerable.
(ii) The gases have regular expansion.
(iii) Gas thermometers have wide range of temperature scale.
(iv) Permanent gases have close resemblance with perfect gas. Therefore, the thermometers filled with permanent gases give the reading close to thermodynamic scale.
(v) Gases have low thermal capacity.
Disadvantages of gas thermometers:
(i) It is not a direct reading thermometer.
(ii) It is not portable.
(iii) It cannot measure accurately the rapidly changing temperatures.
The thermometric liquid used in thermometers should have the following properties:
(i) should have low heat capacity,
(ii) should be good conductor of heat,
(iii) should have large temperature coefficient of expansion,
(iv) should have uniform rate of expansion,
(v) should not be volatile and sticky, and
(vi) should have high boiling point and low freezing point.
Mercury is used as a thermometric liquid because of the following reasons:
(i) it has low heat capacity,
(ii) it has constant value of temperature coefficient of expansion,
(iii) it has high melting point and low freezing point,
(iv) it is good conductor of heat,
(v) it does not stick with glass, and
(vi) it is less volatile and hence has very low vapour pressure
No, water cannot be used as a thermometric liquid because of the following reasons:
(i) has a very high value of specific heat,
(ii) has non-uniform expansion,
(iii) is transparent and is a wet liquid,
(iv) is a bad conductor of heat,
(v) evaporates under vacuum conditions, and
(vi) can only be used in the temperature region 4°C to 100°C.
The quantity of heat which is required to raise the temperature of one mole of gas through one degree Celsius is called the specific heat of a substance.
This heat supplied to gas is not only used to increase the temperature of gas but also used to expand it.
Heat required to increase the temperature of the gas by 1°C is a constant quantity. But the heat required to expand it depends on the manner in which it is heated. That is why gases have infinite number of specific heats depending on the manner in which it is heated up.
(a) Define two principle specific heats of gases.
(b) Why Cp is greater than Cv?
(a) Specific heat at constant volume:
The quantity of heat required to raise the temperature of 1 mole of gas through 1°C at constant volume is known as specific heat at a constant volume.
Specific heat at constant pressure:
The quantity of heat required to raise the temperature of 1 mole of gas through 1°C at constant pressure is called specific heat at constant pressure.
(b) When gas is heated at constant volume, the whole of the heat supplied is used to increase the temperature of gas because gas is not allowed to expand and work done against expansion is zero.
When gas is heated at a constant pressure, it expands, and gas has to do work against the external pressure.
Therefore, in addition to the heat which is required to raise the temperature, heat is also required to expand the gas. So, at constant pressure, we require more heat than at constant volume.
(a) Can Cp < Cv?
(b) Can a system which is not in thermal equilibrium be in thermal equilibrium at a later time? Give an example.
(a) Yes, when water is heated from 0°C to 4°C then Cp < Cv , because water contracts when heated from 0°C to 4°C. Water exhibits anomalous behaviour.
(b) Yes, when we add a hot piece of metal into water, the temperature of the metal piece starts decreasing and that of water starts increasing. After some time temperature of both becomes equal and hence thermal equilibrium is attained.
(a) Define latent heat.
(b) What are dimensions and units of latent heat?
(a) Latent heat is the heat required to transit the phase of unit mass of a substance at a constant temperature.
Latent heat of fusion: It is the quantity of heat required to transit the phase of 1 g of solid into liquid at constant temperature.
Latent heat of vaporization: It is the quantity of heat required to convert 1 g of liquid into vapours at constant temperature.
(b) Dimensions of latent heat are [M 0L2T–2]
Unit of latent heat is J/kg in SI system and cal/g in cgs system.
Define specific heat and heat capacity.
Specific heat:
Specific heat is the quantity of heat required to raise the unit temperature of unit mass of the substance by 1-degree celsius of temperature.
If dH heat is required to raise the temperature of m gram of substance through dT degree Celsius, then
dH = mS(dT)
where,
S is the specific heat of the substance.
Mathematically it is given by,
Unit of specific heat is cal/g/K or J/kg/K.
Given,
Distance between thermometer ice point and steam point =
Separation between two divisions of calibrations = 0.5 mm
∴
i.e. least count of thermometer is
Relation between Fahrenheit scale and Reaumur scale is given by,
Let at temperature T, the reading on both Reaumur scale and Fahrenheit scale be the same.
∴
Thus,
The relation between Fahrenheit and Kelvin is given as,
Let at temperature T, Fahrenheit scale and Kelvin scale readings coincide.
∴
Thus,
So at this temperature, the Fahrenheit and the kelvin scale coincide.
Consider one mole of an ideal gas enclosed in a cylinder fitted with the movable frictionless piston.
Let the gas be heated at constant volume.
On supplying dQ amount of heat, let the temperature of the gas increase by dT.
By first law of thermodynamics,
dQ = dU + dW = dU + PdV CpdT = Cv x dT + RdT
At a constant pressure,
P.dV = R.dT
where Cp, Cv and R are in same units.
Hence, proved.
Let T be the temperature on the faulty thermometer when the true thermometer reads 40°C.
∴
Therefore, temperature on the faulty scale will be
The relation between Celsius and Kelvin is given by,
Let
∴
The relation between Celsius and Fahrenheit is,
Let
∴
Thus,
That is, the temperature is and the temperature of human body is
Let the latent heat of vaporisation of a liquid of mass m at its boiling temperature be L.
Let Vℓ be the volume of liquid and Vv be the volume of vapours and let evaporation take place at constant pressure P.
According to the first law of thermodynamics,
This is the required expression for change in internal energy.
We have,
where. is the pressure of bath whose temperature is to be measured.
We know,
, is the temperature of the bath when pressure is 120 cm of Hg.
We know that pressure is directly proportional to the temperature on absolute scale.
Temperature of triple point of water,
Let T be the temperature of dry ice on Kelvin scale.
So, we have
, is the temperature of the dry ice.
Heat required to raise the temperature of n1 moles of gas A at constant volume through 1°C is given by,
Resistance thermometer is a thermometer, which determines the temperature by studying the variation in resistance of metal due to change in temperature.
Resistance of metals approximately increases linearly with temperature.
By every one-degree rise or fall in temperature, the resistance of conductor increases or decreases by some constant multiple value of resistance of zero degree centigrade.
∴
...(1)
When t = 100,
...(2)
Dividing (1) by (2), we get
where,
Cv is the molar specific heat at constant volume.
In terms of γ, the value of Cv can be written as
We have,
Ratio of specific heat, y = 1.4
R=2 cal/mole/K
Now, using the formula,
Solving(1) and (2), we get
CP = 7 cal/mole/k, is the specific heat at constant volume.
and Cv = 5 cal/mole/K, is the specific heat at constant pressure.
Work done on or by the gas during any process is given by,
where,
P is the pressure acting on the system, and
dV is the change in the volume.
For an isothermal expansion of gas, work done is given by,
where,
R is the universal gas constant,
T is the temperature of the gas,
Vf volume to which the gas expands, and
Vi initial volume of the gas.
Work done by gas during adiabatic process is,
The water at the bottom of the Niagara Falls, which are 50 m high, should be warmer than that at the top. Explain why it is and calculate the temperature rise.
The heat capacity of 1 mol water (which weighs 0.018 kg) is 80 J/K.
The acceleration due to gravity is 9.8 m/s2.
(a) What do the horizontal regions AB and CD represent?
(b) If CD = 9AB, then what interpretation can you make?
(c) What does the reciprocal of slope of BC represent?
(a) The melting of solid is represented by the horizontal portion AB and the magnitude of AB is equal to the latent heat of fusion in kcal/kg.
The horizontal portion CD represents the boiling of liquid and the magnitude of CD is equal to latent heat of vaporisation in k cal/kg.
(b) CD = 9AB implies that the latent heat of vaporisation is nine times the latent heat of fusion.
(c) The curve BC represents the vapour state of substance.
We know,
dQ = msdT
So, the slope of the graph i.e. represents the heat capacity of the liquid.
Let m be the mass of water required to just melt the ice.
By law of calorimetry,
Heat lost by water = Heat gained by ice
Therefore,
m x l x 20 = 800 x 0.5 x 12 +800 x 80 20m = 68800
m = 3440gm
= 3.44 kg
Weight of the copper ball = 400 g
Weight of water = 1000 gm
Temperature = 20o C
Increased temperature = 80o C
Let T be the temperature of furnace.
According to law of calorimetry,
Heat lost by copper = Heat gained by water.
That is,
400 x 0.1 x ( T– 80 ) = 1000 x 1 x 60 T - 80 = 1580
T = 15800C
The amount of work done to convert 1 gram water into steam at 1 atmospheric pressure is given by,
Therefore, the internal energy increases by 500 Cal.
Given,
Heat required to raise the temperature, QP = 70 cal
Number of moles of gas, n = 2 mole
CP = 7 cal/mol/K
Therefore,
Cv = 5 cal/mol/K
Therefore, the quantity of heat required to increase the temperature at constant volume is,
Evaporation | Boiling |
1. The process takes place at all temperature. | 1. Boiling occurs only at boiling temperature. |
2. The process usually occurs on the surface of the matter. | 2. Boiling occurs below the surface. that is, it is a bulk phenomenon. |
Define relative humidity in terms of pressure.
(a) Why is one’s breath visible in winter but not in hot summer?
(b) How does water get cooled in earthen pots?
(a) There are water vapours in the exhaled breath which condense into small droplets of water in winter and become visible.
But in summer the condensation does not take place.
(b) There are pores in the earthen pots and from these pores, water oozes out and evaporation takes place. Now, since cooling is caused after evaporation, therefore, water gets cooled down.
An isobaric process is in which the pressure remains constant but volume and temperature are allowed to change.
i.e. ∆P = 0
Ideal gas equation is given by,
For an isobaric process, P is constant.
Therefore the equation of state for isobaric process reduces to,
P versus V graph is as shown below.
Now, according to first law of thermodynamics,
Here, a part of heat absorbed by gas, increases the internal energy and a part of the heat is used to do work for expansion.
(a) What is an isochoric process? What is its equation of state? Plot P versus V graph.
(b) Discuss the isochoric process using first law of thermodynamics.
(c) What is the work done by or on the gas in isochoric process?
(a) The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points?
(b) There were two fixed points in the original Celsius scale i.e. 0°C and 100°C. On the absolute scale one of the fixed point is the triple point of water. What is the other fixed point on Kelvin scale?
(c) Absolute temperature and temperature on Celsius scale are related as
Why do we have 273.15 in the relation and not the 273.16?
(a) The triple point of water is a unique temperature and does not change with pressure.
The melting point and boiling point of water vary with pressure.
That is, there is no fixed melting and boiling point of water.
(b) The second fixed point on the absolute scale is zero on the absolute scale.
(c) We have 273.15 in the relation, because the triple point of water is 0.01 °C.
Given,
Room temperature, T = 16°C
Dew point temperature, T0 = 7.4°C
SVP at 7°C = 7.5mm of Hg
SVP at 8°C = 8mm of Hg
SVP at 16°C = 13.5mm of Hg
For a small change in temperature, it can be assumed that saturated vapour pressure changes linearly.
The saturated vapour pressure increases by 0.5 mm for one degree rise in temperature.
Therefore increase in saturated vapour pressure for 0.4°C increase in temperature will be 0.2mm of mercury.
Hence, the saturated vapour pressure at dew point i.e. 7.4°C is 7.7 mm of mercury.
Now the relative humidity is,
(a) What are the limitations of first law of thermodynamics?
(b) Two identical samples of a gas are expanded so that volume is increased to twice. One is expanded isothermally and second adiabatically. In which sample, pressure is greater? Explain.
Limitations of the First Law of Thermodynamics are:
(a) First law of thermodynamics does not tell us about the direction in which a given process can take place and up to what extent it can take place.
(b) Let P and V be the pressure and volume of gas.
Let Pi be the pressure of gas when expanded isothermally.
Pa be the pressure of gas when expanded adiabatically.
∴ ∴
Here, we can see,
Therefore, .
So, pressure is greater in the isothermal process.
No, convection only occurs in fluids but not in solids.
How can we measure high temperature like that of the sun and stars?
Strike off the alternatives not applicable, in an isothermal process:
(i) Temperature (remains/does not remain) constant.
(ii) The internal energy of gas (changes/does not change).
(iii) Work done by the gas is (equal/not equal) to heat added.
(iv) The gas (remains/does not remain) in thermal equilibrium with surroundings.
(i) Does not remain
(ii) changes
(iii) not equal
(iv) does not remain.
Two samples of an ideal gas are compressed to half of the initial volume. One sample is compressed isothermally and second adiabatically.
(a) In which sample pressure is high?
(b) On which sample more work will be done?
There are three ways by which heat is transmitted from one body to another:
(i) Conduction: This is the phenomenon in which thermal energy is transferred from hotter part of the body to colder part or vice versa. There is no contact without any transference of material particles.
(ii) Convection: In this process, heat is transferred from one point to another point by actual movement of heated particles.
(iii) Radiation: In this process of transmission heat travels directly from one place to another place without heating any intervening media.
The laws of thermal conduction under steady state the rate of transfer of heat under steady state:
(i) is directly proportional to temperature difference (T2 – T2) across the ends of conductor,
(ii) is directly proportional to area of cross section A of the conductor and
(iii) is inversely proportional to the length £ of the conductor.
Combining all these laws, we get
where K is constant of proportionality and is called thermal conductivity of conductor.
S.I. unit of thermal conductivity is Watt/m/K or J/s/m/K.
CGS unit of K is cal/cm/sec/K or ergs/cm/sec/K.
Dimensional formula of thermal conductivity is [M1L1T–3K–1]
This heat is used in following parts:
(i) A part of total heat received is retained by the cross-section Qa.
(ii) A part of heat (Q1) is transferred to next cross-section towards cold end side. As a result, there is the flow of heat.
Since the temperature of various parts of the rod is increasing, the state is called transient state.
Steady state: As the temperature increases, the rate of loss of heat also increases.
After some time,
Q3= Qa and Q2 =0.
At this stage, the temperature of cross section attains a constant value. This stage is called steady state.
Temperature gradient: At a steady state, the temperature at a given cross-section is constant.
But, the temperature actually decreases along the length of the rod in a direction of flow of heat and is different in different directions.
The rate of change of temperature with distance in the direction of heat flow is called temperature gradient.
The temperature gradient at any point is represented by dθ/dx.
In transient state, the rate at which the rod gets heated is not the same for all the materials.
Heating depends not on the thermal conductivity but also depends on the thermal capacity per unit volume of material.
The rate at which the temperature of any part of rod increases is called thermometric conductivity or diffusivity. Thermal conductivity is equal to the ratio of thermal conductivity to the thermal capacity per unit volume of material.
i.e., for a given conductor, the rate of flow of heat is directly proportional to the temperature difference across the length of is conductance.
The reciprocal of thermal conductance i.e. l/KA is called thermal resistance.
The unit of thermal resistance is J–1 Ks.
The dimensional formula is [M–1L–2T–3K1]
An ideal gas is taken through as shown in figure. Find the work done by gas in the process
What is the net heat absorbed by the gas in cyclic process?
Given here,
Heat
Work done,
According of first law of thermodynamics,
Since increase in internal energy of gas is zero, therefore increase in temperature is also zero.
Explain the phenomenon of transfer of heat by means of convection.
Thermal radiations is the heat energy emitted by a body in the form of radiation.
Following are the properties of thermal radiations:
(i) These radiations are electromagnetic in nature.
(ii) These radiations travel in a straight line with the speed of light.
(iii) These radiations can travel through vacuum as well as in medium.
(iv) In a homogenous media, these radiations travel equally in all the direction.
(v) These radiations undergo the phenomenon of reflection, refraction, etc.
Define the terms:
(i) Absorptance,
(ii) Reflectance
(iii)Transmittance.
(i) Absorbtance: It is defined as the ratio of the amount of thermal radiation absorbed by the body in a certain time to the total amount of thermal radiation incident on it at the same time.
(ii) Reflectance: It is the ratio of the amount of thermal radiation reflected from a body in a certain time to the total amount of thermal radiation incident on it in the same time.
(iii) Transmittance: It is the ratio of the thermal radiation transmitted by it in a certain time to the total amount of thermal radiation incident on it in the same time.
The spectral absorptive power of a body for a wave of wavelength X is defined as the ratio of the heat energy absorbed to the heat energy incident upon per unit time within the unit wavelength range (λ–1/2)to(λ+1/2) around X.
Specral absorptive power is denoted by aλ.
Kirchoff's Law states that the ratio of the emissive power to absorptive power for the radiation of a given wavelength is same for all substances at the same temperature. This constant is equal to the emissive power of perfectly black body of the same wavelength and the same temperature.
The following observations can be explained using Kirchhoff’s law:
(i) In a solar spectrum, a number of dark lines called Fraunhoffer lines are observed. But during solar eclipse, same bright lines are observed as that of Fraunhoffer dark lines. Kirchhoff’s law explains this observation.
(ii) The outer wall of thermos flask is made shining, so that neither it absorbs more heat from surrounding (contents in the flask) nor it radiates more heat to surrounding (space outside) thereby keeping the temperature of contents in flask constant.
(iii) Mark an ink dot on a china ware. You will notice the ink dot dark and china ware shines. If it is heated to high temperature and taken to dark room, the dot begins to shine brightly than chine ware. This indicates that good absorber is a good emitter. This application is also explained by the Kirchoff's law.
When a system undergoes a cyclic process, it is finally brought back to initial state in all respects.
Suppose, an ideal gas is contained in a cylinder fitted with a massless and frictionless piston.
Let the gas be initially in the state represented by point A on the P–V graph. Let it be taken from state A to C along path ABC and brought back to A via path CDA.
Case 1: Work done by gas is equal to area ABCKLA.
Case 2: Work done on the gas is equal to area CDALKA.
Now net work done by gas,
According to first law of thermodynamics,
In cyclic process the change in internal energy is zero.
So,
Net heat absorbed during a cyclic process is equal to work done by gas.
Therefore, the area of the loop on the indicator diagram is equal to both work done and the heat absorbed in the cyclic process.
Let T be the temperature at the interface of two slabs
The rate of flow of heat through slab A is,
The rate of flow of heat through slab B is,
As two slabs are in series, therefore the rate of flow of heat through them will be same.
The rate of flow of heat through composite slab is,
A machine that converts heat energy into mechanical energy is called heat engine.
Heat engine consists of three essential parts:
(i) Source: It is a hot body at high temperature having infinite heat capacity.
(ii) Working substance: It is an ideal gas.
(iii) Sink: It is cold body at low temperature having infinite heat capacity.
The block diagram for the engine is as shown:
The working substance is carried through a series of cyclic process, in which it absorbs some heat from source.
A part of the heat is converted into mechanical work and rest of heat is rejected to sink.
In a cyclic process, the working substance absorbs heat from source and rejects Q2 heat to the sink.
The change in internal energy of gas is zero.
Therefore, using first law of thermodynamics, the net amount of work done by an ideal engine is,
The efficiency of heat engine is defined as the ratio of work obtained to the heat taken up from the source. Efficiency is denoted by .
i.e.
The working substance is taken through a cycle of four operations known as Carnot’s cycle in a Carnot engine.
Four operations of Carnot cycle are:
(i) isothermal expansion,
(ii) adiabatic expansion,
(iii) isothermal compression, and
(iv) adiabatic compression.
Initially the working substance is heated to a temperature of source T1.
The initial thermodynamic state (P1,V1,T1) of working substance.
(i) Isothermal expansion: The cylinder is placed on source and working substance is allowed to expand slowly so that it expands isothermally.
Let the thermodynamic states of working substance change from (P1,V1,T1) to (P2,V2,T1).
Temperature remains constant as it expands isothermally.
This process is depicted on PV diagram by curve AB.
Let Q1 be the heat absorbed by the gas and W1 be the work done by it.
(ii) Adiabatic expansion: Now, the cylinder is placed on an insulated pad and allowed to expand quickly so that the temperature of working substance falls to the temperature of sink T2.
The thermodynamic states of working substance in this process changes from (P2,V2,T1) to (P3,V3,T2).
This process on PV diagram is represented by curve BC.
In this process, the heat absorbed by gas is zero.
Work done on the gas is W2.
(iii) Isothermal compression: When an isothermal compression is carried out, the cylinder is placed on the sink and working substance is compressed slowly so that it compresses isothermally.
The thermodynamic states of working substance change from (P3,V3,T2) to (P4,V4,T2).
This process is shown on PV diagram by curve CD.
Let Q2 be the heat evolved and W3 be the work done on the working substance.
(iv) Adiabatic compression: Finally, the cylinder is placed on an insulated pad and compressed quickly so that the working substance returns to initial state (P1,V1,T1).
This process on PV diagram is illustrated by curve DA.
In this process, the heat evolved by gas is zero.
Work done on the gas is W4.
Net amount of work done by the working substance in 1 complete cycle.
In isothermal and adiabatic expansion, the work W1 and W2 is done by the gas.
Using sign convention W1 and W2 are positive.
In isothermal and adiabatic compression, the work W3 and W4 is done on the gas.
Therefore W3 and W4 are negative.
Thus net work done by working substance in one complete cycle is,
W = W1 + W2 + (-W3) + (-W4)
From equation (2) and (4),
W2 = W4
Therefore,
W = W1 - W3 = Q1 - Q2 ... (5)
That is in a Carnot's heat engine, the mechanical work obtained during each cycle is equal to the area of the loop on PV diagram.
Efficiency of Carnot’s cycle.
The efficiency of Carnot heat engine is given by,
Now, it has been observed that at a steady state, the quantity of heat flowing through every section is same.
i.e., the rate of flow of heat at point P, Q and Y is same.
Quantity of heat that flows from section P to Q in one second is given by,
where,
θ1 and θ2 be the reading of thermometers T1 and T2 respectively.
Amount of heat gained by water in one second at end Y is,
where,
m is the mass of water collected in time t,
s be the specific heat and 3 and θ4 be the reading of thermometers T3 and T4.
Now, equating (1) and (2), we get
The various processes which take place in the working of the refrigerator are:
(i) Adiabatic compression of working substance (ideal gas) outside the room of refrigerator.
(ii) Cooling of adiabatically compressed working substance.
(iii) Adiabatic expansion of cooled working substance inside the room of refrigerator.
Carnot engine cannot be realised in practice because of the following reasons:
(i) source and sink of infinite thermal capacity cannot be realised in actual practice,
(ii) we cannot have perfect insulators and perfect conductors in actual practice.
iii) compression and expansion cannot be carried out very slowly for the cycle to be reversible.
Temperature of source,
Temperature of sink,
The efficiency of Carnot engine is given by,
The air above the water surface in a pond is at temperature –θ°C as a result of which the water in the pond starts freezing. Find the time taken by the pond to freeze from thickness x1 to x2. Assume the required data.
Here,
Lower temperature,
Temperature,
The efficiency of engine is,
Since,
Therefore, the quantity of heat received by engine from source per second is,
Now the heat rejected to sink per second,
Mass of ice melting per second, m = 1.5 gm/s
Latent heat of ice, L = 80 cal/gm = 80x4.2 J/gm
Therefore rate of flow of ice through slabs is,
The efficiency of engine P is,
The efficiency of engine Q is,
Here,
∴
This is the required temperature at which the efficiency of both engines is same.
Given,
Heat released,
Since,
∴
Let at temperature the engine's performance be 150J.
Thus,
and
Now
The hot reservoir should be raised to a temperature of 596.84 K.
Given here,
Heat released, Q = 1000cal = 4200J
Efficiency,
Therefore,
Work done,
Let the man climb up to height h.
Therefore,
i.e.,
Power developed = 45 HP = 33570 J/s
Coal consumed per hour = 30kg
Heat produced per hour =
Heat produced per sec =
Efficiency,
Time taken to freeze the lake from thickness x1 to x2 is given by,
Given that,
The coefficient of performance of refrigerator is,
Also,
∴ Work done by the compressor of refrigerator to remove 200k cal of heat is given by,
Initial thickness of layer of ice, x1 = 3 cm
Final thickness of ice, x2 = 3.2 cm
Temperature of air, = -25o C
Density of ice, = 900 kg/m3
Thermal conductivity, K = 0.005 cal/msK
Time taken to freeze the lake from thickness to x1 is given by,
Given,
Initial temperature,
Final temperature,
Power, P = 240 W
The coefficient of performance of refrigerator,
Therefore,
Quantity of heat removed from refrigerator per second is,
Quantity of heat rejected to room per second,
Quantity of heat rejected to room per minute,
Kirchoff's law states that the ratio of the emissive power to absorptive power for the radiation of a given wavelength is constant for all substances, at the same temperature. This constant is equal to the emissive power of a perfectly black body at the same temperature and corresponding to the same wavelength.
Proof:
Let eλ and aλ be the monochromatic emittance and absorptance of the body for wavelength at temperature T.
Let Q be the amount of radiation incident per unit area on the body in the wavelength range (λ–1/2) to (λ + 1/2 ).
Therefore,
Amount of radiation absorbed by body per second per unit area = aλ Q.
The body is in thermal equilibrium with the enclosure.
Therefore, the amount of radiations absorbed by body per sec per unit area in unit wavelength region is equal to the amount of radiations emitted by body per sec per unit area in the same unit wavelength region.
Therefore, for black body
Interpretations from this graph is as follows:
(i) At a given temperature, the energy distribution is not uniform in spectral radiations of a hot black body.
(ii) At a given temperature, emissive power increases first with increases in wavelength. Emissive power is maximum, at a particular wavelength and then it decreases with further increase in wavelength.
(iii) At every temperature, there exists one wavelength at which emissive power of the black body is maximum. This wavelength is called wavelength of maximum emission.
(iv) With increase in temperature, the emissive power corresponding to each wavelength increases.
(v) The wavelength of maximum emission decreases with the increase in temperature.
(vi) The area under each curve represents the total energy emitted by black body per second per unit area.
It is found that the area under the curve varies directly with fourth power of absolute temperature of black body.
i.e. E ∝ T4.
Here,
Maximum wavelength,
According to Wien's displacement law,
, is the temperature of the star.
Given that,
According to Stefan's law,
∴
This is the ratio of loss of heat.
Given,
Wavelength at which intensity of emission is maximum,
Wavelength of the distant star,
According to Wien's displacement law,
∴
That is,
A solid sphere of radius R, made from a material of density ρ and specific heat s is heated to temperature T and placed in an enclosure at temperature T0 Show that the:
(i) rate of loss of heat varies directly with square of radius and independent of specific heat and density of the material,
(ii) rate of cooling varies inversely with radius, specific heat and density.
i) According to Stefan’s law, the rate of loss of heat by a body is,
From the above relation, it is clear that the rate of loss of heat varies directly with the square of radius and independent of specific heat of material and density of material.
ii) Now, the rate of cooling varies inversely with radius, specific heat, and density.
Hence, the result.
The surface temperature of sun is given by,
Putting the values, we get
Let CHt and CHO be the r.m.s velocity of H2 gas at a different temperature.
That is,
Gunny bags have pores in them which will stop the flow of heat by the process of convection. Also, gunny bags are bad conductors of heat. Therefore, ice is packed in gunny bags.
A new quilt is warmer because the air which is trapped inside the cotton or woollen clothing acts as an insulator and does not allow the transfer of heat. However, when the quilt gets older, the air spacing of the cotton or wool gets compressed. It no longer remains a good insulator. Hence, the person does not feel warm.
Eskimos make igloos which are double walled houses. Two walls are used so that the air which is trapped in between, prevent conduction of heat from inside of the house to outside of the house. Hence, people inside would feel warm.
The thermal conductivity of copper is about 400 times greater than that of glass. That is, the heat flowing through the copper moves about 400 times faster than it moves through the glass.
And the glass holds it's heat longer resulting in the slow transfer of heat to the skin.
Therefore, the piece of iron appear hotter than glass on touching.
Here, we have
M = 4
T = 127 + 273
R =8.31 J mole-1 K-1
Average K.E per mole of helium =
Average K.E of 1 g of helium =
=
Two tumblers have stuck together into each other. How will you loosen them using hot water and cold water?
Two tumblers which are stuck together can be separated using hot and cold water. Firstly, fill the glass on the to with cold water. The cold temperature will make the glasses contract. With cold water in the top glass, pour hot water in the glass which is at the bottom. The heat will make the bottom glass expand. Spontaneous contraction and expansion of the glasses will pull them apart.
Consider 1 mole of an ideal gas at absolute temperature T, of volume V and molecular weight M.
Let N be the Avogadro's number and m be the mass of each molecule of the gas.
Then,
M = m X N
Let, C be the r.m.s velocity of the gas molecules, then pressure P exerted by ideal gas is,
P = or PV =
Now, using the gas equation
PV = RT
Therefore,
Therefore, average K.E. of translation of one mole of the gas =
i.e.,
That is, the average K.E of the gas is independent of the nature of the gas and is dependent only on the temperature of the gas.
Why light gases like hydrogen and helium are absent in the earth’s atmosphere?
The gas molecules with more mass will stay close to the surface of the planet. And the light gases like hydrogen and helium will escape from the surface of the earth. Therefore, light gases are absent in the atmosphere of the earth.
A glass is a brittle material. As soon as droplet comes in contact with hot chimney then the water droplet absorbs heat from glass surface to evaporate and thus creates a cooling effect on the glass. This rapid cooling of the region causes it to develop a crack and the glass shatters.
The internal energy of an ideal gas is a function of temperature only and is independent of pressure and volume of the gas.
The internal energy of real gas depends on the pressure and volume of the gas.
Advantages of mercury over other liquids for ordinary thermometric use is given by:
1.The mercury is a good conductor of heat and expand rapidly and thus the temperature changes quickly.
2.Mercury has a low heat capacity.
3.The mercury is opaque to light and can be seen clearly.
4.The mercury does not stick with the walls of container.
5.Mercury has a freezing point of -39o C and boiling point of 357.
The heat capacity at constant pressure CP is greater than the heat capacity at constant volume CV , because when heat is added at constant pressure, the substance expands and work.
A.
(a) The Young’s modulus of rubber is greater than that of steel.
Tips: -
Because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.A.
The stretching of a coil is determined by its shear modulus.Tips: -
The stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.Given,
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = d/2 = 0.125 cm
Length of the steel wire, L1 = 1.5 m
Length of the brass wire, L2 = 1.0 m
Total force exerted on the steel wire is,
F1 = (4 + 6) g
= 10 × 9.8
= 98 N
Young’s modulus for steel is given by,
Tension force acting on each wire is the same. Hence, the extension is the same for each wire.
Since, the wires are of the same length, the strain will also be same.
Young's modulus is given by,
Given,
Cross-sectional area of wire A = a1 = 1.0 mm2 = 1.0 × 10–6m2
Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2
a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area = F / a
Explain why:
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
Mercury molecules, which make an obtuse angle with glass have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.
On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.
Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.
When air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.
According to equation of continuity,
Area
For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other. Therefore velocity of water increases.
The small opening of a syringe needle controls the velocity of the blood oozing out. This is as a result of the equation of continuity.
At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.
When a fluid flows out from a small hole in a vessel receives a backward thrust.
A fluid flowing out from a small hole has a large velocity according to the equation of continuity.
Area × Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system
A spinning cricket ball has two simultaneous motions types of motion- rotatory and linear.
These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. Therefore, the ball does not follow a parabolic path.
From the law of continuity, we have
A1 = Area of pipe 1,
A2 = Area of pipe 2,
V1 = Speed of fluid in pipe 1, and
V2 = Speed of fluid in pipe 2.
When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more.
According to Bernoulli's principle, if velocity is more, pressure is less.
Pressure is directly proportional to height.
Hence, the level of water in pipe 2 is less. Therefore, figure (a) is not possible.
Tips: -
Given,
Angle of contact between mercury and soda lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 × 10–3 m
Surface tension of mercury at the given temperature, s = 0.465 N m–1
Density of mercury, ρ =13.6 × 103 kg/m3
Dip in the height of mercury = h
Acceleration due to gravity, g = 9.8 m/s2
Relation between surface tension, angle of contact and dip in height is given by,
Temperature | Pressure thermometer A | Pressure thermometer B |
Triple-point of water | 1.250 × 105 Pa | 0.200 × 105 Pa |
Normal melting point of sulphur | 1.797 × 105 Pa | 0.287 × 105 Pa |
Length of the steel tape at temperature T equal to 27°C, l = 1 m = 100 cm
At temperature T1 equal to 45°C, the length of the steel rod, l1 = 63 cm
Coefficient of linear expansion of steel, α = 1.20 × 10–5 K–1
Let l2 be the actual length of the steel rod and l' be the length of the steel tape at 45°C.
l' = l + αl(T1 - T)
∴ l' = 100 + 1.20 × 10-5 × 100(45 - 27)
= 100.0216 cm
Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as,
l2 = (100.0216 / 100) × 63
= 63.0136 cm
Therefore, the actual length of the rod at 45.0°C is 63.0136 cm.
Its length at 27.0°C is 63.0 cm.
Gas | Molar specific heat(CV) |
Hydrogen | 4.87 |
Nitrogen | 4.97 |
Oxygen | 5.02 |
Nitric Oxide | 4.99 |
Carbon Monoxide | 5.01 |
Chlorine | 6.17 |
The above-listed gases are diatomic. Diatomic gases have other degree of freedom apart from the translational degree of freedom.
Inorder to increase the temperature of these gases, heat must be supplied to them. this heat increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.
For only rotational motion,
Molar specific heat of the diatomic gas is = (5/2) R
=
Apart from Chlorine in the above list, for all gases molar specific heat is (5/2) R.
At room temperature, Chlorine has vibrational modes apart from rotational and translational modes of motion.
The P-T phase diagram for Carbon dioxide is as shown below:
a) At the triple point, that is temperature = -56.6 o C and pressure = 5.11 atm, solid, liquid and vapour phase of carbon dioxide exist in equilibrium.
b) Fusion and boiling point of carbon dioxide will decrease when pressure decreases.
c) For carbon dioxide, the critical temperature is 31.1° C and critical pressure is 73.0 atm. If the temprature of carbon dioxide is more than 31.1° C, it can not be liquified, however large pressure we may apply.
d) Carbon dioxide will be
(a) a vapour, at =70° C under 1atm.
(b) a solid, at -6° C under 10 atm
(c) a liquid, at 15° C under 56 atm.
The P-T phase diagram of CO2 is as shown below:
a) Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solid phase, so carbon dioxide will condense directly into the solid without becoming liquid.
b) Since the pressure 4 atm is less than 5.11 atm the carbon dioxide will condense directly into solid without becoming liquid.
c) When a solid CO2 at 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temprature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the T-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of CO2 at 10 atm.
d) Since 70° C is higher than the critical temprature of CO2, so the CO2 gas can not be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.
Given,
Side of the given cubical ice box, s = 30 cm = 0.3 m
Thickness of the ice box, l = 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m = 4 kg
Time gap, t = 6 h = 6 × 60 × 60 s
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermocole, K = 0.01 J s–1 m–1 K–1
Heat of fusion of water, L = 335 × 103 J kg–1
Let m’ be the total amount of ice that melts in 6 h.
The amount of heat lost by the food,
Area of cross-section of the piston, A = 425 cm2 = 4.25
Maximum load on piston, F = 3000 kg = 3000 X 9.8 N
Therefore,
Maximum pressure on the piston on which car is,
Mean free path = 1.11 × 10–7 m
Collision frequency = 4.58 × 109 s–1
Successive collision time ≈ 500 × (Collision time)
Pressure inside the cylinder containing nitrogen,
P = 2.0 atm
= 2.026 × 105 Pa
Temperature inside the cylinder, T = 17°C =290 K
Radius of a nitrogen molecule, r = 1.0 Å
= 1 × 1010 m
Diameter, d = 2 × 1 × 1010 = 2 × 1010 m
Molecular mass of nitrogen, M = 28.0 g
= 28 × 10–3 kg
The root mean square speed of nitrogen is given by the relation,
Collision frequency =
= 508.26 / (1.11 × 10-7)
= 4.58 × 109 s-1
Collision time is given as,
T =
=
= 3.93 × 10-13 s
Time taken between successive collisions,
T ' = l / vrms
= 1.11 × 10-7 / 508.26
= 2.18 × 10-10 s
Therefore,
Hence, the time taken between successive collisions is 500 times the time taken for a collision.
Answer the following:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T =
A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
Use the formula v = to explain why the speed of sound in air:
(b) increases with temperature,
t(s) | 0 | T/8 | 2T/8 | 3T/8 | 4T/8 | 5T/8 | 6T/8 | 7T/8 |
y (cm) | ![]() |
3 | ![]() |
0 | -![]() |
-3 | -![]() |
0 |
Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m,
(b) 0.5 m,
(c) λ/2
(d) 3λ/4
Given here,
Speed of P waves, vP = 8.0 km/s
Speed of S waves, vs = 4.0 km/s
Let, t be the time taken by the first wave P to reach from point where earth quake occurs.
Time taken,
t = 4 min = 4 x 60 s = 240 s
And, let x be the distance upto which the earthquake occurs is given by,
x = vpt
= 8 km/s x 240 s
= 1920 km
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:
A.
Arithmetic mean time of an oscillating simple pendulum
Mean deviation of a simple pendulum
Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum =
A pendulum clock loses 12 s a day if the temperature is 408C and gains 4 s a day if the temperature is 208C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:
25 C; α=1.85×10−5/ °C
60 °C; α=1.85×10−4/ °C
30°C; α=1.85×10−3/°C
55°C; α=1.85×10−2/°8C
A.
25 C; α=1.85×10−5/ °C
Time period of pendulum,
Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10-5/C
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively):
B.
For the polytropic process, specific heat for an ideal gas.
‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:
A.
As, T will be maximum temperature where product of pV is maximum
Equation of line AB, we have
The period of oscillation of a simple pendulum is . The measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:
2%
3%
1%
5%
B.
3%
Time period,
thus, changes can be expressed as
According to the question, we can write
Again time period
A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its Bob, the time period changes TM. If Young's modulus of the material of the wire is Y, then 1/Y is equal to (g = gravitational acceleration)
A.
Time period,
When additional mass M is added to its Bob where Δ is increase in length
As we know that,
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume and pressure.
If the shell now undergoes an adiabatic expansion the relation between T and R is
T ∝ e-R
T ∝ e-3R
T ∝ (1/R)
T ∝(1/R3)
C.
T ∝ (1/R)
According to given equation,
A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both the cases body is brought from an initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:
ln2,4ln2
ln2,ln2
ln2,2ln2
2ln2,8ln2
B.
ln2,ln2
Since entropy is a state function, therefore a change in entropy in both the processes must be same .
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is:
C.
For an adiabatic process TVγ-1 = constant
We know that average time of collision between molecules
Where n= number of molecules per unit volume
vrms = rms velocity of molecules
Thus, we can write
n =K1V-1 and Vrms = K2T1/2
Where K1 and K2 are constants.
For adiabatic process TVγ-1 = constant. Thus we can write
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?
A.
During oscillation, the motion of a simple pendulum KE is maximum of mean position where PE is minimum.
The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 )
2.2 x 108 Pa
2.2 x 109 Pa
2.2 x 107 Pa
2.2 x 106 Pa
A.
2.2 x 108 Pa
According to Hooke's law, i.e,
If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for the tensile case.
Give, length of a steel wire (L) = 10 cm,
Temperature (θ) = 100oC
As length is constant,
∴
Now, pressure = stress =y x strain
[Given, Y = 2 x 1011N/M2 and α = 1.1 x 10-5 K-1]
= 2 x 1011 x1.1 x 10-5 x 100
=2.2 x 108
On the heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R and the surface tension of water is T, value of r just before bubbles detach is (density of water is ρw)
A.
The bubble will detach if,
Buoyant force ≥ Surface tension force
One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:
The change in internal energy in the process AB is -350 R.
The change in internal energy in the process BC is -500 R.
The change in internal energy in the whole cyclic process is 250 R.
The change in internal energy in the process CA is 700 R.
D.
The change in internal energy in the process CA is 700 R.
According to first law of thermodynamics,
(i) change in internal energy from A to B i.e,
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)
16 cm
22 cm
38 cm
6 cm
A.
16 cm
Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected..
For air trapped in tube, p1V1 = p2V2
p1 = patm = pg76
V1 = A.8 [ A = area of cross section]
p2 = patm - ρg(54-x) = ρg(22+x)
V2 = A.x
ρg76 x 8A = ρg (22+x) Ax
x2+22x-78x 8
by solving, x = 16
A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in the same direction, then
amplitude of motion is 3a
time period of oscillations is 8τ
amplitude of motion is 4a
time period of oscillations is 6τ
D.
time period of oscillations is 6τ
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
12
8
6
4
C.
6
Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists:
List I | List II | ||
A. | Infared waves | 1. | To treat muscular strain |
B. | Radio waves | 2. | For broadcasting |
C. | X -rays | 3. | To detect fracture of bones |
D. | Ultraviolet | 4. | Absorbed by the ozone layer of the atmosphere |
A | B | C | D |
4 | 3 | 2 | 1 |
A | B | C | D |
1 | 2 | 4 | 3 |
A | B | C | D |
3 | 2 | 1 | 4 |
A | B | C | D |
1 | 2 | 3 | 4 |
D.
A | B | C | D |
1 | 2 | 3 | 4 |
Infrared rays are used to treat muscular strain.
Radio waves are used for broadcasting
X- rays are used to detect fracture of bones.
Ultraviolet rays are absorbed by ozone.
A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at the equilibrium position. The extension x0 of the spring when it is in equilibrium is
C.
At equilibrium ΣF = 0
The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it will decrease to α times its original magnitude, where α equals
0.7
0.81
0.729
0.6
C.
0.729
Amplitude of damped oscillator
The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is
povo
4povo
B.
Heat is extracted from the source in path DA and AB is
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
C.
FBD of piston at equilibrium
⇒ Patm A + mg = P0A
FBD of piston when piston is pushed down a distance x
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, the density of liquid is ρ and L is its latent heat of vaporisation.
ρL/T
T/ρL
2T/ρL
D.
2T/ρL
Decrease in surface energy = Heat required in vapourisation.... (i)
Decrease in surface energy = T x ΔA
= 2T x 4πrdr [ For two surface]
heat required in vapourisation = Latent heat
= ML = V x ρL
= (4πr2dr) ρL
Now from eq (i)
2T x 4πrdr = 4πr2ρL
r = 2T/ρL
Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in the figure. Efficiency of this cycle is nearly:(Assume the gas to be close to ideal gas)
15.4%
9.1%
10.5%
12.5%
A.
15.4%
The efficiency of a process is defined as the ratio of work done to energy supplied.
Here,
Where Cp and Cv are two heat capacities (molar)
If a simple pendulum has the significant amplitude (up to a factor of 1/e of original) only in the period between t = Os to t = τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:
0.693/b
b
1/b
2/b
D.
2/b
For damped harmonic motion,
ma = - kx - mbv
or
ma + mbv +kx = 0
Solution to above equation is
Where amplitude drops exponentially with time
Average time τ is that duration when amplitude drops by 63%. i.e., becomes Ao/e
Thus,
The transverse displacement y(x,t) of a wave on a string is given by
This represents a
wave moving in +x direction with speed
wave moving in -x direction with speed
standing wave frequency
Standing wave of frequency
B.
wave moving in -x direction with speed
Given wave is,
Two particles are executing simple harmonic motion of the same amplitude Aand frequency ω along the x - axis. Their mean position is separated by distance X0 (X0 >A). If the maximum separation between them is (X0 + A), the phase difference between their motion is
π/3
π/4
π/6
π/2
A.
π/3
x1 = A sin( ωt + Φ1)
x2 = A sin( ωt + Φ2)
100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)
8.4 kJ
84 kJ
2.1 kJ
4.2 kJ
A.
8.4 kJ
ΔQ = M,S,ΔT
= 100 × 10-3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
ΔV = 8.4 kJ
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat γ.It is moving with speed v and it suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by
C.
As no heat is lost, therefore,
Loss of kinetic energy = Gain of internal energy of gas
i.e.,
A mass M, attached to a horizontal spring, executes SHM with an amplitude A1. When the mass M passes through its mean position than a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio of (A1/A2) is
C.
At mean position, Fnet = 0
Therefore, By conservation of linear momentum,
Mv1 = (M+m)v2
Mω1A1 = (M+m)ω2A2
⇒
Water is flowing continuously from a tap having an internal diameter 8 × 10-3 m. The water velocity as it leaves the tap is 0.4 ms-1. The diameter of the water stream at a distance 2 × 10-1 m below the tap is close
to:
7.5 x 10-3 m
9.6x 10-3 m
3.6x 10-3 m
5.0x 10-3 m
C.
3.6x 10-3 m
From Bernoulli's theorem
A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?
C.
ρoil < ρ < ρwater
Oil is the least dense of them, so it should settle at the top with water at the base. Now, the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So, it will stay at the oil -water interface.
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is
4
3
2
1
C.
2
From FBD of sphere, using Lami theorem
F/Mg = Tan θ .... (i)
when suspended in liquid as θ remains unchanged,
Two conductors have the same resistance a 0°C but their temperature coefficients o resistance are α1 and α2. The respective temperature coefficients of their series parallel combinations are nearly
A.
Re = Ro +Ro
2R(1 + αsΔT)
= R( 1 +α1ΔT) + R(1+α2ΔT)
= αs = α1+α2 / 2
Rp = R x R / R+R
The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y= 0.02(m) sin The tension in the string is
4.0 N
12.5
0.5 N
6.25 N
D.
6.25 N
The tension in the string
on comparing with the standard equation,
y = a sin (ω t = kx)
T = μv2 = μω2/k2 = 0.04
= 0.04 (2π/0.004)2/(2π/0.50)2
= 6.25
The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be
2.5 x 1025
-2.5 x 1025
-1.61 x 1023
1.38 x 1023
B.
-2.5 x 1025
Using ideal gas equation
PV = nRT
(N is number of moles)
P0V0 = niR × 290 ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR × 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)
putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025
Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that
Cp – Cv = a for hydrogen gas
Cp – Cv = b for nitrogen gas
The correct relationship between a and b is
a = 14 b
a = 28 b
a = 1/14b
a= b
A.
a = 14 b
CP – CV = R
where CP and CV are molar specific heat capacities
As per the question
a= R/2
b= R/28
a= 14b
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:(Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C
1250°C
825°C
800°C
885° C
D.
885° C
Heat given = Heat taken
(100) (0.1)(T – 75) = (100)(0.1)(45) + (170)(1)(45)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C
A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T.Time taken for 10 complete oscillations is :
6.98 s
8.76 s
6.65 s
8.89 s
C.
6.65 s
I = 7.5 × 10–6 kg – m2
M = 6.7 × 10–2 Am2
By substituting value in the formula
T = .665 sec
for 10 oscillation, time taken will be
Time = 10 T = 6.65 sec
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides.K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by
3α/PK
3PKα
P/3αK
P/αK
C.
P/3αK
Due to thermal expansion
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer?
(speed of light = 3 × 108 ms–1)
17.3 GHz
15.3 GHz
10.1 GHz
12.1 GHz
A.
17.3 GHz
Doppler effect in light (speed of observer is not very small compared to speed of light)
A jar filled with two non-mixing liquids 1 and 2 having densities ρ1 and ρ2 respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in the figure.
Which of the following is true for ρ1, ρ2 and ρ3?
ρ3 < ρ1 < ρ2
ρ1 < ρ3 < ρ2
ρ1 < ρ2 < ρ3
ρ1 < ρ3 < ρ2
D.
ρ1 < ρ3 < ρ2
As liquid 1 floats above liquid 2,
ρ1 < ρ2
The ball is unable to sink into liquid 2, ρ3 < ρ2
The ball is unable to rise over liquid 1
ρ1 < ρ3 Thus, ρ1 < ρ3 < ρ2
A capillary tube (A) is dropped in water. Another identical tube (B) is dipped in a soap water solution.Which of the following shows the relative nature of the liquid columns in the two tubes?
C.
Capillary rise h = 2T cosθ /ρgr. As soap solution has lower T, h will be low.
A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be
8 cm
10 cm
4 cm
20 cm
D.
20 cm
Water will rise to the full length of capillary tube
Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity v. The retarding viscous force acting on the spherical ball is
directly proportional to R but inversely proportional to v.
directly proportional to both radius R and velocity v.
inversely proportional to both radius R and velocity v.
inversely proportional to R but directly proportional to velocity v.
B.
directly proportional to both radius R and velocity v.
Retarding force acting on a ball falling into a viscous fluid
F = 6πηRv
where R = radius of the ball
v = velocity of ball and
η = coefficient of viscosity
∴ F ∝ R and F ∝ v
Or in words, retarding force is proportional to both R and v
If two soap bubbles of different radii are connected by a tube,
air flows from the bigger bubble to the smaller bubble till the sizes are interchanged.
air flows from bigger bubble to the smaller bubble till the sizes are interchanged
air flows from the smaller bubble to the bigger.
there is no flow of air.
C.
air flows from the smaller bubble to the bigger.
The excess pressure inside the soap bubble is inversely proportional to the radius of soap bubble i.e. P ∝1/r, r being the radius of the bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.
Two non-mixing liquids of densities and n
(n >1) are put in a container. The height of each liquid is h. a solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to,
{1+(n-1)p}
{1+(n+1)p}
C.
{1+(n-1)p}
The situation can be depicted as follows:
According to Archimedes principle,
Weight of the cylinder = (upthrust)1 + (upthrust)2
i.e.,
The charge following through a resistance R varies with time t as Q = at - bt2, where a and b are positive constants. The total heat produced in R is,
a3R/3b
a3R/2b
a3R/b
a3R/6b
D.
a3R/6b
Given,
Charge, Q = at - bt2 ... (i)
We know that,
Coefficient of linear expansion of brass and steel rods are and
. Lengths of brass and steel rods are l1 and l2 respectively. If (I2 - I1) is maintained same at all temperatures, which one of the following relations holds good?
C.
Coefficient of linear expansion of brass =
Coefficient of linear expansion =
Length of brass and steel rods are l1 and l2 respectively.
Given,
Increase in length (l2'-l1' ) is same for all temperature.
So,
l2'-l1' = l2 - l1
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is,
100 cm
150 cm
200 cm
66.7 cm
B.
150 cm
b) The smallest length of the air column is associated with the fundamental mode of vibration of the air column.
Lmin =
The next higher length of the air column is,
The molecules of a given mass of a gas have r.m.s. velocity of 200 m/s at 27o C and 1.0 x 105 Nm-2 pressure. When the temperature and pressure of the gas are respectively, 127o C and 0.05 x 105 Nm-2, the RMS velocity of its molecules in m/s is,
A.
It is given that,
vrms = 200 m/s
T1 = 300 K
P1 = 105 N/m2
rms velocity of gas molecules, vrms
For two different cases,
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced its half. Then,
compressing the gas through adiabatic process will require more work to b done
compressing the gas through isothermally or adiabatically will require the same amount of work.
which of the case (whether compression through isohermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.
compressing the gas isothermally will require more work to be done
B.
compressing the gas through isothermally or adiabatically will require the same amount of work.
Plotting P-V graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume.
A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4 x 105 J/kg and g =10 N/kg]
544 km
136 km
68 km
34 km
B.
136 km
As per conservation of energy, energy gained by the ice during its fall from height h is given by,
Given, only one-quarter of its energy is absorbed by the ice.
So,
The approximate depth of an ocean is 2700 m.The compressibility of water is 45.4 x 10-11 Pa-1 and density of water is 103 kg/m3. What obtained at the bottom of the ocean?
0.8 x10-2
1.0 x 10-2
1.2 x 10-2
1.4 x 10-2
C.
1.2 x 10-2
Given d = 2700m
Compressibility =454 x 10-11 per pascal
The pressure at the bottom of ocean is given by
= 103 x 10 x 2700 = 27 x 106 Pa
So,fractional compression =compressibilty x pressure
= 45.4 x 10-11 x 27 x 106 = 1.2 x 10-2
The two ends of a metal rod are maintained at temperatures 100o C and 110o C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200o C and 210o C, the rate of heat flow will be
44.0 J/s
16.8 J/s
8.0 J/s
4.0 J/s
D.
4.0 J/s
Here, Δ T1 = 110-100 = 10o C
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. 10o C. So, the same rate of heat will flow in the second case.
Hence,
The ratio of the specific heats in terms of degrees of freedom (n) is given by
C.
The specific heat of gas at constant volume in terms of degree of freedom 'n' is
A particle is executing SHM along a straight line. Its velocities at distance x1 and x2 from the mean position are v1 and v2, respectively. Its time period is
B.
Let A be the amplitude of oscillation then,
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is
80 cm
100 cm
120 cm
140 cm
C.
120 cm
The fundamental frequencies of closed and open organ pipe are given as,
Given the second overtone (i.e. third harmonic) of open pipe is equal to the fundamental frequency of close pipe
Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be
380 J
500 J
460 J
300 J
C.
460 J
Since, initial and final points are same
A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is
. The ratio
is,
A.
Wavelength of the transverse pulse is, ... (i);
where v is the velocity of the wave and f is the frequency of the wave.
We know that,
T is the tension in the spring is the mass per unit length of the rope
From eqns. (i) and (ii), we have
For two different cases, we have
A refrigerator works between 4o C and 30o C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 Joules)
23.65 W
236.5 W
2365 W
2.365 W
B.
236.5 W
Given,
Temperature of the source, T = 30o C = 30 + 273 = 303 K
Temperature of sink, T2 = 4o C = 4 + 273 = 277 K
We know,
where, Q2 is the amount of heat drawn from the sink at T2,
W is the work done on the working substance,
Q1 is the amount of heat rejected to source at room temperature T1.
That is,
An ideal gas goes from state A to state B via three different processes as indicated in the p-V diagram.
If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and ΔU1, ΔU2, ΔU3 indicate the change in internal energy along the three processes respectively, then
Q1> Q2> Q3 and ΔU1=ΔU2= ΔU3
Q3> Q2> Q1 and ΔU1=ΔU2= ΔU3
Q1= Q2= Q3 and ΔU1=ΔU2= ΔU3
Q3> Q2> Q1 and ΔU1>ΔU2> ΔU3
A.
Q1> Q2> Q3 and ΔU1=ΔU2= ΔU3
For all process 1, 2 and 3
ΔU = UB -UA is same
Therefore, ΔU1 = ΔU2 = ΔU3
Now, ΔQ = ΔU+ΔW
Now, ΔW = work done by the gas
A train moving t a speed of 220 ms-1 towards a stationary object. emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is
(speed of sound in air is 330 ms-1)
3500 Hz
4000 Hz
5000 Hz
3000 Hz
C.
5000 Hz
From Doppler's shift, we know for this case
The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos (t)
where, X is the displacement at time t is the frequency of the oscillation
Which one of the following graphs shows correctly the variation of a with t?
A.
Putting values in the given expression,
x = A cos (t)
At t = 0, x =+ A
At, t = , x = A cos
Again, at t = T/2, x = A cos = A cos
= -A
As per the above results, only graph (i) will satisfy the above results.
If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?
Q/4πR2σ
(Q/4πR2σ)-1/2
(4πR2Q/σ)1/4
(Q/4πR2σ)1/4
D.
(Q/4πR2σ)1/4
From Stefan' law
E =σT4
so, the rate energy production
Q=E x A
Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is . Which of the following graphs is a straight line?
B.
According to Hooke's law,
Young's Modulus is given by, ... (i)
V = A xl = constant ... (ii)
From equations (i) and (ii), we have
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
Energy = 4VT is released.
Energy = 3VT is absorbed
Energy = 3 VT is released
Energy is neither absorbed nor released.
C.
Energy = 3 VT is released
If the surface area changes, it will change the surface energy also.
When the surface area decreases, it means energy is released ad vice-versa.
Change in surface energy is ... (i)
Let, there be n number of drops initially.
So, ... (ii)
Volume is constant.
So, ... (iii)
From equations (ii) and (iii), we have
Since, R > r, A is negative.
That is, the surface area is decreased.
Hence, energy must be released.
Energy released =
In the above expression, (-)ve sign shows that amount of energy is released.
Steam at 100o C is passed into 20 g of water at 10o C. When the water acquires a temperature of 80o C, the mass of water present will be,
24 g
31.5 g
42.5 g
22.5 g
D.
22.5 g
Heat lost by the steam = heat gained by water
Let, m' be the amount of heat converted into water.
m' x L = ms
m' x 540 = 20 x 1 x (80 - 10)
m' =
Therefore, net amount of water = 20 + 2.5 = 22.5 g
Certain quantity of water cools from 70o C to 60o C in the first 5 min and to 54o C in the next 5 min. The temperature of the surroundings is,
45o C
20o C
42o C
10o C
A.
45o C
Assume, the temperature of the surrounding be to C.
Case 1:
The average of 70 and 60 is 65. ... (i)
Case 2: ... (ii)
Dividing equations (i) and (ii), we get
On solving, we get
Temperature, t = 45o C
A monoatomic gas at a pressure p, having a volume V expands isothermally to a volume 16 V. The final pressure of the is (take ),
64 p
32 p
16 p
C.
For isothermal expansion,
pV = p' x 2 V []
p' = p/2
For adiabatic expansion,
A thermodynamic system undergoes cyclic process ABCDA as shown in the figure. The work done by the system in the cycle is:
po Vo
2 po Vo
zero
D.
zero
Work done in the cyclic process = Area bound by the closed configuration
= Area of closed configuration
The mean free path of molecules of a gas, (radius r) is inversely proportional to,
r3
r2
r
B.
r2
Mean free path, l =
That is,
If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by,
n = n1 + n2 + n3
A.
A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hr. He finds that traffic has eased and a car moving ahead of him at 18 km/hr is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be,
1332 Hz
1372 Hz
1412 Hz
1454 Hz
C.
1412 Hz
Both the observer and the source are moving.
Therefore, using the formula of apparent frequency,
Liquid oxygen at 50 K is heated to 3000 K at a constant pressure of 1 atm. The rate of heating is constant. which one of the following graphs represents the variation of temperature with time?
A.
Graph (a) shows the variation temperature with time. At first temperature will increase then there will be state change from liquid to gas.
One mole of an ideal gas goes from an initial state A to final state B via two processes. It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct p-V diagram representing the two processes is
D.
According to the question, first gas goes from volume V to 3V and after this volume is reduced from 3V to V at constant pressure. In the isothermal expansion, the p-V curve is a rectangular hyperbola.
A thermodynamic system is taken through the cycle ABCD as shown in a figure. Heat rejected by the gas during the cycle is
2 pV
4 pV
1/2 pV
pV
A.
2 pV
For given cyclic process, ΔU = 0
therefore, Q =W
Also, W = - area enclosed by the curve
= AB x AD
= (2p-p) (3V-V)
=-p x 2V
therefore,
Heat rejected = 2pV
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
Length = 50 cm, diameter = 0.5 mm
Length = 100 cm, diameter = 1 mm
Length = 200 cm, diameter = 2 mm
Length = 300 cm, diameter = 3 mm
A.
Length = 50 cm, diameter = 0.5 mm
The wettability of a surface by a liquid depends primarily on
viscosity
surface tension
density
angle of contact between the surface and the liquid
D.
angle of contact between the surface and the liquid
The value of angle of contact determines whether a liquid will spread on the surface or not.
The piece of iron is heated in a flame. If first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using,
Stefan's law
Wien's displacement law
Kirchoff's law
Newton's law of cooling
B.
Wien's displacement law
The equation of Wien's displacement law, i.e.,
A gas is taken through the cycle A → B → C → A, as shown, What is the net work done by the gas?
2000 J
1000 J
Zero
-2000 J
B.
1000 J
Net work done = Area enclosed in pV curve i.e. Δ ABC
= = 1000 J
In the given (V-T )diagram, what is the relation between pressures p1 and p2?
p2 =p1
p2>p1
p2<p1
Cannot be predicted
C.
p2<p1
The slope of the graph directly proportional to 1/p
The amount of heat energy required raising the temperature of 1g of helium at NTP, from T1K and T2K is
3/8 NaKB(T2-T1)
3/2 NaKB(T2-T1)
3/4 NaKB(T2-T1)
3/8 NaKB(T2/T1)
A.
3/8 NaKB(T2-T1)
We Know that,
If we study the vibration of a pipe open at both ends, then the following statements is not true
open end will be anti-node
Odd harmonics of the fundamental frequency will be generated
All harmonics of the fundamental frequency will be generated
Pressure change will be maximum at both ends
D.
Pressure change will be maximum at both ends
A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 5.13 Hz. The unknown frequency is
250 Hz
246 Hz
240 Hz
260 Hz
A.
250 Hz
Here,
Hence, the unknown frequency is 254 Hz.
Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r, Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become.
B.
As Fe =mg tanθ
We have Fe =tan θ1 ad F'e = tan θ
therefore,
Out of the following functions representing motion of a particle which represents SHM
Only (IV) does not represent SHM
(I) and (III)
(I) and (II)
Only (I)
C.
(I) and (II)
For a simple harmonic motion
Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air.
increases by a factor 20
increases by a factor 10
decreases by a factor 20
decrease by a factor 10
A.
increases by a factor 20
Velocity of sound v = nλ
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
1:2
1:4
1:5
1:1
A.
1:2
Here, KE1 = 1 - 0.5 = 0.5 eV
KE2 = 2.5 - 0.5 = 2 eV
During an isothermal expansion, a confined ideal gas does - 150 J of work against its surroundings. This implies that
300 J of heat has been added to the gas
no heat is transferred because the process is isothermal
150 J of heat has been added to the gas
150 J of heat has been removed from the gas
C.
150 J of heat has been added to the gas
From the first law of thermodynamics
ΔU = Q + W
For isothermal process, ΔU = 0
therefore, Q = - W
Given W = - 150
Therefore, Q = + 150
When Q is positive, the heat is added to the gas.
If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by
1,-1,-1
-1,-1,1
-1,-1,-1
1,1,1
A.
1,-1,-1
According to the principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct if the dimension of all the terms occurring on both sides of the equations is same.
Given critical velocity of liquid flowing through a tube are expressed as
x+ y = 0, - x-3y+z = 1, -x = -1
z = -11 x = 1, y = -1
4.0 g of a gas occupies 22.4 L at NTP. The specific heat capacity of the gas at constant volume is 5.0 JK-1 mol-1. If the speed of sound in this gas at NTP is 952 ms-1, then the heat capacity at constant pressure is.
(Take gas constant R = 8.3 Jk-1 mol-1)
8.0 JK-1mol-1
7.5 JK-1mol-1
7.0 JK-1mol-1
8.5 JK-1mol-1
A.
8.0 JK-1mol-1
Given, M = 4 gm
V= 22.4 L,
CV = 5 JK-1 mol-1
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is
155 Hz
205 Hz
10.5 Hz
105 Hz
D.
105 Hz
Given l = 75 cm, f1 = 420 Hz and f2 = 315 Hz
As two consecutive resonant frequencies for a string fixed at both ends will be +
Thus, lowest resonant frequency of a string is 105 Hz.
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20o C, th temperature of the surroundings to which it rejects heat is
31o C
41o C
11o C
21o C
A.
31o C
COfficient of performance (β) of a refrigerator is defined as the ratio of the quantity of heat removed per cycle to the work done on the substance per cycle to remove this heat.
Given, coefficient of performance of a refrigerator, β = 5
Temperature of surface i.e., inside freezer,
T2 = - 20o C = - 20 +273 = 253 K
The temperature of surrounding i.e heat rejected outside T1 =?
So,
The value of the coefficient of volume expansion of glycerin is 5 x 10-4 K-1. The fractional change in the density of glycerin for a rise of 400C in its temperature is
0.015
0.020
0.025
0.010
B.
0.020
Given the value of coefficient of volume expansion of glycerin is 5 x 10-4 K-1
As, original density of glycerine
Thus, fractional change in the density of glycerine for a rise of 40o C in its temperature,
Water rises to a height 'h' in a capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then
water rises upto the tip of the capillary tube and then starts overflowing like a fountain
water rises upto the top of capillary tube and stays there without overflowing
water rises upto a point little below the top and stays there
Water does not rise at all
A.
water rises upto the tip of the capillary tube and then starts overflowing like a fountain
It is given that water rises to a height 'h' in a capillary tube. so, the length of the capillary tube above the surface water is made less than 'h' then the height of water > length of capillary tube ⇒ So, the liquid will be staying there.
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be
β2/α2
α/β
β2/α
2πβ/α
D.
2πβ/α
For a particle executing SHM, we have maximum acceleration,
An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?
Adiabatic
Isobaric
Isochoric
Isothermal
A.
Adiabatic
Given, ideal gas is compressed to half its initial volume i.e.,
Vo = V/2
The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero ie.e Wisochoric = 0
As we know, work done on the gas = Area under curve i.e,
Wadiabatic > Wisothrmal > Wisochoric
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60o with the source observer is at rest. the apparent frequency observed by the observer (velocity of sound in air 330 ms-1) is
100 Hz
103 Hz
106 Hz
97 Hz
B.
103 Hz
Given, as a source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance. Such that, source is moving with a speed of 19.4 m/s at angle 60o with source- observer line as shown in figure
The apparent frequency heard by observer
t = 0
C.
For a perpendicular vector, we have A. B = 0
A mass of diatomic gas at a pressure of 2 atm is compressed adiabatically so that its temperature rise from 27oC to 927o C. The pressure of the gas is final state is
28 atm
68.7 atm
256 atm
8 atm
C.
256 atm
T1 = 273 + 27 = 300k
T2 = 273+927 = 1200 K
Gas equation for adiabatic process
The displacement of a particle along the x-axis is given by x = a sin2 ωt. The motion of the particle corresponds to
simple of harmonic motion of frequency ω/π
simple harmonic motion of frequency 3ω/2π
non simple harmonic motion
simple harmonic motion of frequency ω/2π
D.
simple harmonic motion of frequency ω/2π
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time t?
Q/4
Q/16
2Q
Q/2
B.
Q/16
In steady state, the amount of heat flowing from one face to the other face in time t is given by
In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated (ECE of chlorine is 0.367 x 10-6 KgC-1)
1.76 x 10-3 kg
9.67 x 10-3 kg
17.61 x 10-3 kg
3.67 x 10-3 kg
C.
17.61 x 10-3 kg
Mass of the substance deposited at the cathode
m= zit
An alpha nucleus of energy mv2/2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
1/Ze
v2
1/m
1/v4
C.
1/m
An α - a particle of mass m possesses initial velocity when it is at large distance from the nucleus of an atom having atomic number Z. At the distance of closest approach, the kinetic energy of α - particle is completely converted into potential energy. Mathematically,
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in the thermodynamical process, which of the following is true?
ΔU = - ΔW, in a adiabatic process
ΔU = ΔW, in an isothermal process
ΔU = ΔW, in an adiabatic process
ΔU = -ΔW, in an isothermal process
A.
ΔU = - ΔW, in a adiabatic process
From the law of thermodynamics
ΔQ = ΔU + ΔW
For a adiabatic process ΔQ = 0
ΔU = - ΔW
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by
σr2T4/R2
σr2T4/4πR2
σr4T4/r4
A.
σr2T4/R2
If r is the radius of the star and T its temperature, then the energy emitted by the star per second through radiation in accordance with Steafan's law will be given by
AσT4 = 4πr2σ T4
In reaching a distance R this energy will spread over a sphere of radius R; so the intensity of radiation will be given by
If Cp- Cv denote the specific heats (per unit mass) of an ideal gas of molecular weight M
Here R is the molar gas constant
Cp-CV = R/M2
Cp-CV = R
Cp-CV = R/M
Cp-CV = MR
C.
Cp-CV = R/M
Cp-CV = R
A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is
Mv2/T
Mv2 / 2T2
Mv2/T2
Mv2 / 2T
D.
Mv2 / 2T
The kinetic energy of particle = Mv2 / 2
Power = Energy /Time
P = Mv2 /2T
A wave in a string has an amplitude of 2 cm. The wave travels in the +ve direction of x -axis with a speed of 128 ms-1 and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is
y = (0.02) m sin (7.85 x +1005t)
y = (0.02) m sin (15.7 x -2010t)
y = (0.02) m sin (15.7 x + 2010t)
y = (0.02) m sin (7.85 x - 1005t)
D.
y = (0.02) m sin (7.85 x - 1005t)
Find the parameters and put in the general wave equation.
here, A = 2 cm
direction = + ve x direction
v = 128 ms-1
and 5 λ = 4
A block body t 227o C radiates heat at the rate of 7 cal cm-2 s-1. At a temperature of 727o C, the rate of heat radiated in the same units will be
60
50
112
80
C.
112
Apply Stefan's law of radiation.
The internal energy change in a system that has absorbed 2 Kcal of heat and done 500 J of work is
8900 J
6400 J
5400 J
7900 J
D.
7900 J
Heat given to a system (ΔQ) is equal to the sum of the increase in the internal energy (Δu) and the work done (ΔW) by the system against the surrounding and 1 cal = 4.2. J
According to first law of thermodynamics
ΔU = Q-W
= 2 x 4.2 x 1000-500
= 8400-500
=7900 J
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
mv3 /2
mv3
mv2/2
m2v2 /2
A.
mv3 /2
Let m is the mass per unit length rate of mass per sec = mx/t = mv
Rate of KE = (mv)v2/2 = mv3/2
In a thermodynamic process which of the following statements is not true?
In an adiabatic process, the system is insulated from the surroundings
In an isochoric process pressure remains constant
In an isothermal process, the temperature remains constant
In an adiabatic process pVγ = constant
B.
In an isochoric process pressure remains constant
For an adiabatic process, there should not be any exchange of heat between the system and its surroundings. All walls of the container must be perfectly insulated. In adiabatic changes, gases obey Poisson's law, ie pVγ = constant. In an isochoric process volume remains constant and for isobaric process pressure remains constant.
Each on the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 gm-1. When both strings vibrate simultaneously the number of beats is
5
7
8
3
B.
7
The number of beats will be the difference of frequencies of the two strings.
Sodium has body centred packing.Distance between two nearest atoms is 3.7 A. The lattice parameter is
6.8 A
4.3 A
3.0 A
8.6 A
B.
4.3 A
The neighbour distance of a body centred cubic cell, where a is the lattice parameter.
A block of mass M is attached to the lower end of a vertical strong. The string is hung from a ceiling and has to force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be
Mg/k
2Mg//k
4 Mg/ k
Mg/2k
B.
2Mg//k
Use of the law of conservation of energy. Let x be the extension in the spring.
Applying conservation of energy
The driver of a car travelling with speed 30 ms-1 towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 ms-1, the frequency of reflected sound as heard by driver is
550 Hz
555.5 Hz
720 Hz
500 Hz
C.
720 Hz
Apply Doppler's effect.
Whenever there is a relative motion between a source of the sound and the observer (listener), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by the source.
On a new scale of temperature (which is linear), an called the W scale, the freezing and boiling point of water are 39o W and 239o W respectively. what will be the temperature of 39o C celsius scale?
78o W
117o C
200o C
139o W
B.
117o C
In general, whenever we are to go from any known scale to any unknown scale, then we follow the equation
(temperature on known scale) - (LFP for known scale)
______________________________________________
(UFP - LFP ) Known (temerature on unknown scale)
= - (LEP for unknown scale)
________________________________________
(UFP-LFP) unknown
or
t= 117o W
LFP = Lower fixed point
UFP = Upper fixed point
If Q, E andW denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then
W =0
Q = W = 0
E=0
Q =0
C.
E=0
In a cyclic process, the system is not isolated from the surrounding.
In a cyclic process, a system starts from one point and ends at the same point. In this case, the change in the internal energy must again be zero and therefore the thermal energy added to the system must equal the work done during the cycle. That is, in a cyclic process.
ΔU = 0 or E = 0
Two simple harmonic motions of angular frequency 100 and 1000 rad s-1 have the same displacement amplitude. The ratio of their maximum acceleration is
1:10
1:102
1:103
1:10
B.
1:102
Acceleration of simple harmonic motion is
An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 200 C? The temperature boiling water is 100o C.
6.3 min
8.4 min
12.6 min
4.2 min
A.
6.3 min
Heat is evolved due to joule's effect is used up in boiling water.
such as
Vlt = msΔt
t = msΔt/VI
Putting under given values
I= 4A, V= 220 volt, m =1 kg,
Δt = (100-20)oC, s = 4200 J/kgo C
t = 1 x 4200 x 80 / 220 x 4 = 6.3 min
A point performs simple harmonic oscillation of period T and the equation of motion is given x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
T/8
T/6
T/3
T/12
D.
T/12
Velocity is the time derivative of displacement.
Performing SHM
Thus, at a T/12 velocity of the point will be equal to half of its maximum velocity.
The wave described by y = 0.25 sin (10 π x -2 πt), where x and y are in metre and t in second, is a wave travelling along the
- ve x direction with frequency 1 Hz
+ve x direction with frequency π Hz and wavelength λ = 0.2 m
+ve x direction with frequency 1 Hz and wavelength λ = 0.2 m
-ve x direction with amplitude 0.25 m and wavelength λ = 0.2 m
C.
+ve x direction with frequency 1 Hz and wavelength λ = 0.2 m
The sign between two terms in the argument of sine will define its direction.
Writing the given wave equation
y= 0.25 sin (10 πx- 2πt) ... (i)
The minus (-) between (10πx) and (2πt) implies that the waves is travelling along the positive x-direction.
Now comparing eq (i) with standard wave equation
y = a sin (kx -ωt) ..... (ii)
we have
a= 0.25m, ω = 2π, k = 10 π m
therefore,
2π/T = 2π
f= 1Hz
λ = 2π/k = 2π/10π = 0.2 m
Therefore, the wave is travelling along +ve x direction wth frequency 1 Hz and wavelength 0.2m
At 10o C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110o C this ratio is
x
383x/283
10x/110
283x/383
D.
283x/383
Writing ideal gas law
PV = nRT
If the cold junction of a thermocouple is kept at 0o C and the hot junction is kept at To C, then the relation between neutral temperature (Tn) and temperature of inversion (Ti) is:
Tn = 2Ti
Tn = Ti -T
Tn = Ti + T
A.
It is found that temperature of inversion (Ti) is as much above the neutral temperature (Tn) as neutral temperature is above the temperature of the cold junction (T), ie,
Ti -Tn = Tn-T
Ti = 2Tn-T
But, here the cold junction is kept at 0o C, hence
T = 0
Thus, Ti = 2Tn
Or Tn = Ti/2
Nickel shows the ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show:
paramagnetism
anti -ferromagnetism
no magnetic property
diamagnetism
A.
paramagnetism
Nickel exhibits ferromagnetism because of a quantum physical effect called exchange couplingin which the electron spins of one atom interact with those of neighbouring atoms. The result is aligment of the magentic dipole moments of the atoms, inspite of the randomizing tendency of atomic collisions. The presistent aligment is what gives ferromagnetic material their permanent magnetism.
If the temerature of ferromagnetic material is raised above a certain critical value, called the Curie temperature, the exchange coupling ceases to be effective. Most such material then become simply tend to align with an external field but much more weakly, and thermal agitation can now more easily distrupt the alignmet.
A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
T/4
T/8
T/12
T/2
C.
T/12
Let displacement equation of particle executing SHM is
y = a sin ωt
As particle travels half of the amplitude from the equilibrium position, so
y= a/2
therefore,
Hence, the particle travels half of the amplitude from the equilibrium in T/12 sec.
The particle executing simple harmonic motion has a kinetic energy Ko cos2 ωt. The maximum values of the potential energy and the toatal energy are respectively:
0 and 2Ko
Ko/2 and K0
Ko and 2Ko
Ko and Ko
D.
Ko and Ko
In simple harmonic motion, the total energy of the particle is constant at all instants which are totally kinetic when the particle is passing through the mean position and is totally potential when the particle is passing through the extreme position.
The variation of PE and KE with time is shown in the figure, by the dotted parabolic curve and solid parabolic curve respectively.
Figure indicated that maximum values of total energy KE and PE of SHM are equal.
Now, EK = Ko cos2 ωt
therefore, (EK)max = Ko
So, (EP)max = Ko
and (E)Total = Ko
A mass of 2.0 Kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan?
Take g = `10 m/s2
8.0 cm
10.0 cm
any value less than 12.0 cm
4.0
B.
10.0 cm
Let he minimum amplitue of SHM is a.
Restoring force on spring
F = ka
Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a.
ka = mg
a = mg/k
Here, m =2 kg, k = 200 N/m, g = 10 m/s2
therefore, a = 2 x 10 / 200 = 10/100 m.
10 cm
A black body is at 727o C. It emits energy at a rate which is proportional to:
(727)2
(1000)4
(1000)2
(727)2
B.
(1000)4
Amount of heat energy radiated per secnd by unit area of black body is directly proportional to fourth power of absoulte temperature.
Accordign to steafan's law,
Where σ is constatn of proportionality and called Steafan's constant. Its value is
5.67 x 10-8 Wm-2 K-4.
hence, E = (727+273)4
E = (1000)4
An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62o C, its efficiency is doubled. The temperature of the source is:
124o C
37o C
62o C
99o C
D.
99o C
Efficiency of engine is given by
For a cubic structure which one of the following relations indicating the cell characteristic is correct?
D.
In cubic crystal, the crystal axes are perpendicular to one another (α= β =γ=90) and the repetitive interval is the same along the three axes (a= b=c).
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:
0.5π
π
0.707 π
zero
A.
0.5π
The displacement equation of particle executing SHM is
x = a cos ( ωt +Φ) .. (i)
velocity, v = dx/dt = - aω sin (ωt + Φ) .. (ii)
and acceleration,
Fig (i) is a plot of eq (i) with Φ = 0. Fig (ii) shows Eq. (ii) also with Φ = 0 fig (iii) is a plot of Eq (iii) . it should be noted that in the figures the curve of v is shifted (to the left) from the curve of x by one - equate period (T/4).Similarly, the acceleration curve of A is shifted (to the left) by T/4 relative to the velocity curve of v. This implies that velocity is 90o C (0.5 π) out phase with the displacement and the acceleration is 90o C (0.5 π) out of phase with the velocity but 180o π out phase with displacement.
A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically in the spring. so that the spring is compressed by a distance d. The net work done in the process is:
B.
Work done is equal to change in energy of the body.
The situation is shown in the figure. When mass m falls vertically on spring, then spring si compressed by distance d.
Hence, network done in the process is
W= Potential energy stored in the spring + Loss of potential energy of mass
In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per min is liberated? ECE of chlorine is 0.367 x 10-6 kg/C
17.6 mg
21.3 mg
24.3 mg
13.6 mg
A.
17.6 mg
The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity i.e., charge passed through the electrolyte.
Power consumed in electrolysis, P = 100 W
Voltage applied, V = 125 V
So, current in the solution,
According to first law of Faraday, mass liberated at an electrode is directly proportional to charge passed through the electrolyte i.e.,
where z is a constant called 'electro chemical equivalent' (ECE).
Also, Q = it
Given,
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then:
B.
Force applied on the body will be equal to upthrust for vertical oscillations.
Let block is displaced through x m, then weight of displaced water or upthrust (upwards)
where A is area of cross-section of the block and is its density. This must be equal to force (=ma) applied, where m is mass of the block and
is acceleration.
or
This is the equation of simple harmonic motion.
Time period of oscillation
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency?
275 K
175 K
250 K
225 K
C.
250 K
The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e.,
Here, is the temperature of source and T2 is the temperature of sink
As given,
and
So,
Let temperature of the source be increased by x K, then efficiency becomes
A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at
4000 Å
5000 Å
6000 Å
3000 Å
D.
3000 Å
The product of wavelength corresponding to maximum intensity of radiation and temperature of the body in Kelvin is constant.
According to Wien's law
where is wavelength corresponding to maximum intensity of radiation and T is temperature of the body in Kelvin.
Given, T = 1227 + 273 = 1500 K,
T' = 1227 + 1000 + 273 = 2500 K
A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be
f/4
8f
f/2
C.
In a series LC circuit, frequency of LC oscillations is given by
or
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8cm the potential energy stored in it is
4U
8U
16U
U/4
C.
16U
Let extension produced in a spring be x initially in stretched condition spring will have potential energy
where k is spring constant or force constant.
The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant pressure to that at constant volume is
7/5
8/7
5/7
9/7
A.
7/5
We have given molar specific heat at constant pressure
Mayor's relation can be written as:
Molar specific heat at constant volume = Gas constant
i.e., CP - CV = R
Hence, required ratio is
Two sound waves with wavelengths 5.0 m and 5.5 m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second
12
0
1
6
D.
6
Let
The relation between frequency , wavelength and velocity is given by
...(i)
The frequency corresponding to wavelength,
The frequency corresponding to wavelength
Hence, no of beats per second
A transverse wave propagating along x-axis is represented by:
where x is in metres and t is in seconds. The speed of the wave is
8 m/s
D.
8 m/s
The standard transverse wave propagating along x-axis can be written as
The given equation is
The standard wave equation can be written as,
...(2)
where is amplitude, k the propagation constant and
the angular frequency,
comparing the Eqs. (i) and (ii), we have
Speed of transverse wave
The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room A?
2
4
1
A.
2
Sabine's formula for reverberation time is
where V is volume of hall in m3. total absorption of the hall (room)
Here s1, s2, s3 ...... are the surface areas of the absorbers and a1, a2, a3.......are their respective absorption coefficients.
Which one of the following statements is true?
Both light and sound waves in air are transverse
The sound waves in air are longitudinal while the light waves are transverse
Both light and sound waves in air are longitudinal
Both light and sound waves can travel in vacuum
B.
The sound waves in air are longitudinal while the light waves are transverse
In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse wave, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave itself. Light waves being electromagnetic are transverse waves.
A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is
650 kg m–3
425 kg m–3
800 kg m–3
928 kg m–3
D.
928 kg m–3
hoil ρoil g = hwater ρwater g
140 × ρoil= 130 × ρwater
A steam of a liquid of density ρ flowing horizontally with speed v rushes out of a tube of radius r and hits a vertical wall nearly normally. Assuming that the liquid does not rebound from the wall, the force exerted on the wall by the impact of the liquid is given by
πrρv
πrρv2
πr2ρv
πr2ρv2
D.
πr2ρv2
Cross-sectional area A = πr2
Volume of liquid flowing per second = AV = πr2v
Mass of the liquid flowing out per second = πr2vρ
Initial momentum of liquid per second = mass of liquid flowing x speed of liquid
= πr2vρ x v = πr2v2ρ
Since the liquid does not rebound after impact, the momentum after impact is zero.
Therefore, the rate of change of momentum = πr2v2ρ
According to Newton's second law, the force exerted on wall = rate of change of momentum
= πr2ρv2
A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at one litre per second. On increasing the pressure. this increases to two litres per second. The firman has now to
push forward twice as hard
push forward four times as hard
push backward four times as hard
push backward twice as hard
B.
push forward four times as hard
The rate of change of mass, dm/dt = avρ
Therefore, F =vdm/dt
= (avρ)v = av2ρ
A volume of liquid flowing per second = av
This is proportional to velocity. By doubling this volume of liquid flow per second the force becomes four times. when liquid flows forward, the hosepipe tens to come backwards. So to keep it intact, it should be pushed forward. Thus, the hosepipe should be pushed forward four times.
A liquid is allowed into a tube of truncated cone shape. Identify the correct statement from the following.
The speed is high at the wider end and low at the narrow end
The speed is low at the wider end and high at the narrow end.
The speed is same at both ends in a streamline flow.
The liquid flows with the uniform velocity in the tube.
B.
The speed is low at the wider end and high at the narrow end.
For an incomressible liquid equation of continuity.
AV = = constant or A ∝(1/V)
Therefore, at the wider end speed below and at the narrow end speed will be high.
Mercury boils at 367° C. However, mercury thermometers are made such that they can measure temperature upto 500°C. This is done by
maintaining vacuum above mercury column in the stem of the thermometer.
filling nitrogen gas at high pressure above the mercury column.
filling oxygen gas at a high pressure above mercury column.
filling nitrogen gas at a low pressure above the mercury column.
B.
filling nitrogen gas at high pressure above the mercury column.
If we fill nitrogen gas at a high pressure above mercury level, the boiling point of mercury is increased which can extend to the range upto 500°C
Two identical glass spheres filled with air are connected by a horizontal glass tube. The glass tube contains a pellet of mercury at its mid-point. Air in one sphere is at 0°C and the other is at 20°C. If both the vessels are heated through 10°C, then neglecting the expansions of the bulbs and the tube.
the mercury pellet gets displaced towards the sphere at a lower temperature.
The mercury pellet gets displaced towards the sphere at a higher temperature
The mercury pellet does not get displaced at all.
the temperature rise causes the pellet to expand with any displacement.
C.
The mercury pellet does not get displaced at all.
Let n1 and n2 be the number of moles in the bulbs at 0°C and 20°C, respectively, then,
pV = 273n1R = 293n2R
when the vessels are heated, let the volume of low temperature be V1 and that of the other be V2. Since pressure are same, hence
V1p = 283 n1R and V2p = 303 n2R
From Eqs. (i) and (ii), we get
Thus, the mercury pellet remains at the same position.
A girl is swinging on a swinging plate form in sitting position when the girl stands up then the period of swinging?
Will not change
Will depend on height of girl
Will be longer
Will be shorter
D.
Will be shorter
When the girl stands up, the centre of mass gets higher which effect on point of oscillation. As a result effective length will decrease. The period of will varies with square root of the effective length, time period will decrease.
we know this equation,
The displacement of the motion of a particle is represent by a equation . The motion of the particle is
SHM with amplitude
SHM with amplitude A+B
SHM with amplitude A
Oscillatory but not in SHM
A.
SHM with amplitude
A linear combination of sine and cosine functions
....1
then equation (1) becomes
it is clear that the equation number (2) in simple harmonic motion with the amplitude squaring and adding (2) and(3),we get
A ball falling in a lake depth 200 m shows a decrease of 0.1 % in its volume at the bottom then the bulk modulus of the material of the ball is
A.
The pressure exerted by water on the ball
Given:- =(density of water) ,
h=200m
p=2009.8=19.6
=
Bulk modulus B==
The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs.If p is positive, ∆V is negative.Thus for a system in equilibrium, the value of bulk modulus B is always positive. So B is positive
B =19.6
A fixed volume of iron is drawn into a wire of length l. The extension x produced in this wire by a constant force F is proportional to
l
C.
Young's mosulous= ...(l-length ,-change in length)
Graph between l and is straight line.
A capillary tube of radius r can support a liquid of weight 6.28 x 10-4 N. If the surface tension of the liquid is 5 x 10-2N/m. The radius of capillary must be
m
1.5
A.
Given:- weight =6.2410-4N,
W=T.
r==
r = 2m
A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is
48N
81N
24N
16N
B.
81N
String is vibrating in fundamental mode,it's frequency
'n' is given by ,
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.170.The frequency of the wave will be
2.94Hz
0.73Hz
0.36Hz
1.47Hz
D.
1.47Hz
Maximum displacement from mean occurs at.
For a motion in SHM particle taken th of periodic time to reach maximum displacement.
frequency n ===1.47Hz
A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C. It will take
50 sec
90 sec
60 sec
48 sec
D.
48 sec
The rate of cooling α excess of temperature
Newton's of cooling
A cylinder contains 10kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the cylinder, if final pressure is 2.5×106 N/m2 will be( temperature of gas is constant)
15.2 kg
3.7 kg
zero
7.5 kg
D.
7.5 kg
Initial mass of the gas m1=10kg
Initial pressure P1=10N/m2
Final pressure P2 =2.5×106 N/m2
According to kinetic theory of gases
A resonance air column of length 40 cm resonates with tunning fork of frequency 450Hz. Ignoring end corrections , the velocity of sound in air will be
1020m/s
920m/s
820m/s
720m/s
D.
720m/s
We know that the minimum length of the air column tube corresponds to the fundamental mode of vibrations. Hence by ignoring end corrections
Length of column =40cm=0.4m
Frequency of tuning fork = 450 Hz
Using the formula n=
or v= 450×4×0.4=720m/s
If vibration of string is to be two times, then tension in the string must be made
Eight times
Four times
Twice
Half
B.
Four times
Here initial vibration n1=n , final vibration n2 =2n
Initial tension T1 =T
Vibration frequency of string
The kinetic energy of one molecule of a gas at normal temperature and pressure will be (k = 8.31 J/mole K)
1.7×103 J
10.2×103 J
3.4×103 J
6.8×103 J
C.
3.4×103 J
According to kinetic theory, K.E of 1g-mole of an ideal gas
where R is universal gas constant, T and n are the temperature of the gas and the number of moles respectively.
Hence K.E at normal temperature 0o C =273 K
The radius R of the soap bubble is doubled under isothermal condition. If T be the surface tension of soap bubble. The work done in doing so is given by
A.
Surface tension is force per unit length ( or surface energy per unit area) acting in the plane of interface between the plane of the liquid and any other of the substance.
Molecules inside the liquid. Forces on a molecule due to others are shown above.
An ice-cube of density 900 kg /m3 is floating in water of density 1000 kg / m3. The percentage of volume of ice-cube outside the water is
20%
35%
10%
25%
C.
10%
Density is quantity of mass per unit volume.
A liquid is largely incompressible and its density is nearly constant at all pressures The percentage of the volume of ice cube outside the water
A tank is filled with water upto height H. When a hole is made at a distance h below the level of water. What will be the horizontal range of water jet?
A.
Applying Bernoulli's theorem
A charged oil drop is suspended in uniform field of 3 x 104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 x 10-15 kg and g = 10 m/s2 )
3.3 × 10-18 C
3.2 × 10-18 C
1.6 × 10-18 C
4.8 × 10-18 C
A.
3.3 × 10-18 C
In steady state,
electric force on drop = weight of drop
q E = mg
A frame made of metallic wire enclosing a surface area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by
100%
75%
50%
25%
C.
50%
Surface energy = surface tension × surface area
E = T × 2A
New surface energy
% decrease in surface energy
% decrease in surface energy = 50%
The potential energy of a molecule on the surface of a liquid compared to one inside the liquid is
zero
lesser
equal
greater
D.
greater
When the surface area of a liquid is increased, molecules from the interior of the liquid rise to the surface. As these molecules reach the surface work is done against the cohesive force. This work is stored in the molecules in the form of potential energy. Thus, the potential energy of the molecules lying in the surface is greater than that of the molecules in the interior of the liquid.
The above figure is an intermolecular cohesive force in a bulk of liquid with a free surface.
Aerofils are so designed that the speed of air
on top side is more than on lower side
on top side is less than on lower side
is same on both sides
is turbulent
A.
on top side is more than on lower side
The aerofoils are so designed that
Pupper side < Plower side
So that the aerofils get a lifting force in upward direction
According to Bernoulli's theorem, where the pressure is large, the velocity will be minimum or vice-versa.
Thus vupper side > vlower side
Two glass plates are separated by water. If surface ension of water is 75 dyne/cm and area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is
102 dyne
104 dyne
105 dyne
106 dyne
C.
105 dyne
The shape of water layer between the two plates is shown in figure
Thickness d of the film = 0.12 mm
= 0.012 cm
Radius R of the cylindrical face =
Pressure difference across the surface
Area of each plate wetted by water = A
Force F required to separate the two plates is given by
F = pressure difference × area
=
Putting the given values, we get
F =
F = 105 dyne
The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is
30 N
150 N
300 N
600 N
C.
300 N
Pressure at neck of bottle
P1 =
=
Similarly, the pressure at bottom of bottle
P2 =
P2 =
According to Pascal's law, liquids transmits pressure equal in all directions.
F2 =
F2 = 300N
A square wire frame of size L is dipped in a liquid. On taking out a membrane is formed. If the surface tension of liquid is T, then force acting on a frame will be
2T L
4T L
8T L
16T L
C.
8T L
Since, the wire frame is dipped in liquid, therefore its membrane has two free surfaces.
Total length of square wire frame in contact of membrane
= 2 x perimeter of square
= 2 × 4L
= 8L
Hence, force acting on a frame
F = T l
= T x SL
F = 8TL
64 spherical rain drops of equal size are falling vertically through air with a terminal velocity 1.5 ms-1 .If these drops coalesce to form a big spherical drop, then terminal velocity of big drop is
8 ms-1
16 ms-1
24 ms-1
32 ms-1
C.
24 ms-1
Volume of big drop = 64 x volume of a small drop
⇒ R = 4 r
The terminal velocity of spherical rain drop
⇒ v ∝ r2
∴ v2 = 16 v1
= 16 × 1.5
v2 = 24 ms-1
The equation of a wave is y = 5 sin ; where x is in cm and t is in second. The maximum velocity of the wave will be
1 ms-1
2 ms-1
1.5 ms-1
1.25 ms-1
D.
1.25 ms-1
Equation of wave
y = 5 sin
The standard equation of a wave in the given form
y = sin
Comparing the given equation with the standard equation, we get
and
ω = 25
Therefore, maximum velocity of particles of the medium,
vmax =
= 5 × 125 cm s-1
vmax = 1.25 ms-1
If there were no gravity, which of the following will not be there for a fluid?
Viscosity
Surface tension
Pressure
Archimedes' upward thrust
D.
Archimedes' upward thrust
Archimedes' upward thrust will be absent for a fluid, if there were no gravity.
The cylindrical tube of a spray pump has a cross-section of 8 cm2 , one end of which has 40 fine holes each of area 10-8 m2. If the liquid flows inside the tube with a speed of 0.15 m min-1 the speed with which the liquid is ejected through the holes is
50 m/s
5 m/s
0.05 m/s
0.5 m/s
B.
5 m/s
One of the fundamental principles used in the analysis of uniform flow is known as continuity of flow. This principle is derived from the fact that mass is always conserved in fluid systems regardless of the pipeline complexity or direction of flow.
According to the equation of continuity,
Q = A1 V1 = A2 V2
a v = constant ... (i)
Where Q is the volumetric flow rate
A is the cross-sectional area of flow
V is the mean velocity
Area of cross secrction of the spray pump,
a1 = 8 cm2 = 8 × 10-4 m2
number of holes n = 40
v1 = 0.15 m min-1 =
∴ For tube,
= a1 v1
For holes,
= a2 v2
According to law of continuity from equation (i)
∴ a2 v2 = a1 v1
∴ 40 × 10-8 × v =
⇒ v =
⇒ v = 5 m/s
Horizontal tube of non-uniform cross-section has radii of 0.1 m and 0.05 m respectively at M and N. For a streamline flow of liquid, the rate of liquid flow is
changing continuously with time
greater at M than at N
greater at N than at M
same at M and N
C.
greater at N than at M
The mass flow per unit time passing through each cross-section does not change, even if the pipe diameter changes. This is the law of conservation of mass.
A continuity equation in physics is an equation that describes the transport of some quantity. It is perticularly simple and powerful when applied to conserved quantity.
According to the equation of continuity
ρ A v = constant
If the fluid is incompressible e.g water, with ρ being effectively constant, then
A v = constant
Where A is the area of cross-section, v is the velocity and ρ is the density of a fluid.
Av gives the volume flux or flow rate remains constant throughout the pipe of flow.
The velocity of flow will increase if cross-section decreases and vice-versa.
i.e A1 V1 = A2 V2
⇒ Av = constant
Therefore, the rate of liquid flow will be greater at N than at M.
If two soap bubbles of different radii are connected by a tube, then
air flows from bigger bubble to the smaller bubble till sizes become equal
air flows from bigger bubble to the smaller bubble till sizes are interchanged
air flows from smaller bubble to bigger
there is no flow of air
C.
air flows from smaller bubble to bigger
The excess pressure inside the soap bubble is inversely proportional t radius of soap bubble i.e
where r is the radius of the bubble.
It flows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus if these two bubbles are connected by a tube, air will flow from smaller bubble to a bigger bubble and the bigger bubble grows at the expense of the smaller one. Air flows from smaller bubble to bigger as the pressure in a smaller bubble is higher.
A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If T is the surface tension of soap solution, the energy spent in the blowing
D.
Initially area of soap bubble
Under isothermal condition radius becomes 2 r, Then, area
A2 =
=
Increase in surface area
=
Energy spent
W = T × ΔA
= T . 24
W = 24
A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water
V
A.
This difference in pressure between the top and the bottom of the object produces an upward force on it. This is called upthrust.
According to Archimedes Principle, the upthrust on an object in a fluid is equal to the weight of the fluid displaced, so the volume of the object multiplied by the density of the fluid.
When a body (sphere) is half immersed, then
Upthrust = weight of sphere
∴
When body (sphere) is fully immersed then,
Upthrust = weight of sphere + weight of water pourd in sphere
V' =
The speed of air flows on the upper and lower surfaces of a wing of an aeroplane are v1 and v2 respectively. If A is the cross section area of the wing and ρ is the density of air, then the upward life is
C.
Due to the specific shape of wings when the aeroplane runs air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called dynamic lift act on the plate .
Upward lift = pressure difference × area of wing
=
The work done in increasing the size of a soap film for 10cm × 6cm to 10cm × 11cm is 3 × 10-4 J. The surface tension of the film is
1.0 × 10-2 N/m
6.0 × 10-2 N/m
3.0 × 10-2 N/m
1.5 × 10-2 N/m
C.
3.0 × 10-2 N/m
Given:-
Increasing size ΔA = (10 × 10-2 × 11 × 10-2 - 10 × 10-2 × 6 × 10-2 )
ΔA = ( 10 × 11 - 10 × 6 )
Work done = area increased × surface tension
surface tension
ST =
(since soap film has two surfaces)
ST = 3 × 10-2 N/m
A balloon contains 500 m3 of helium at 27o C and 1 atmosphere pressure. The volume of the helium at - 3° C temperature and 0.5 atmosphere pressure will be
1000 m3
900 m3
700 m3
500 m3
B.
900 m3
According to Boyle's law
V ∝
The volume of a given amount of gas is proportional to the ratio of its Kelvin temperature and its pressure
As the temperature goes up and vice versa.
Using gas equation law
⇒
⇒ V2 = 900 m3
An ice-cube of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of volume ofice-cube outside the water is
20%
35%
10%
25%
C.
10%
The percentage of volume of ice cube outside the water is
=
=
= 10 %
The lower end of capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?
5.9
4.9
2.9
1.9
C.
2.9
hHg - height of the mercury
hw - height of water
hHg =
and hw =
where, the symbols have their usual meanings
Now,
=
A capillary tube of length L and radius r is connected with another capillary tube of the same length but half the radius in series. The rate of steady volume flow of water through first capillary tube under a pressure difference of p is V. The rate of steady volume flow
through the combination will be (the pressure difference across the combination is p)
17 V
C.
Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is potential difference is replaced by Δp and electrical resistance
R1 = = R
and R2 =
= 16 R
Electric current is replaced by rate of volume flow V'
p = VR1 = VR .....(i)
p' = p = V' Req
= V' (R1 + R2 )
= 17 V'R
From Eqs. (i) and (ii), we get
VR = 17 V'R
⇒ V' =
A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be
D.
Viscous force acting on spherical drop
Fv = 6 rv
∴ Terminal velocity v =
where, η = coefficient of viscosity of liquid
σ = density of material of spherical drop
ρ = density of liquid
Power imparted by viscous force = Rate of production of heat
P =
= Fv . v
= 6 ηrv2
= 6 ηrv .
⇒ ∝ r5
A solid sphere of volume V and density p floats at the interface of two immiscible liquids of densities and p, respectively. If ρ1 < ρ < ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquid is
B.
V = Volume of solid sphere
Let V1 = Volume of the part of the sphere immersed in a liquid of density p1 and
V2 = Volume of the part of the sphere immersed in liquid of density p2
The law of flotation says that for a floating object the weight of the object equals the weight of the fluid.
According to law of flotation,
Vρ g = V1 p1 g + V2 p2 g ....(i)
V1 ( ρ ρ1 ) g = V2 ( p2 ρ )
⇒
Assertion: A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink.
Reason: The buoyancy of an object depends both on the materials and shape of the object.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
C.
If assertion is true but reason is false.
When a needle is placed carefully on the surface of water, it floats on the surface of water due to the surface tension of water, which does not allow the needle to sink.
In case of a steel ball, the surface tension of the water is not sufficient to keep it floating, so it sinks down.
A cylindrical tank is filled with water to level of 3 m. A hole is opened at height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. The square of the speed with which water is coming out from the orifice is (Take g = 10 m s-2)
50 m2 s-2
40 m2 s-2
51.5 m2 s-2
50.5 m2 s-2
A.
50 m2 s-2
Suppose A be the area of cross section of tank, a be the area of hole, ve be the velocity of
efflux, h be the height of liquid above the hole,
A
Let v be the speed with which the level decreases in the container. Using equation of
continuity, we get
ave = Av
⇒
Using Bernoulli's theorem, we have
P0 + hρg + = P0 +
ρ = density of fluid
P = pressure
⇒ hρg +
⇒ =
=
⇒ = 50 m2 s-2
Viscous force exerted by the liquid flowing between two plates in a streamline flow depends upon the
velocity gradient in the direction perpendicular to the plates
area of plate
co-efficient of viscosity of the liquid
all of these
D.
all of these
Viscous force exerted by the liquid flowing between two plates is given by
F =
So Viscous force exerted by the liquid flowing between two plates in a streamline flow depends upon the all the three factors mentioned in option.
Sponsor Area
Sponsor Area