#GrowWithGreen

Grade XIPhysics

Exam Preparation Booklet

Chapter Wise - Important Questions and Solutions

Questions

Units and Measurements

Q1. After reading the physics book, Anamika recalled and noted down the expression for

the speed of efflux of a liquid of density ρ as

However, he forgot to write ρ in the relation. Where should Anamika put the missing ρ ?

(3 marks)

Q2. Three capacitors C 1 , C 2 and C 3 are connected in series and its equivalent capacitance is

C eq .

If C eq = 0.15 ± 0.02 µF

C 1 = 0.35 ± 0.07 µF and

C 3 = 0.78 ± 0.13 µF

Find capacitance of C 2 with error limits.

(3 marks)

Motion in a Straight Line

Q1. The acceleration-time graph of a car is represented in the given figure.

Write the time intervals for which the car has moved with

(i) constant acceleration

(ii) zero acceleration

(2 marks)

Q2. A body starts from rest with varying acceleration, . Obtain the change in

position of the body.

(2 marks)

Q3. The minimum distance required by a truck, moving at a speed of 35 km/h, to stop is 40

m. Initially, if the truck was travelling at a speed of 70 km/h, then what would have

been the minimum stopping distance?

(2 marks)

Q4. Derive the equations of motion for uniformly accelerated motion.

(3 marks)

Motion in a Plane

Q1. A particle undergoes uniform circular motion along a radius of 5 cm and with a speed

given as v = 2 t 2 + 7.5 t cm/s.

Calculate

(a) the tangential acceleration at t = 2 seconds

(b) the total acceleration at t = 2 seconds

(4 marks)

Q2. Find the value of α so that vector is perpendicular to the vector .

(2 marks)

Q3. A projectile is fired at an inclination so as to have maximum horizontal range. Show

that the maximum horizontal range is four time the maximum height attained by the

projectile.

(2 marks)

Q4. A man is walking with a speed 2m/s from east to west. Rain falls vertically with a speed

of 10m/s. What is the direction in which he should hold his umbrella?

(2 marks)

Laws of Motion

Q1. A block of mass 5 kg is placed on an adjustable inclined plane. The block just begins to

slide down if the angle of inclination is made 60° with the ground.

Calculate

(a) the coefficient of static friction

(b) the force of static friction

(4 marks)

Q2. Three blocks of different masses, connected by light and inextensible strings, are pulled

by a force F = 20 N as shown below.

Calculate:

1. acceleration of each block.

2. tension in each string.

(4 marks)

Q3. Find the mass of the object in the following arrangement.

(3 marks)

Q4. Newton’s second law of motion is the basic law of motion. Explain

(1 mark)

Q5. A gun containing a bullet weighs M . The mass of each bullet is m . What will be the

recoil velocity of the gun, if the bullet is fired will speed v ?

(2 marks)

Q6. What are the centripetal and frictional forces acting on a car moving with speed v, if

such banking is done?

(2+1 marks)

Work, Energy and Power

Q1. A particle is subjected to force F x that varies with position as shown in the figure. Find

the work done by the force on the particle as it moves

(a) from x = 0 m to x = 4 m

(b) from x = 4 m to x = 8 m

(c) from x = 8 m to x = 12 m

(d) What is the total work done by the force over distance x = 0 m to x = 12 m.

(2 marks)

Q2. (i) Define conservative force.

(ii) What is the equivalence of mass and energy?

(iii) An electron is moving through a potential of 15 eV. What is the mass of the

electron?

(3 marks)

Q3. State and prove the work−energy theorem.

(2 marks)

Q4. A variable force, F acts on an object where:

F = (3 x 2 − 2 x + 4) N

How much work is done by this force in moving the object form x = 2 m to x = 4 m?

(2 marks)

Q5. Derive an expression for final velocities of two point objects colliding elastically with

each other.

(3 marks)

System of Particles and Rotational Motion

Q1. Three identical thin rods, each of length L and mass M are welded perpendicular to

each other as shown in the figure. The system is rotated about an axis passing through

the end of one rod and is parallel to another. Find the moment of inertia of the system.

(4 marks)

Q2. A wheel starts with 3 rad/s and rotates with a constant angular acceleration 4 rad/s 2 .

Find

(a) its angular displacement at t = 2s

(b) its angular speed at t = 3s

(c) number of revolutions in 2s

(3 marks)

Q3. Consider a seesaw of length l as shown in the figure. It has two blocks at the end points

of masses m and M (M > m ) respectively. At a moment, system rotates with an angular

speed ω. Find an expression for the magnitude of the system’s angular momentum.

(Rod of seesaw is mass less).

(2 marks)

Q4. Two particles A and B of mass 3 kg and 5 kg respectively are projected from same

ground level as show in the figure. Find the velocity of the centre of mass of the system.

(4 marks)

Q5. Obtain an expression for the position vector of the centre of mass of a system of two

particles of masses m 1 and m 2 . r 1 and r 2 are the position vectors of the two masses.

(3 marks)

Q6. Prove that the radii of gyration of a hollow sphere and a solid sphere having the same

radius 'r' about an axis passing through their centers and perpendicular to their plane are

in the ratio .

(2 marks)

Gravitation

Q1. The ratio of time periods of two satellites orbiting a planet is 2 : 3. Calculate the ratio of

their orbital velocities.

(2 marks)

Q2. A planet made entirely of iron has a radius of 7.5 × 10 7 m. What would be the value of

acceleration due to gravity on its surface? [Density of iron = 7850 km/m 3 ]

(2 marks)

Q3. Show that the escape speed from the moon is about five times smaller than that from the

earth. The radii of earth and moon are 6370 km and 1737 km respectively.

(2 marks)

Q4. Deduce the total energy expression of the satellite of mass m moving in a circular orbit

around the Earth? Explain the significance of this energy being negative.

(3 marks)

Mechanical Properties of Solids

Q1. Label the given stress-strain curve for a metal wire.

(1 mark)

Q2. How much pressure should be applied on a litre of water to compress it by 0.2%?

(Given that: Bulk modulus of water = 2.2 × 10 9 N/m 2 )

(2 marks)

Q3. State Hooke's law. What are the conditions in which this law is valid? Find the

expression for young's modulus of material of a wire of length l , radius of cross-section

‘ r ’ loaded with a body of mass M producing an extension Δ l in it.

(2 marks)

Q4. Find the tensile stress and elongation in a metal wire of length 6 m and the diameter of

cross- section is 28 mm on which a load of 20 kg is suspended.

(Young's modulus of the metal = 1.0 × 10 11 N/m 2 , take g = 10 ms –2 )

(2 marks)

Mechanical Properties of Fluids

Q1. An insect walks on water surface with the help of its legs. Each leg has an

approximately spherical shape of radius 2.0 × 10 −5 m. The mass of the insect is 50 ×

10 −7 kg. Find the angle at which its six legs are supported on the surface of water.

Assume the water temperature is 20°C. Surface tension of water is 0.075 N/m.

(3 marks)

Q2. (i) Name the principle which helps us to determine the speed of efflux.

(ii) A water tank has a small hole in its side at a height 25 cm from the bottom. The

height of the water column in the tank is 1 m. The air pressure inside the tank above

the water column is 1.05 × 10 5 Pa. Calculate the velocity with which water will

come out from the small hole.

(1+2 marks)

Q3. A hydraulic lift has two pistons of radii 10 cm and 25 cm each. How much force should

one exert on the piston with a smaller radius in order to use the machine to lift a car of

mass 2310 kg?

(2 marks)

Q4. Using the law of conservation of energy, derive Bernoulli’s equation for steady fluid

flow?

(4 marks)

Thermal Properties of Matter

Q1. What is the relation between the coefficients of linear expansion, area expansion and

volume expansion. What would be the change in volume of a glass rod, which is initially 75 cm long and 2 cm thick, if its temperature is increased by 50°C. [ ∝ glass = 9 × 10 −6 °C −1 ]

(2 marks)

Q2. What do the flat portions in the given graph signify?

(2 marks)

Q3. 720 g of ice at −10°C is supplied with heat energy of 210 kJ. What will be its final state

and temperature? (Heat of fusion of ice = 333 kJ/kg; specific heat capacity of ice =

2220 J/kg/k)

(3 marks)

Q4. The given figure shows a pine wall of thickness d p and a brick wall of thickness d b (= 3

d p ). The pine wall and brick wall sandwich two layers of unknown material having

identical thicknesses and thermal conductivities. The thermal conductivity of pine is k p

and that of brick is k b (= 15 k p ). After the steady state is reached, the following

measurements of interface temperatures are obtained.

T 1 = 25°C

T 2 = 20°C

T 5 = −10°C

Find the value of T 4 .

(2 marks)

Thermodynamics

Q1. Calculate the total work done by a sample of an ideal gas if it is heated by applying an

energy of 200 J. [Take γ = 1.5] (2 marks)

Q2. (i) Establish the relation,

(ii) Why is greater than ? (2+1 marks)

Q3. What is Carnot theorem? Describe Carnot cycle and derive an expression for efficiency of the Carnot engine.

(5 marks)

Q4. Derive the expressions for the work done in an adiabatic and isothermal process and show that the p-v curve of adiabatic process is always steeper than the isothermal process.

(5 marks)

Kinetic Theory

Q1. Calculate the specific heat capacity of one gram mole of a solid containing 6.02 × 10 23

number of atoms. (3 marks)

Q2. Using , explain the rise in temperature on heating a gas with the help of the kinetic theory of gases.

(2 marks)

Oscillations

Q1. The cone of the loudspeaker of a music player vibrates in SHM at a frequency of 300

Hz where the amplitude of the cone is 6 × 10 –4 m and x = A at t = 0.

(A) Find the velocity and acceleration as a function of time.

(B) Find the velocity and acceleration at t = 2.4 ms

(4 marks)

Q2. A block of mass 2 kg is attached to the ceiling by two springs parallel to each other of

same spring constant 30 N/m. Find the frequency of the vibration.

(2 marks)

Q3. For a damped oscillator m = 250 g, k = 85 N/m, and b = 70 g/s, how long does it take

for the mechanical energy of the oscillator to drop to one half of its initial value?

(4 marks)

Q4. A block of mass 700 g is fastened to a spring whose spring constant is k = 56 N/m. The

block is kept on a frictionless table and pulled to a displacement of x = 15 cm from its

equilibrium position and released from rest at t = 0.

(i) What is the amplitude of oscillation?

(ii) What is the initial phase of motion?

(iii) What is the displacement function x ( t ) for the spring-block system?

(3 marks)

Waves

Q1. Two guitar strings having identical wavelengths of 0.850 m are tuned to 450 Hz. The

tension in one of the strings is increased by 2.5% when they are stuck. Find the beat

frequency between the fundamentals of the two strings.

(2 marks)

Q2. A firehouse has a siren on the roof which makes sound of frequency of 802 Hz. The air

near the siren is blowing with speed of 10 ms –1 .

(a) Find the wavelength of the sound travelling in the medium.

(b) Fire-fighters are approaching near the siren at speed of 10 ms −1 .

What frequency do they hear? Speed of sound in still air = 343 ms −1 .

(2 marks)

Q3. The function of a transverse pulse in a string at t = 0 is given by

The pulse is travelling in the positive x-direction with a speed of 5ms –1 . Write the

function .

(1 mark)

Q4. A travelling harmonic wave on a string is expressed as

Calculate

(i) amplitude, frequency, and wavelength of the wave

(ii) displacement and velocity of the wave at x = 2 cm and t = 1 s

(4 marks)

Q5. When a speeding car goes past a police radar van, it measures a drop of 14% in the

pitch of the sound produced by the horn. The velocity of sound in air is 325 m/s.

Calculate the speed of the car.

(3 marks)

Q6. Describe standing wave and normal modes in a stretched spring.

(5 marks)

Solutions

Units and Measurements

Ans 1.

The given relation is

Physical relations should be dimensionally correct. Also, two physical quantities add

only when their dimensions are the same.

If we divide 2(P − Pa) by density ρ , then dimension of both the quantity will become

the same. This can be shown as

Also, dimension of v = [M 0 L 1 T −1 ]

And dimension of

Hence, the correct expression for the speed of efflux is

Ans 2. We can rearrange the given equation to ascertain the value of C 2 .

And the corresponding error equation would be

Now, equation (1) can be used to calculate the capacitance C 2 , so

or the capacitance C 2 would be, C 2 = 0.39 µF ...(3)

Now, the error in measurement of C 2 can be calculated using (2), so

So, the required value of C 2 is 0.39 ± 0.19 µF

Motion in a Straight Line

Ans 1. (i) The motion of a car with constant acceleration is represented by a straight line

parallel to the t -axis on an acceleration-time graph. It can be observed in the

given graph that a - t curve is parallel to the t -axis in the intervals 18 s − 20 s.

Hence, the car has moved with constant acceleration in the interval 18 s − 20 s.

(ii) When a car is moving with zero acceleration, its motion can be represented by a

straight line on the t -axis in an acceleration-time graph. It can be observed that

the curve is a straight line on the t -axis itself in the interval 10 s − 18 s. Hence,

the car has moved with zero acceleration between the interval 10 s − 18 s.

Ans 2. Given that,

varying acceleration,

For velocity v , it can be written as

For displacement dx , it can be written as

Ans 3. From the third equation of motion, we have:

If, u = 70 km/h, s = ?

Again, we have:s

Hence, the minimum stopping distance would have been 160 m.

Ans 4. For uniformly accelerated motion,

If initial velocity is taken as v 0 , final velocity as v, and t 1 − t 2 = t , then . Or,

(First equation of motion)

This relation can be represented graphically as shown below:

The area under the curve of this graph gives displacement ‘ x ’.

Total Area = Area of triangle QRT + Area of OPQT

But, v = v 0 + at.

Thus, equation (i) can also be written as:

Motion in a Plane

Ans 1. (a) The tangential acceleration is given as

or

So, at = 4 t + 7.5

Now, at t = 2 secs.

at = 4 × 2 + 7.5

So, tangential acceleration will be

at = 15.5 cm/s 2

(b) Now, the radial acceleration is given as

Now, at t = 2 seconds

Thus, a T = 105.8 cm/s 2

Thus, the total acceleration will be

Ans 2. Suppose

s

Given,

So, = AB cos 90°

⇒ 3 − 2α = 2 = 0

⇒ 5 − 2α = 0

Ans 3. Horizontal range,

The maximum height attained, for θ = 45°

From (i) and (ii), R max = 4H max

Ans 4.

Velocity of man,

Velocity of rain

Man should hold the umbrella at angle θ as shown in figure.

Negative sign show that angle is made with negative x -axis, and slope is downward.

Laws of Motion

Ans 1. Let us construct an appropriate diagram illustrating the concerned situation.

(a) Now, under equilibrium condition along the y direction

N = mg cos θ ......(1)

= 5 × 9.8 × cos 60°

or N = 24.5 N .......(2)

Now, along the x direction

fs = mg sin 60°

Now as fs = μs N, we have

μs N = mg sin 60°

or

Thus, the coefficient of static friction is μ s = 1.732

(b) The force of static friction will be

fs = μ s N = 1.732 × 24.5 Or f s = 42.43 N

Ans 2. We shall construct a free body diagram for the given situation (forces are considered

only along x− axis)

(a) Now, each block will undergo same acceleration as they are imparted with the

same force.

According to Newton’s second law,

F = ma

Or

Here,

m = m 1 + m 2 + m 3

= 2kg +5kg + 7kg = 14kg

F = 20 N

Thus, acceleration

Or

a = 1.43 m/s 2

Now, let us break up the free body diagram to find individual tensions

Now, from the above figure, we get

For block 3,

F − T 1 = m 3 × a

or T 1 = F − m 3 a

= 20 − (7 × 1.43)

Thus, T 1 = 9.99 N

For block 2

T 1 − T 2 = m 2 × a

T 2 = T 1 − m 2 a

= 9.99 − (5 × 1.43)

Thus, T 2 = 2.84 N

Ans 3. Here, we shall draw a free body diagram of the given situation.

Now, under equilibrium condition along x- axis

T = T 2 cos 60 = 30 N

Or

and along y- axis

W = T 2 sin 60 = mg

or

Thus, the mass of the object will be

m = 5.30 kg

Ans 4. Newton’s second law of motion gives the magnitude of force. According to this law,

force is the rate of change of momentum. The law expresses the requirement for a

force to accelerate an object. Also, it gives an estimate of force and its output. Hence,

the law is the basic law of motion.

Ans 5. Let the recoil velocity of the gun be ‘ V ’.

Initial momentum = 0 (before firing)

∴ Final momentum = 0 (after firing)

Final momentum of gun = ( M − m ) V

Final momentum of bullet = mv

Final momentum of gun and bullet = ( M − m ) V + mv

( M − m ) V + mv = 0

Ans 6. (a)

From the free body diagram of the banked road and the car, we can see that:

Thus, Minimum force of friction = 0

Then,

On dividing equation (3) by (2), we obtain:

Hence, the track must be banked at an angle of . (b) Let the mass of the car be ‘ m ’.

The centripetal force acting on the car will be . Hence, the frictional force is zero.

Work, Energy and Power

Ans 1. = Area under force displacement curve

(a) For the region 0 ≤ x ≤ 4 m

(b) For the region 4m ≤ x ≤ 8 m

W = 3 × 4 = 12 J

(c) For the region 8 m ≤ x ≤ 12 m

(d) For the region 0 ≤ x ≤ 12 m

W = 6 J + 12 J + 6 J = 24 J

Ans 2. (i) If the work done by a force on an object is independent of the path taken, then

the force is known as conservative force. Spring force and gravitational force are

the conservative forces.

(ii) The equivalence of mass and energy suggests that mass is equivalent to energy

and energy is equivalent to mass. Mass ( m ) and energy ( E ) are related as

(iii) It is given that,

Energy possessed by the electron, E = 15 eV = 15 × 1.6 × 10 −19 J

Speed of light, c = 3 × 10 8 m/s

The mass-energy equivalence relation is given as

Ans 3. According to the work-energy theorem, if an external force acts on a rigid object, then

the mechanical work done by the external force is equal to the change in the kinetic

energy of the object.

On using the third equation of motion, we obtain:

v 2 − u 2 = 2 as

On multiplying both sides by ,

Hence, Work done = Final kinetic energy − Initial kinetic energy.

Ans 4. Work done in producing infinitesimal displacement, dW = Fdx

Hence, 40J of work is done.

Ans 5.

We know that linear momentum and kinetic energy are conserved in an elastic

collision,

m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2

m 2 ( v 2 − u 2 ) = m 1 ( u 1 − v 1 ) ... (1)

Also,

On dividing equation (2) by (1),

Now, on solving, we obtain:

Similarly,

System of Particles and Rotational Motion

Ans 1.

For the rod along x-axis

For the rod (along y-axis) parallel to the axis of rotation

For the rod along z-axis

Consider a infinitesimal amount of material between z and z + dz has mass

and is at distance from the axis of rotation.

So,

Ans 2. Given that, ω 0 = 3 rad/ss

α = 4 rad/s 2 (a) t = 20s

Using

θ = 14 rad

(b) t = 3s

Using ω = ω 0 + α t, we get,

= 3 + 4 × 3 = 15 rad/s

ω =15 rad/s

(c) Number revolutions in 2 s,

Ans 3.

Moment of inertia of system about point 0

Given angular speed of system = ω

∴ Angular momentum of the system is given by

L = Iω

Ans 4.

Angle with x -axis

Ans 5. The given situation can be represented as

r cm = Position vector of the centre of mass of the system

The net force experienced by the system is equal to the vector sum of the forces

experienced by two masses.

Total mass of the system = m 1 + m 2

Let a 1 , a 2 , and a cm are the accelerations of mass m 1 , mass m 2 , and centre of mass

respectively. Hence, it can be written

Substituting

This is the position vector of the centre of mass of the given system.

Ans 6. We know that, the moment of inertia a given as

I = MR 2 = MK 2

Here, K is the radius of gyration

Now, for a hollow sphere, the momentum of inertia is

and for a solid sphere, the moment of inertia is

Now, the ratio of the two radii of gyration will be

Gravitation

Ans 1. According to Kepler’s third law,

Where,

T→ time period

R→ orbital radius

Now, we have

We are given that T 1 : T 2 = 2 : 3

So,

Now, we know that the orbital velocity can be written as

So,

So, the ratio of orbital velocities will be

Ans 2. The acceleration due to gravity is given as

Where,

M and R are the mass and radius of the planet respectively.

Now, we now that mass (M) = density (D) × volume (V)

or

Or M = 1.386 × 10 28 kg

Now,

Thus, the value of acceleration due to gravity will be g = 164.26 m/s 2

Ans 3. Escape speed from the surface of a planet is given by the relation,

For the Earth,

= 9.8 m/s 2 = R E = 6370 km

∴ Escape speed of the earth,

For the moon,

R M = 1737 km ∴ Escape speed of the earth,

Ans 4. Consider a satellite in a circular orbit of radius (R E + h) from center of the earth.

where, R E = radius of the earth and, h = distance of satellite from surface of the earth.

If m is the mass and v is speed of the satellite. Then the centripetal force required for

the orbit is,

directed towards the center. This centripetal force is provided by the gravitational

force, i.e.

Comparing equation (1) and (2).

We get,

Now kinetic energy of the satellite,

Now potential energy from the center of the earth is,

where, M E = mass of the earth

The negative total energy indicates that the motion of the satellite is bound towards

the Earth.

Mechanical Properties of Solids

Ans 1.

Ans 2. Change in volume = 0.2% = 0.002 l

Hence, 4.4 × 10 6 N/m 2 pressure should be applied on a litre of water to compress it by

0.2%.

Ans 3. Hooke’s law states that strain produced on a string is directly proportional to the stress applied to it. This law is obeyed by many materials as long as the load doesn’t exceed the material's elastic limit. This implies Hooke’s law is applicable until the stress-strain curve is linear (i.e. when they are directly proportional to each other). Stress α strain Stress = k × strain Where, k is the proportionality constant, and is known as the modulus of elasticity Following are the three different types of moduli: 1. Young’s Modulus of Elasticity : It is the ratio of the normal stress to the longitudinal strain given by:

Where, F - Force applied here F = mg r - Radius of the wire l - Original length Δ l - Change in length Unit → Nm −2 or Pascal (denoted by Pa)

Ans 4.

Here the restoring force will have magnitude equal to the weight of the load that is

F = mg

so F = mg

= 92.76 10 4 Nm -2

Elongation

Mechanical Properties of Fluids

Ans 1. The upward surface tension force is equal to the downward gravitational pull.

Now, the surface tension force acts all around a circle of radius r , at angle θ .

The weight mg is balanced by the vertical components of force.

Now, the length which is in contact with the water, L ≈ 2π r (as the circumference is

very small)

So, the net upward force due to surface tension is

F y ≈ (S cos θ ) L

F y ≈ 2π r S cos θ … (1)

The insect has six legs to support its balance, each supports one-sixth of the weight of the insect. So,

Ans 2. (i) The speed of efflux from the side of a container is given by the application of the

Bernoulli’s principle.

(ii) It is given that,

Height of the small hole from the bottom = 25 cm = 0.25 m

Height of the water column = 1 m

Hence, depth at which small hole is located, h = 1 − 0.25 = 0.75 m

Water density, ρ = 10 3 kg/m 3

Air pressure inside the tank above the water column, P = 1.05 × 10 5 Pa

Atmospheric pressure, P a = 1.01 × 10 5 Pa

Acceleration due to gravity, g = 9.8 m/s 2

Speed of efflux, v , is given as

Hence, the speed of efflux of water is 3.94 m/s.

Ans 3. According to Pascal’s law, the pressure on the smaller piston due to the force exerted

is transmitted undiminished throughout the liquid.

Let the magnitude of force applied on the smaller piston be F 1 .

Let the magnitude of force obtained for lifting the car be F 2.

Thus,

where, A 1 = π (10) 2 cm 2

A 2 = π (25) 2 cm 2

F 2 = Weight of the car

= 2310 × 9.8

= 22638 N.

or,

Hence, one needs to exert a force of 3622.08 N to lift the car.

Ans 4. Let us consider steady flow of a liquid from region 1 to region 2:

Here, P 1 = Pressure at region 1 on entry of liquid

A 1 = Area of cross-section in region1

h 1 = Height of region 1 from zero level

v 1 = Velocity with which liquid enters region1

P 2 = Pressure at region 2 on exit of liquid

A 2 = Area of cross-section in region 2

h 2 = Height of region 2 from zero level

v 2 = Velocity with which liquid exits region 2

Now, let us consider the flow of liquid for an infinitesimal time interval, Δ t .

Work done on fluid at region1

From the equation of continuity of steady fluid flow,

It is the same volume of fluid that exits region 2 in the same time internal Δ t .

Therefore, work done on fluid at region 2

Total work done on fluid = W 1 − W 2 = ( P 1 − P 2 ) Δ V

From the law of conservation of energy, part of this work done goes in increasing the

gravitational potential energy of the fluid as it flows from height h 1 to height h 2, and a

part of it is used in changing its kinetic energy. Also, we assume that there is no

dissipation of energy.

Increase in gravitational potential energy

where, ρ = Density of fluid

g = Acceleration due to gravity

Change in kinetic energy

Thus,

or,

This is Bernoulli’s equation for steady fluid flow.

Thermal Properties of Matter

Ans 1. If a body has the coefficient of linear expansion α, coefficient of area expansion β and

coefficient of volume expansion γ then

β = 2α

γ = 3α

We shall use the following equation for superficial expansion.

ΔV = 3 ∝ V 0 ΔT

Here,

ΔV → change in volume

V 0 → original volume

and ΔT = 50°C

Now, by plugging in the values, we get

So, the change in volume of the glass rod will be ΔV = 0.318 cm 3

Ans 2. The flat portions in the given graph occur during the melting or boiling of water.

During melting, the heat supplied to ice is used for overcoming the force of attraction

between the molecules of ice. Owing to this, the heat supplied to ice does not cause

any rise in temperature. As a result, we get a flat-shaped graph during melting.

Similarly, during evaporation, the heat supplied to water is used for overcoming the

force of attraction between the molecules of water. Owing to this, the heat supplied to

water does not cause any rise in temperature. As a result, we get a flat-shaped graph

during boiling.

Ans 3. Heat required by 720 g of ice at −10°C to raise its temperature to 0°C ,

Q 1 = C ice m ( T f − T i )

Where C ice = Specific heat capacity of ice = 2220 J/kg/k

m = Mass of ice = 0.72 kg

T f = Final temperature = 0°C

T i = Initial temperature = 10°C

Therefore, Q 1 = 0.72 × 2220 × 10 = 15984 J = 15.984 kJ

Since the ice is supplied 210 kJ of heat, it has (210 − 15.984) kJ of energy left for it to

melt.

Q rem = Remaining heat energy = 210 − 15.984 = 194.016 kJ

Mass of ice that can melt using this energy

Where L F = Heat of fusion of ice

Or,

Thus, we see that all the ice cannot melt with only 210 kJ of energy. What we obtain

is a freezing mixture of ice and water, whose temperature is 0°C.

Mass of ice remaining = 720 − 582.6 = 137.37 g

The final state thus consists of 582.6 g of water and 137.37 g of ice at 0°C.

Ans 4. Let the area of the walls be A .

Rate of heat conduction through pine wall, H p = k p A ( T 1 − T 2 ) = k p A (25 − 20) = 5 k p A

Rate of heat conduction through brick wall, H b = k b A ( T 4 − T 5 ) = 15 k p A ( T 4 + 10)

Since the conduction is said to be steady, H p = H b .

Or, 5 k p A = 15 k p A ( T 4 + 10)

1 = 3 ( T 4 + 10)

Or,

Thermodynamics Ans 1. Now, we calculate the total work done ( d W) by using the first law of thermodynamics

shown below

d W = d Q − d U

Here, d Q = n C P (T 2 − T 1 )

and dU = n C V (T 2 − T 1 )

or

Thus, Now, d W = d Q − d U

Or by plugging in the values, we get,

Thus, total work done will be d W = 66.66 J

Ans 2. (i) First law of thermodynamics is given as

Where,

= Amount of heat supplied to a system

= Change in internal energy of the system

= Work done = p

= Specific heat capacity at constant volume

= Specific heat capacity at constant pressure

Specific heat is given by the relation,

At constant pressure p , specific heat is given as

(ii) A part of C P goes on increasing the volume of the gas and the remaining

increases the temperature of the gas while C V increases only the temperature of

the gas. Hence, C P is greater than C V .

Ans 3. Carnot's theorem states that

No heat engine working at a given temperature has greater efficiency than that of a

reversible engine working at the same temperature under same conditions.

● Working between the two given temperatures of the hot ( T 1 ) and cold ( T 2 )

reservoirs, efficiency of no engine can be more than that of a Carnot engine.

● Efficiency of the Carnot engine is independent of the nature of the working

substance

● For a Carnot cycle

The sequence of steps constituting one cycle of a Carnot engine is called a Carnot

cycle. It consists of four steps

● Isothermal expansion ● Adiabatic expansion ● Isothermal compression ● Adiabatic compression

The p-v curve describing the carnot cycle is shown as:

The initial pressure, volume and temperature of the ideal gas enclosed in the cylinder

P 1 ,V 1 and T 1 . From the curve it is observed that the gas first expands isothermally to

the state having pressure, volume and temperature to be P 2 , V 2 and T 1 . From this state

it compressed isothermally to the state having pressure, volume and temperature to be

P 3 , V 3 and T 2 .From this state it is compressed adiabatically to the state having

pressure, volume and temperature to be P 1 , V 1 and T 1 .

The efficiency of the carnot engine can be found by calculating the work done in

individual cycle.

Isothermal expansion of gas

Heat absorbed ( Q 1 ) by the gas from the reservoir is the work done.

( W 1 → 2 ) by the gas,

● Step 2 → 2 s

Adiabatic expansion of gas

Work done by the gas,

● Step 3 → 4

Isothermal compression

Heat released ( Q 2 ) by the gas to the reservoir is the work done ( W 3 → 4 ) on the

gas by the environment.

● Step 4 → 1

Adiabatic compression

Work done on the gas,

● Total work done, W

● Efficiency (η) of a Carnot engine,

From the adiabatic processes in equations (2) and (4), we get

On putting the values of equation (6) in equation (5), we get

Ans 4. Isothermal Process

It is the process that occurs at a constant temperature that is PV = Constant

Isothermal changes take place in an ideal gas (at temperature T )

Consider μ moles of an ideal gas enclosed in a cylinder with perfectly conducting

walls and the piston that is perfectly frictionless and conducting

Initially,

Pressure = P 1

Volume = V 1

Finally,

Pressure = P 2

Volume = V 2

At the intermediate stage,

Pressure = P

Work done for the entire process

● There is no change in the internal energy of an ideal gas in isothermal process.

● In isothermal expansion ( V 2 > V 1 ), the gas absorbs heat and does work ( W > 0)

● In isothermal compression ( V 2 < V 1 ), work is done on the gas by the

environment and heat is released.

Adiabatic Process ● During this process, no heat enters or leaves the thermodynamic system during

the change.

● System is insulated from its surroundings.

PV γ = Constant (1) γ = Ratio of specific heats

● If an ideal gas undergoes a change adiabatically,

Initial state = P 1 , V 1 Final state = P 2 , V 2 Then,

The following graph shows the P - V curves of an ideal gas for two adiabatic processes connecting two isotherms.

Work done, From equation (1), we get

= Constant ×

is a constant

● Work done by the gas ( W > 0); so T 2 < T 1

● Work done on the gas ( W < 0); so T 2 > T 1

The slope of an isothermal or adiabatic curve is given by

For an isothermal change, PV= C

Differentiating both sides

P.dV + V.dP = 0

V.dP = − PdV

For adiabatic process, PV γ = C

Differentiating both sides

P γ V ( γ–1 ) dV + V γ .dP = 0

V γ dP = − PV (γ –1 ) dV

comparing the slopes of both the curves

for adiabatic process

for isothermal process

we know that γ >1 so the p-v curve for adiabatic process will be more steeper

than that of isothermal curve.

Kinetic Theory

Ans 1. We know that from the law of equi-partition of energy the total internal energy will be

in a three dimensional consideration.

U = 3 × K B T × N A = 3 RT

So, according to the first law of thermodynamics

ΔQ = ΔU + ΔW

or

ΔQ = ΔU + PΔV

Now, if the substance is a solid, it is considered that its physical dimensions will not

change. So, ΔV = 0

Thus

ΔQ = ΔU

or

ΔQ = Δ (3RT) = 3RΔT

Now, the specific heat is given as

or

So

C = 3R

Now, as R = 8.31 J/K/mole

The specific heat of solids will be

C = 3 × 8.31 = 24.93 J/K/mole

Ans 2. According to the kinetic theory of gases, pressure and volume are related as:

Thus, we see that temperature is a measure of the total kinetic energy of the gas i.e.,

when a gas is heated, the heat causes the gas particles to move more rapidly, thereby

increasing their average kinetic energy. This rise in the kinetic energy is causes a rise

in the temperature of the gas.

Oscillations

Ans 1. As sold in precious example

Angular frequency of motion, ω = 2πf = 1884 rads –1

So x = (6 × 10 –4 m) cos (1884t)

(A) Velocity,

Acceleration, a − (1.1304) × (1884) cos (1884t)

= –2129.6736 cos (1884t)

(B) At t = 2.4 × 10 –3 s, we have

Velocity, v = − 1.1304 sin (1884 × 2.4 × 10 –3 rad) ms −1

= 1.11 ms –1

Acceleration, a = − 2129.6736 cos (1884 × 2.4 × 10 −3 rod) = 404 ms −2

Ans 2.

Suppose the block is initially at equilibrium, then F 1 + F 2 = mg

Now the spring is lowered by a distance x from the equilibrium, then

F 1 + F 2 = mg − kx − kx

F 1 + F 2 = mg − 2kx

Comparing this equation with the general form of SHM, we get that the effective spring

constant of this system is 2k. Thus, frequency of vibration

Ans 3. (a)

(b) Mechanical energy at time t is given by,

To find the value of t at which energy we have:

By taking natural log on both sides, we obtain:

Ans 4. (i) The time-varying quantity of displacement function , (ω t + Φ) is called the

'phase' of a motion.

Since the surface is frictionless and the block is released at 15 cm from it's

equilibrium position with zero kinetic energy and maximum elastic potential

energy, the block will have zero kinetic energy, whenever it is 15 cm from the

equilibrium position. This means that the displacement will never be greater

than 15 cm.

Thus, the amplitude is 15 cm.

(ii) At time t = 0, the block is located at x = x 0.

On substituting these values in the equation, x ( t ) = x 0 sin (ω t + Φ) We have, 1 = sin Φ

Or, Φ rad

Hence, the initial phase of the motion is Φ rad.

(iii) Displacement function x ( t ) = x 0 sin (ω t + Φ),

where, Φ rad and,

therefore, x ( t ) = 15 sin (8.94 t )

Waves

Ans 1. We have

Thus, f 2 = 1.0124f 1

= 1.0124 × 450 H 2

Thus, the beat frequency is

f beat = 455.59 Hz−450 Hz

f beat = 5.59 Hz

Ans 2. (a) Sound moving upwind with speed

= 333 ms −1

Similarly, sound moving downwind with speed

v d = 343 ms −1 −10 ms −1 = 353 ms −1

(b) When the fire-fighters come from upwind direction, the relative speed of the

source is zero.

So,

When the fire-fighters come from downwind direction, the relative speed of

source is 20 ms –1 .

So,

Ans 3. After time t, the displacement

Thus, putting the expression for x ’ , in the given equation, we get

This is a required function y ( x, t )

Ans 4. (i) The general equation of a travelling wave is given as

Where,

y ( x , t ) = Displacement of the wave at distance x and time t

A = Amplitude of the wave

ω = Angular frequency = 2π ν

ν = Frequency of oscillation

k = Propagation constant =

λ = Wavelength of the wave

Φ = Phase angle

Comparing equation (1) with the given equation, we obtain

A = 1 m

ω = 2π ν = 10 rad/s

∴

k = = 4.5 × 10 −3 m −1

(ii) The travelling harmonic wave on a string is expressed as

For x = 0.02 m and t = 1 s, the given equation reduces to

Negative sign indicates that the point is below the x -axis.

Velocity of the wave is given as

Ans 5. When the car is moving towards the van, the change in horn frequency is given as

When the car crosses the van, the change in horn frequency is given as

On dividing these relations, we obtain

Where,

v 0 = Velocity of sound in air = 325 m/s

v s = Velocity of the car

The pitch (frequency) of the horn sound has dropped by 14% i.e.

Ans 6. Standing Wave and Normal Modes in a Stretched String

● Steady wave pattern in a string is called stationary wave.

● Incident and reflected waves are represented as

y = a sin ( ωt − kx ) (1)s

y ′ = − a sin ( ωt + kx ) (2)

● Applying principle of superposition,

y = y 1 + y 2

● Amplitude = 2 a sin kx (varies from point to point)

● String vibrates in phase with different amplitudes at different points. They give

rise to standing waves.

● Nodes − points of zero amplitude

● Antinodes − points of largest amplitude

● System has a fixed set of natural frequencies called normal modes of oscillation.

● Positions of nodes is given as,

● Distance between two successive nodes

● Position of antinodes is given as,

● Distance between two successive antinodes

● Boundary condition of string

L = Length of string

● ∴ From equation (3),

● Corresponding frequencies,

Vibrating String

Mode of vibration Harmonic Tone Nodes Antinode Frequency

First or fundamental First Fundamental 2 1

Second Second First 3 2

n th n th ( n − 1) tones n + 1 n

● Fundamental mode − Lowest possible natural frequency of a system