Solved Paper−3 Class 11, Physics

Time: 3 hours Max. Marks 70 General Instructions 1. All questions are compulsory. Symbols have their usual meaning. 2. Use of calculator is not permitted. However you may use log table, if required. 3. Draw neat labeled diagram wherever necessary to explain your answer. 4. Q.No. 1 to 7 are of very short answer type questions, carrying 1 mark each. 5. Q.No.8 to 19 are of short answer type questions, carrying 2 marks each. 6. Q. No. 20 to 27 carry 3 marks each. Q. No. 28 to 30 carry 5 marks each.

1. Name the fundamental force having longest range of operation. 2. Define AU & express it in meters. 3. Draw position time graph for motion with negative acceleration

4. Find the angles between vectors ˆˆ ˆ2A i j k= + −�

& and ˆˆ ˆ 2B i j k= − + −�

.

5. An artificial satellite orbiting the earth in very thin atmosphere loses its energy

gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

6. Give the location of the centre of mass of a cylinder of uniform mass density 7. What is the temperature of the triple-point of water on an absolute scale whose unit

interval size is equal to that of the Fahrenheit scale ? 8. Explain the parallax method for measuring distance of earth from sun when radius

of earth is known.

9. A particle starts from the origin at t = 0 s with a velocity of and moves in

the x-y plane with a constant acceleration of .

(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time? 10. Explain why passengers are thrown forward from their seats when a speeding bus

stops suddenly?

11. On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and

C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

12. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and

radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

13. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

(i) (ii) t (iii) (iv) 14. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is

pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

15. n an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150

°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

16. Use the formula to explain why the speed of sound in air

(a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. 17. A simple pendulum of length l and having a bob of mass M is suspended in a car.

The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

18. A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

19. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

20. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–

1 to 3.5 m s–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

21. The position of a particle is given by

Where t is in seconds and the coefficients have the proper units for r to be in

metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

22. (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

23. A body of mass 0.5 kg travels in a straight line with velocity where

. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

24. A wire stretched between two rigid supports vibrates in its fundamental mode with

a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

25. The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

26. In changing the state of a gas adiabatically from an equilibrium state A to another

equilibrium stateB, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

27. Estimate the average thermal energy of a helium atom at (i) room temperature (27

°C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

28. (i) Two billiard balls each of mass 0.05 kg moving in opposite directions with

speed 6 m s–1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

(ii) A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

29. (i) Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

(ii) Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure.

30. (i) Assuming the earth to be a sphere of uniform mass density, how much would

a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

(ii) Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Solved Paper−3 Class 11, Physics

Solutions

1: Gravitational & Electromagnetic force. 2: AU is the average distance of the Sun from the Earth = 1.496 × 1011 m 3:

4: _ _ _& _ _ _ _angle between A B is given as� �

( ) ( )ˆ ˆˆ ˆ ˆ ˆ2 2cos

ˆ ˆˆ ˆ ˆ ˆ2 2

1 2 2cos

1 4 1 1 1 41

cos2

60o

i j k i j k

i j k i j kθ

θ

θ

θ

+ − − + −=

+ − − + −

− + +⇒ =

+ + + +

⇒ =

=

i

5: When an artificial satellite, orbiting around earth, moves closer to earth, its

potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

6: The centre of mass (C.M.) is a point where the mass of a body is supposed to be

concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.

7: Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:

Let TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on

absolute scale. Both the temperatures can be related as:

It is given that: TK1 – TK = 1 K Subtracting equation (i) from equation (ii ), we get:

Triple point of water = 273.16 K

∴Triple point of water on absolute scale = = 491.69

8: To measure the distance D of a faraway planet S by the parallax method, we

observe it from two different positions (observatories) A and B on the Earth, separated by distance AB = b at the same time as shown in Fig. 2.2.

We measure the angle between the two directions along which the planet is viewed at these two points. The ∠ASB in Fig. 2.2 represented by symbol θ is called the parallax angle or parallactic angle.

As the planet is very far away, b/D<< 1 and therefore, θ is very small. Then we

approximately take AB as an arc of length b of a circle with centre at S and the distance D as D = b/θ

9: Velocity of the particle,

Acceleration of the particle Also,

But,

Integrating both sides:

Where,

= Velocity vector of the particle at t = 0

= Velocity vector of the particle at time t

Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r

Since the motion of the particle is confined to the x-y plane, on equating the

coefficients of , we get:

(a) When x = 16 m:

∴y = 10 × 2 + (2)2 = 24 m (b) Velocity of the particle is given by:

10: When a speeding bus stops suddenly, the lower portion of a passenger’s body,

which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

11: Velocity of car A, vA = 36 km/h = 10 m/s

Velocity of car B, vB = 54 km/h = 15 m/s Velocity of car C, vC = 54 km/h = 15 m/s Relative velocity of car B with respect to car A, vBA = vB – vA = 15 – 10 = 5 m/s Relative velocity of car C with respect to car A, vCA = vC – (– vA) = 15 + 10 = 25 m/s At a certain instance, both cars B and C are at the same distance from car A i.e., s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m = Hence, to avoid an accident, car B must cover the same distance in a maximum of

40 s. From second equation of motion, minimum acceleration (a) produced by car B can

be obtained as:

12: Mass of the hollow cylinder, m = 3 kg

Radius of the hollow cylinder, r = 40 cm = 0.4 m Applied force, F = 30 N The moment of inertia of the hollow cylinder about its geometric axis: I = mr2 = 3 × (0.4)2 = 0.48 kg m2

Torque, = 30 × 0.4 = 12 Nm For angular acceleration , torque is also given by the relation:

Linear acceleration = rα = 0.4 × 25 = 10 m s–2

13: (ii) t

Mass of the body = m Acceleration of the body = a

Using Newton’s second law of motion, the force experienced by the body is given by the equation:

F = ma Both m and a are constants. Hence, force F will also be a constant. F = ma = Constant … (i) For velocity v, acceleration is given as,

Power is given by the relation: P = F.v Using equations (i) and (iii ), we have: P ∝ t

Hence, power is directly proportional to time.

14: Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m Area of the copper piece: A = l × b = 19.1 × 10–3 × 15.2 × 10–3 = 2.9 × 10–4 m2 Tension force applied on the piece of copper, F = 44500 N Modulus of elasticity of copper, η = 42 × 109 N/m2

Modulus of elasticity, η

= 3.65 × 10–3

15: Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T1 = 150°C Final temperature of the metal, T2 = 40°C Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g Volume of water, V = 150 cm3 Mass (M) of water at temperature T = 27°C: 150 × 1 = 150 g Fall in the temperature of the metal: ∆T = T1 – T2 = 150 – 40 = 110°C Specific heat of water, Cw

= 4.186 J/g/°K Specific heat of the metal = C Heat lost by the metal, θ = mC∆T … (i) Rise in the temperature of the water and calorimeter system: ∆T′’ = 40 – 27 = 13°C Heat gained by the water and calorimeter system: ∆θ′′ = m1 Cw∆T’ = (M + m′) Cw ∆T’ … (ii ) Heat lost by the metal = Heat gained by the water and colorimeter system mC∆T = (M + m’) Cw ∆T’ 200 × C × 110 = (150 + 25) × 4.186 × 13

If some heat is lost to the surroundings, then the value of C will be smaller than the

actual value.

16: (a) Take the relation:

Now from the ideal gas equation for n = 1: PV = RT For constant T, PV = Constant Since both M and γ are constants, v = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

(b) Take the relation:

For one mole of an ideal gas, the gas equation can be written as: PV = RT

P = … (ii ) Substituting equation (ii ) in equation (i), we get:

Where, Mass, M = ρV is a constant γ and R are also constants

We conclude from equation (iv) that . Hence, the speed of sound in a gas is directly proportional to the square root

of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

(c) Let be the speeds of sound in moist air and dry air respectively.

Let be the densities of moist air and dry air respectively. Take the relation:

Hence, the speed of sound in moist air is:

And the speed of sound in dry air is:

On dividing equations (i) and (ii ), we get:

However, the presence of water vapour reduces the density of air, i.e.,

Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in

a gaseous medium, the speed of sound increases with humidity.

17: The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration Where, v is the uniform speed of the car R is the radius of the track Effective acceleration (aeff) is given as:

Time period, Where, l is the length of the pendulum

∴Time period, T

18: Work done by the steam engine per minute, W = 5.4 × 108 J

Heat supplied from the boiler, H = 3.6 × 109 J

Efficiency of the engine =

Hence, the percentage efficiency of the engine is 15 %. Amount of heat wasted = 3.6 × 109 – 5.4 × 108 = 30.6 × 108 = 3.06 × 109 J Therefore, the amount of heat wasted per minute is 3.06 × 109 J.

19: Temperature of the helium atom, THe = –20°C= 253 K

Atomic mass of argon, MAr = 39.9 u Atomic mass of helium, MHe = 4.0 u Let, (vrms)Ar be the rms speed of argon. Let (vrms)He be the rms speed of helium. The rms speed of argon is given by:

… (i) Where, R is the universal gas constant TAr is temperature of argon gas The rms speed of helium is given by:

… (ii ) It is given that: (vrms)Ar = (vrms)He

= 2523.675 = 2.52 × 103 K Therefore, the temperature of the argon atom is 2.52 × 103 K.

20: 0.18 N; in the direction of motion of the body

Mass of the body, m = 3 kg Initial speed of the body, u = 2 m/s Final speed of the body, v = 3.5 m/s Time, t = 25 s Using the first equation of motion, the acceleration (a) produced in the body can be

calculated as: v = u + at

As per Newton’s second law of motion, force is given as: F = ma = 3 × 0.06 = 0.18 N Since the application of force does not change the direction of the body, the net

force acting on the body is in the direction of its motion.

21: (a)

The position of the particle is given by:

Velocity , of the particle is given as:

Acceleration , of the particle is given as:

(b) 8.54 m/s, 69.45° below the x-axis

The magnitude of velocity is given by:

The negative sign indicates that the direction of velocity is below the x-axis.

22: (a) 100 rev/min

Initial angular velocity, ω1= 40 rev/min Final angular velocity = ω2 The moment of inertia of the boy with stretched hands = I1 The moment of inertia of the boy with folded hands = I2 The two moments of inertia are related as:

Since no external force acts on the boy, the angular momentum L is a

constant. Hence, for the two situations, we can write:

(b) Final K.E. = 2.5 Initial K.E.

Final kinetic rotation, EF

Initial kinetic rotation, EI

The increase in the rotational kinetic energy is attributed to the internal

energy of the boy.

23: Mass of the body, m = 0.5 kg

Velocity of the body is governed by the equation, Initial velocity, u (at x = 0) = 0

Final velocity v (at x = 2 m) Work done, W = Change in kinetic energy

24: (a) Mass of the wire, m = 3.5 × 10–2 kg

Linear mass density, Frequency of vibration, ν = 45 Hz

∴Length of the wire, The wavelength of the stationary wave (λ) is related to the length of the wire

by the relation:

For fundamental node, n = 1: λ = 2l

λ = 2 × 0.875 = 1.75 m The speed of the transverse wave in the string is given as: v = νλ= 45 × 1.75 = 78.75 m/s (b) The tension produced in the string is given by the relation: T = v2µ = (78.75)2 × 4.0 × 10–2 = 248.06 N

25: Acceleration due to gravity on the surface of moon, = 1.7 m s–2

Acceleration due to gravity on the surface of earth, g = 9.8 m s–2 Time period of a simple pendulum on earth, T = 3.5 s

Where,

l is the length of the pendulum

The length of the pendulum remains constant.

On moon’s surface, time period,

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

26: The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

This is an adiabatic process. Hence, change in heat is zero. ∴ ∆Q = 0 ∆W = –22.3 J (Since the work is done on the system) From the first law of thermodynamics, we have: ∆Q = ∆U + ∆W Where, ∆U = Change in the internal energy of the gas ∴ ∆U = ∆Q – ∆W = – (– 22.3 J) ∆U = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is: ∆Q = 9.35 cal = 9.35 × 4.19 = 39.1765 J Heat absorbed, ∆Q = ∆U + ∆Q ∴∆W = ∆Q – ∆U = 39.1765 – 22.3 = 16.8765 J Therefore, 16.88 J of work is done by the system.

27: (i) At room temperature, T = 27°C = 300 K

Average thermal energy Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

∴ = 6.21 × 10–21J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.

(ii) On the surface of the sun, T = 6000 K

Average thermal energy

= 1.241 × 10–19 J Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19J.

(iii) At temperature, T = 107 K

Average thermal energy

= 2.07 × 10–16 J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.

28: (i) Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s Magnitude of the initial momentum of each ball, pi = 0.3 kg m/s After collision, the balls change their directions of motion without changing the magnitudes of their velocity. Final momentum of each ball, pf = –0.3 kg m/s Impulse imparted to each ball = Change in the momentum of the system = pf – pi = –0.3 – 0.3 = –0.6 kg m/s The negative sign indicates that the impulses imparted to the balls are opposite in direction.

(ii) Let m, m1, and m2 be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.

Initial momentum of the system (parent nucleus) = 0 Let v1 and v2 be the respective velocities of the daughter nuclei having

masses m1 and m2.

Total linear momentum of the system after disintegration = According to the law of conservation of momentum: Total initial momentum = Total final momentum

Here, the negative sign indicates that the fragments of the parent nucleus

move in directions opposite to each other.

29: (i) (a)

Take the case given in figure (b).

Where, A1 = Area of pipe1 A2 = Area of pipe 2 V1 = Speed of the fluid in pipe1 V2 = Speed of the fluid in pipe 2 From the law of continuity, we have:

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 speed is more, then 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.

(ii) 10.5 m

Density of mercury, ρ1 = 13.6 × 103 kg/m3 Height of the mercury column, h1 = 0.76 m Density of French wine, ρ2 = 984 kg/m3

Height of the French wine column = h2 Acceleration due to gravity, g = 9.8 m/s2

The pressure in both the columns is equal, i.e., Pressure in the mercury column = Pressure in the French wine column

= 10.5 m

Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

30: (i) Weight of a body of mass m at the Earth’s surface, W = mg = 250 N

Body of mass m is located at depth, Where,

= Radius of the Earth Acceleration due to gravity at depth g (d) is given by the relation:

Weight of the body at depth d,

(ii) Orbital period of

Orbital radius of

Satellite is revolving around the Jupiter Mass of the latter is given by the relation:

Where,

= Mass of Jupiter G = Universal gravitational constant

Orbital radius of the Earth,

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.