Solved Paper−2 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 shortest range of operation. 2. Define light year & express it in meters 3. Draw position time graph for motion with positive acceleration

4. If ˆˆ ˆ0.4 0.8i j kλ+ − is a unit vector, then find the value of λ.

5. In an elastic collision of two billiard balls, is the total kinetic energy conserved

during the short time of collision of the balls (i.e. when they are in contact) ? 6. Give the location of the centre of mass of a sphere of uniform mass density 7. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool

from 60 °C to 30 °C. The temperature of the surroundings is 20 °C. 8. A new unit of length is chosen such that the speed of light in vacuum is unity. What

is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

9. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a

constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

10. 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?

11. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

12. In the HCl molecule, the separation between the nuclei of the two atoms is about

1.27 Å (1 Å = 10–10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

13. A body constrained to move along the z-axis of a coordinate system is subject to a

constant force F given by

Where are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

14. 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?

15. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively.

Express these temperatures on the Celsius and Fahrenheit scales.

16. Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact

do not necessarily settle to the mean temperature (T1 + T2)/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the

different parts of a plant from getting too hot) should have high specific heat. 17. The motion of a particle executing simple harmonic motion is described by the

displacement function, x (t) = A cos (ωt + φ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s,

what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

18. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s–1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.

19. 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? 20. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial

acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast. 21. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h

passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s–1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit?

(Take g = 10 m s–2).

22. To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?

(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient.

23. A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of

mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

24. Explain why

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2)/2.

(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

(c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a

desert at the same latitude. 25. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of

1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

26. The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of

which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?

27. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a

speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 × 1030 kg).

28. (i) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant

velocity of 36 km/h, (c) just after it is dropped from the window of a train accelerating with 1 m

s–2, (d) lying on the floor of a train which is accelerating with 1 m s–2, the stone

being at rest relative to the train. Neglect air resistance throughout.

(ii) A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of 10 m s–1, (b) downwards with a uniform acceleration of 5 m s–2, (c) upwards with a uniform acceleration of 5 m s–2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled

down freely under gravity? 29. (a) A string of mass 2.50 kg is under a tension of 200 N. The length of the

stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

(b) A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz.

30. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its

density decreases with height. The precise dependence is given by the so-called law of atmospheres

n2 = n1 exp [-mg (h2 – h1)/ kBT] Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this

relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n2 = n1 exp [-mg NA(ρ - P′) (h2 –h1)/ (ρRT)] Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.

[NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Solved Paper−2 Class 11, Physics

Sulutions 1: Strong nuclear force. 2: light year is distance that light travels with velocity of 3 × 108 m s–1 in 1 year 1 ly= 9.46 × 1015 m 3:

4: ˆˆ ˆ0.4 0.8 _ _ _i j k is unit vetorλ+ −

( ) ( )2 2 2

ˆˆ ˆ0.4 0.8 1

0.4 0.8 1

0.2

i j kλ

λ

λ

⇒ + − =

⇒ + + =

⇒ =

5: No. In an elastic collision, the total initial kinetic energy of the balls will be equal to

the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.

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: According to Newton’s law of cooling, we have:

Where, Temperature of the body = T Temperature of the surroundings = T0 = 20°C K is a constant Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s Integrating equation (i), we get:

The temperature of the body falls from 60°C to 30°C in time = t’ Hence, we get:

Equating equations (ii ) and (iii ), we get:

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

8: Distance between the Sun and the Earth:

= Speed of light × Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8 min 20 s = 500 s

∴Distance between the Sun and the Earth = 1 × 500 = 500 units

9: Length of the string, l = 80 cm = 0.8 m

Number of revolutions = 14 Time taken = 25 s

Frequency, Angular frequency, ω = 2πν

Centripetal acceleration,

The direction of centripetal acceleration is always directed along the string, toward

the centre, at all points.

10: 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.

11: For train A:

Initial velocity, u = 72 km/h = 20 m/s Time, t = 50 s Acceleration, aI = 0 (Since it is moving with a uniform velocity) From second equation of motion, distance (sI)covered by train A can be obtained as:

= 20 × 50 + 0 = 1000 m For train B: Initial velocity, u = 72 km/h = 20 m/s Acceleration, a = 1 m/s2 Time, t = 50 s From second equation of motion, distance (sII)covered by train A can be obtained

as:

Length of both trains = 2*400m = 800m Hence, the original distance between the driver of train A and the guard of train B is

2250 – 1000 - 800 = 450m.

12: The given situation can be shown as:

Distance between H and Cl atoms = 1.27Å Mass of H atom = m Mass of Cl atom = 35.5m Let the centre of mass of the system lie at a distance x from the Cl atom. Distance of the centre of mass from the H atom = (1.27 – x) Let us assume that the centre of mass of the given molecule lies at the origin.

Therefore, we can have:

Here, the negative sign indicates that the centre of mass lies at the left of the

molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.

13: Force exerted on the body,

Displacement, s = m Work done, W = F.s

Hence, 12 J of work is done by the force on the body.

14: Length of the steel wire, L1 = 4.7 m

Area of cross-section of the steel wire, A1 = 3.0 × 10–5 m2 Length of the copper wire, L2 = 3.5 m Area of cross-section of the copper wire, A2 = 4.0 × 10–5 m2 Change in length = ∆L1 = ∆L2 = ∆L Force applied in both the cases = F Young’s modulus of the steel wire:

… (i) Young’s modulus of the copper wire:

Dividing (i) by (ii ), we get:

The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

15: Kelvin and Celsius scales are related as:

TC = TK – 273.15 … (i) Celsius and Fahrenheit scales are related as:

… (ii ) For neon: TK = 24.57 K

∴TC = 24.57 – 273.15 = –248.58°C

For carbon dioxide: TK = 216.55 K ∴TC= 216.55 – 273.15 = –56.60°C

16: (a) When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T1 + T2)/2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

17: Initially, at t = 0:

Displacement, x = 1 cm Initial velocity, v = ω cm/sec. Angular frequency, ω = π rad/s–1 It is given that:

Squaring and adding equations (i) and (ii ), we get:

Dividing equation (ii ) by equation (i), we get:

SHM is given as:

Putting the given values in this equation, we get:

Velocity, Substituting the given values, we get:

Squaring and adding equations (iii ) and (iv), we get:

Dividing equation (iii ) by equation (iv), we get:

18: Speed of wind on the upper surface of the wing, V1 = 70 m/s

Speed of wind on the lower surface of the wing, V2 = 63 m/s Area of the wing, A = 2.5 m2 Density of air, ρ = 1.3 kg m–3 According to Bernoulli’s theorem, we have the relation:

Where, P1 = Pressure on the upper surface of the wing P2 = Pressure on the lower surface of the wing The pressure difference between the upper and lower surfaces of the wing provides

lift to the aeroplane.

Lift on the wing

= 1512.87 = 1.51 × 103 N Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.

19: 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,

20: Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s2 Acceleration due to gravity, g = 10 m/s2 Using Newton’s second law of motion, the net force (thrust) acting on the rocket is

given by the relation: F – mg = ma F = m (g + a) = 20000 × (10 + 5) = 20000 × 15 = 3 × 105 N

21: Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane, v = 720 km/h = 200 m/s Let θ be the angle with the vertical so that the shell hits the plane. The situation is

shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s Time taken by the shell to hit the plane = t Horizontal distance travelled by the shell = uxt

Distance travelled by the plane = vt The shell hits the plane. Hence, these two distances must be equal. uxt = vt

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H)

higher than the maximum height achieved by the shell.

22: Angular speed of the rotor, ω = 200 rad/s

Torque required, τ = 180 Nm The power of the rotor (P) is related to torque and angular speed by the relation: P = τω = 180 × 200 = 36 × 103 = 36 kW Hence, the power required by the engine is 36 kW.

23: Mass of the bullet, m = 0.012 kg

Initial speed of the bullet, ub = 70 m/s Mass of the wooden block, M = 0.4 kg Initial speed of the wooden block, uB = 0 Final speed of the system of the bullet and the block = ν Applying the law of conservation of momentum:

For the system of the bullet and the wooden block: Mass of the system, m' = 0.412 kg Velocity of the system = 2.04 m/s Height up to which the system rises = h Applying the law of conservation of energy to this system: Potential energy at the highest point = Kinetic energy at the lowest point

= 0.2123 m The wooden block will rise to a height of 0.2123 m. Heat produced = Kinetic energy of the bullet – Kinetic energy of the system

= 29.4 – 0.857 = 28.54 J

24: (a) When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T1 + T2)/2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

25: Angular frequency of the piston, ω = 200 rad/ min.

Stroke = 1.0 m

Amplitude, The maximum speed (vmax) of the piston is give by the relation:

Ans: Area of cross-section of the spray pump, A1 = 8 cm2 = 8 × 10–4 m2

Number of holes, n = 40 Diameter of each hole, d = 1 mm = 1 × 10–3 m Radius of each hole, r = d/2 = 0.5 × 10–3 m Area of cross-section of each hole, a = πr2 = π (0.5 × 10–3)2 m2 Total area of 40 holes, A2 = n × a = 40 × π (0.5 × 10–3)2 m2 = 31.41 × 10–6 m2 Speed of flow of liquid inside the tube, V1 = 1.5 m/min = 0.025 m/s Speed of ejection of liquid through the holes = V2 According to the law of continuity, we have:

= 0.633 m/s Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.

27: Yes

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, fg Where, M = Mass of the star = 2.5 × 2 × 1030 = 5 × 1030 kg m = Mass of the body R = Radius of the star = 12 km = 1.2 ×104 m

Centrifugal force, fc

= mrω2 ω = Angular speed = 2πν

ν = Angular frequency = 1.2 rev s–1 fc = mR (2πν)2 = m × (1.2 ×104) × 4 × (3.14)2 × (1.2)2 = 1.7 ×105m N Since fg > fc, the body will remain stuck to the surface of the star.

28: (i) (a) 1 N; vertically downward

Mass of the stone, m = 0.1 kg Acceleration of the stone, a = g = 10 m/s2

As per Newton’s second law of motion, the net force acting on the stone,

F = ma = mg = 0.1 × 10 = 1 N Acceleration due to gravity always acts in the downward direction. (b) 1 N; vertically downward

The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(c) 1 N; vertically downward It is given that the train is accelerating at the rate of 1 m/s2. Therefore, the net force acting on the stone, F' = ma = 0.1 × 1 = 0.1 N

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,' stops acting on the stone. This is

because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.

F = mg = 1 N This force acts vertically downward. (d) 0.1 N; in the direction of motion of the train

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

Acceleration of the train, a = 0.1 m/s2 The net force acting on the stone will be in the direction of motion of

the train. Its magnitude is given by: F = ma = 0.1 × 1 = 0.1 N

(ii) (a) Mass of the man, m = 70 kg

Acceleration, a = 0 Using Newton’s second law of motion, we can write the equation of

motion as: R – mg = ma Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration a = 0 ∴R = mg = 70 × 10 = 700 N

∴Reading on the weighing scale = (b) Mass of the man, m = 70 kg Acceleration, a = 5 m/s2 downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma R = m(g – a) = 70 (10 – 5) = 70 × 5 = 350 N

∴Reading on the weighing scale = (c) Mass of the man, m = 70 kg Acceleration, a = 5 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma R = m(g + a) = 70 (10 + 5) = 70 × 15 = 1050 N

∴Reading on the weighing scale = (d) When the lift moves freely under gravity, acceleration a = g

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma R = m(g – a) = m(g – g) = 0

∴Reading on the weighing scale = The man will be in a state of weightlessness.

29: (a) Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N Length of the string, l = 20.0 m

Mass per unit length, The velocity (v) of the transverse wave in the string is given by the relation:

∴Time taken by the disturbance to reach the other end, t =

(b) Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 103 m/s

Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz The wavelength of sound in the tissue is given as:

30: According to the law of atmospheres, we have:

n2 = n1 exp [-mg (h2 – h1)/ kBT] … (i) Where, n1 is thenumber density at height h1, and n2 is the number density at height h2 mg is the weight of the particle suspended in the gas column Density of the medium = ρ' Density of the suspended particle = ρ Mass of one suspended particle = m' Mass of the medium displaced = m Volume of a suspended particle = V According to Archimedes’ principle for a particle suspended in a liquid column, the

effective weight of the suspended particle is given as: Weight of the medium displaced – Weight of the suspended particle = mg – m'g

Gas constant, R = kBN

… (iii ) Substituting equation (ii ) in place of mg in equation (i) and then using equation (iii ),

we get: n2 = n1 exp [-mg (h2 – h1)/ kBT]

= n1 exp [- (h2 – h1) ]

= n1 exp [- (h2 – h1) ]