Sample Paper−1 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 19are 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 four fundamental forces. 2. Define parsec & express it in meters. 3. Draw position time graph for motion with zero acceleration. 4. What is the direction of zero vector? 5. Define horse power. 6. On what factors does the center of mass of a body depends? 7. Explain why a body with large reflectivity is a poor emitter.

8. A physical quantity P is related to four observables a, b, c and d as follows:

9. are unit vectors along x- and y-axis respectively. What is the magnitude and

direction of the vectors and ? What are the components of a vector

along the directions of and ? 10. A bullet of mass 0.04 kg moving with a speed of 90 m s-1 enters a heavy wooden

block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?

11. A car moving along a straight highway with a speed of 126 km h–1 is brought to a

stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

12. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both

having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

13. A pump on the ground floor of a building can pump up water to fill a tank of volume

30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

14. Four identical hollow cylindrical columns of mild steel support a big structure of

mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

15. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same

length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).

16. Explain why (a) The size of the needle of a syringe controls flow rate better than the thumb

pressure exerted by a doctor while administering an injection

(b) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

17. A rocket is fired from the earth towards the sun. At what distance from the earth’s

centre is the gravitational force on the rocket zero? Mass of the sun = 2 ×1030 kg, mass of the earth = 6 × 1024kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

18. The transverse displacement of a string (clamped at its both ends) is given by

Where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0

×10–2 kg. Answer the following: (a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

(c) Determine the tension in the string. 19. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water

vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.

20. A truck starts from rest and accelerates uniformly at 2.0 m s–2. At t = 10 s, a stone is

dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)

21. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By

adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.

22. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom? 23. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an

uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

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

25. A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the

earth is 1.50 ×108 km away from the sun?

26. Explain why (or how):

(a) In a sound wave, a displacement node is a pressure antinode and vice versa, (b) Bats can ascertain distances, directions, nature, and sizes of the obstacles

without any “eyes”, (c) A violin note and sitar note may have the same frequency, yet we can

distinguish between the two notes,

27. 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.]

28. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

Lowest Point Highest Point (a) mg – T1 mg + T2 (b) mg + T1 mg – T2

(c) mg + T1 – /R mg – T2 + /R

(d) mg – T1 – /R mg + T2 + /R

T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.

29. (i) What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room

temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)

(ii) 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.

30. (i) A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (a) the frequency of oscillations, (b) maximum acceleration of the

mass, and (c) the maximum speed of the mass.

(ii) In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is

(a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Sample Paper−1 Class 11, Physics

SOLUTIONS

1: Gravitational force, electromagnetic force, strong nuclear force, and weak nuclear

force. 2: 1 parsec = 3.08 × 1016 m (Parsec is the distance at which average radius of earth’s

orbit subtends an angle of 1 arc second) 3:

4: Arbitrary. 5: 1 hp = 746 W 6: Geometrical shape of the body& distribution of mass inside the body 7: Because a body with large reflectivity is poor absorber of heat & hence by

Kirchhoff’s law a body with large reflectivity is a poor emitter.

8:

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%,

respectively. What is the percentage error in the quantity P?

Percentage error in P = 13 %

9: Consider a vector , given as:

On comparing the components on both sides, we get:

Hence, the magnitude of the vector is .

Let be the angle made by the vector , with the x-axis, as shown in the following figure.

Hence, the magnitude of the vector is .

Let be the angle made by the vector , with the x- axis, as shown in the following figure.

It is given that:

On comparing the coefficients of , we have:

Let make an angle with the x-axis, as shown in the following figure.

Angle between the vectors

Component of vector , along the direction of , making an angle

Let be the angle between the vectors .

Component of vector , along the direction of , making an angle

10: The retardation ‘a’ of the bullet (assumed constant) is given by

2290 90

67502 2 0.6

ua ms

s−− − ×= = = −

×

The retarding force, by the Second Law of motion, is = 0.04 kg × 6750 ms-2 = 270 N

11: Initial velocity of the car, u = 126 km/h = 35 m/s Final velocity of the car, v = 0 Distance covered by the car before coming to rest, s = 200 m Retardation produced in the car = a From third equation of motion, a can be calculated as:

From first equation of motion, time (t) taken by the car to stop can be obtained as:

12: Let m and r be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis, The moment of inertia of the solid sphere about an axis passing through its centre,

We have the relation:

Where, α = Angular acceleration τ = Torque I = Moment of inertia

For the hollow cylinder,

For the solid sphere,

As an equal torque is applied to both the bodies,

Now, using the relation:

Where, ω0 = Initial angular velocity t = Time of rotation ω = Final angular velocity For equal ω0 and t, we have: ω ∝ α … (ii ) From equations (i) and (ii ), we can write: ωII >ωI

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

13: Volume of the tank, V = 30 m3

Time of operation, t = 15 min = 15 × 60 = 900 s Height of the tank, h = 40 m Efficiency of the pump, = 30% Density of water, ρ = 103 kg/m3 Mass of water, m = ρV = 30 × 103 kg Output power can be obtained as:

For input power Pi,, efficiency is given by the relation:

14: Mass of the big structure, M = 50,000 kg

Inner radius of the column, r = 30 cm = 0.3 m Outer radius of the column, R = 60 cm = 0.6 m Young’s modulus of steel, Y = 2 × 1011 Pa Total force exerted, F = Mg = 50000 × 9.8 N

Stress = Force exerted on a single column = 122500 N

Young’s modulus, Y

Where,

Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2)

= 7.22 × 10–7 Hence, the compressional strain of each column is 7.22 × 10–7.

15: Initial temperature, T1 = 40°C

Final temperature, T2 = 250°C Change in temperature, ∆T = T2 – T1 = 210°C Length of the brass rod at T1, l1 = 50 cm Diameter of the brass rod at T1, d1 = 3.0 mm Length of the steel rod at T2, l2 = 50 cm Diameter of the steel rod at T2, d2 = 3.0 mm Coefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1 Coefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1 For the expansion in the brass rod, we have:

For the expansion in the steel rod, we have:

Total change in the lengths of brass and steel, ∆l = ∆l1 + ∆l2 = 0.2205 + 0.126 = 0.346 cm Total change in the length of the combined rod = 0.346 cm Since the rod expands freely from both ends, no thermal stress is developed at the

junction. 16: (a) The small opening of a syringe needle controls the velocity of the blood

flowing out. This is because 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.

(b) When a fluid flows out from a small hole in a vessel, the 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.

17: Mass of the Sun, Ms = 2 × 1030 kg

Mass of the Earth, Me = 6 × 10 24 kg Orbital radius, r = 1.5 × 1011 m Mass of the rocket = m

Let x be the distance from the centre of the Earth where the gravitational force acting

on satellite P becomes zero. From Newton’s law of gravitation, we can equate gravitational forces acting on

satellite P under the influence of the Sun and the Earth as:

18: (a) The general equation representing a stationary wave is given by the displacement f unction:

y (x, t) = 2a sin kx cosωt This equation is similar to the given equation:

Hence, the given function represents a stationary wave. (b) A wave travelling along the positive x-direction is given as:

The wave travelling along the negative x-direction is given as:

The superposition of these two waves yields:

The transverse displacement of the string is given as:

Comparing equations (i) and (ii ), we have:

∴Wavelength, λ = 3 m It is given that: 120π = 2πν Frequency, ν = 60 Hz Wave speed, v = νλ = 60 × 3 = 180 m/s (c) The velocity of a transverse wave travelling in a string is given by the relation:

Where, Velocity of the transverse wave, v = 180 m/s Mass of the string, m = 3.0 × 10–2 kg Length of the string, l = 1.5 m

Mass per unit length of the string,

Tension in the string = T

From equation (i), tension can be obtained as: T = v2

µ = (180)2 × 2 × 10–2 = 648 N

19: Volume of the room, V = 25.0 m3

Temperature of the room, T = 27°C = 300 K Pressure in the room, P = 1 atm = 1 × 1.013 × 105 Pa The ideal gas equation relating pressure (P), Volume (V), and absolute temperature

(T) can be written as: PV = kBNT Where, KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1 N is the number of air molecules in the room

∴

= 6.11 × 1026 molecules Therefore, the total number of air molecules in the given room is 6.11 × 1026.

20: (a) 22.36 m/s, at an angle of 26.57° with the motion of the truck

(b) 10 m/s2 (a) Initial velocity of the truck, u = 0 Acceleration, a = 2 m/s2 Time, t = 10 s As per the first equation of motion, final velocity is given as: v = u + at = 0 + 2 × 10 = 20 m/s The final velocity of the truck and hence, of the stone is 20 m/s.

At t = 11 s, the horizontal component (vx) of velocity, in the absence of air resistance, remains unchanged, i.e.,

vx = 20 m/s The vertical component (vy) of velocity of the stone is given by the first

equation of motion as: vy = u + ayδt Where, δt = 11 – 10 = 1 s and ay = g = 10 m/s2

∴vy = 0 + 10 × 1 = 10 m/s The resultant velocity (v) of the stone is given as:

Let θ be the angle made by the resultant velocity with the horizontal

component of velocity, vx

= 26.57° (b) When the stone is dropped from the truck, the horizontal force acting on it

becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s2 and it acts vertically downward.

21: No

Range, R = 3 km Angle of projection, θ = 30° Acceleration due to gravity, g = 9.8 m/s2 Horizontal range for the projection velocity u0, is given by the relation:

The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of

45° with the horizontal, that is,

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

Hence, the bullet will not hit a target 5 km away.

22: A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, v = 5 m/s Angle of inclination, θ = 30° Height reached by the cylinder = h (a) Energy of the cylinder at point A:

Energy of the cylinder at point B = mgh Using the law of conservation of energy, we can write:

Moment of inertia of the solid cylinder,

In ∆ABC:

Hence, the cylinder will travel 3.82 m up the inclined plane. (b) For radius of gyration K, the velocity of the cylinder at the instance when it

rolls back to the bottom is given by the relation:

The time taken to return to the bottom is:

Therefore, the total time taken by the cylinder to return to the bottom is

(2 × 0.764) 1.53 s.

23: Mass of the bolt, m = 0.3 kg

Speed of the elevator = 7 m/s Height, h = 3 m Since the relative velocity of the bolt with respect to the lift is zero, at the time of

impact, potential energy gets converted into heat energy.

Heat produced = Loss of potential energy = mgh = 0.3 × 9.8 × 3 = 8.82 J The heat produced will remain the same even if the lift is stationary. This is because

of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

24: (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.

25: Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m

Time period of the Earth = Te Time period of Saturn, Ts = 29. 5 Te Distance of Saturn from the Sun = rs From Kepler’s third law of planetary motion, we have

For Saturn and Sun, we can write

Hence, the distance between Saturn and the Sun is .

26: (a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.

Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

27: 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) ]

28: (a) The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law of motion, the net force acting on the stone

at this point is equal to the centripetal force, i.e.,

… (i) Where, v1 = Velocity at the lowest point

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:

… (ii ) Where, v2 = Velocity at the highest point

It is clear from equations (i) and (ii ) that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

29: (i) Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, ∆T = 45°C Molecular mass of N2, M = 28 Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles,

Molar specific heat at constant pressure for nitrogen,

The total amount of heat to be supplied is given by the relation:

∆Q = nCP ∆T = 0.714 × 29.05 × 45 = 933.38 J Therefore, the amount of heat to be supplied is 933.38 J.

(ii) (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.

30: (i) Spring constant, k = 1200 N m–1

Mass, m = 3 kg Displacement, A = 2.0 cm = 0.02 cm (a) Frequency of oscillation v, is given by the relation:

Where, T is the time period

Hence, the frequency of oscillations is 3.18 cycles per second. (b) Maximum acceleration (a) is given by the relation: a = ω2 A Where,

ω = Angular frequency = A = Maximum displacement

Hence, the maximum acceleration of the mass is 8.0 m/s2. (c) Maximum velocity, vmax = Aω

Hence, the maximum velocity of the mass is 0.4 m/s.

(ii) (a) x = 2sin 20t

(b) x = 2cos 20t

(c) x = –2cos 20t The functions have the same frequency and amplitude, but different initial phases. Distance travelled by the mass sideways, A = 2.0 cm Force constant of the spring, k = 1200 N m–1 Mass, m = 3 kg Angular frequency of oscillation:

= 20 rad s–1 (a) When the mass is at the mean position, initial phase is 0. Displacement, x = Asin ωt = 2sin 20t (b) At the maximum stretched position, the mass is toward the extreme right.

Hence, the initial phase is .

Displacement,

= 2cos 20t (c) At the maximum compressed position, the mass is toward the extreme left.

Hence, the initial phase is .

Displacement,

= –2cos 20t

The functions have the same frequency and amplitude (2 cm), but

different initial phases .