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QUESTION PAPER

XI – PHYSICS

Time: Three Hours Maximum Marks: 70General Instructions

(a) All questions are compulsory.(b) There are 26 questions in total. Questions 1 to 5 carry one mark each,

questions 6 to 10 carry two marks each, questions 11 to 22 carry three marks each, question 23 is a value based question carry 4 marks and questions 25 to 26 carry five marks each.

(c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions.

(d) Use of calculator is not permitted.(e) You may use the following physical constants wherever necessary.

e 1.61019C c 3108 ms1

h 6.61034 JSo 4 107 NA2

kB 1.381023 JK 1

N A 6.0231023 / mole mn 1.61027 kg

1. Why we use a platinum iridium alloy in making prototype meter and kilogram? (1)2. Define coefficient of restitution or coefficient of resilience. (1)

3. Radius of gyration is a constant quantity. (1)

4. How does internal energy of gas changes in an adiabatic process? (1)

5. Give two examples of intensive variables. (1)

6. The length of the rod as measured in an experiment was found to be 3.23 m, 3.25 m, 3.27 m, 3.22. Find the absolute error. (2)

7. Name a polar satellite. State its uses. (2)

8. Two wooden blocks A and B are connected to a chord, passing through a frictionless pulley. Block A is pushed to the left on a surface by a horizontal force. Draw the free body diagram of both the blocks.

(2)

9. If the kinetic energy of one mole of an ideal gas is E=3/2 RT, what will be its Cp? (2)

10. Two point masses of 3 Kg and 2 Kg are attached at the two ends of a horizontal spring with spring constant k=200Nm-m. Find the natural frequency of vibration of the system. (2)

11. If a body A of mass ‘M’ is thrown with velocity u at an angle 300 with the horizontal and another body B is of the same mass be projected with the same velocity at angle 600 to the horizontal, then prove that the ratio of horizontal ranges will be 1:1 and that of maximum height will be 1:3. (3)

12. Calculate the recoil velocity V of the gun of mass M when a bullet of mass m is fired with a horizontal velocity v. (3)

13. Derive equation for loss of kinetic energy in case of a completely inelastic collision in one dimension? (3)

14. Consider two bodies X and Y with mass m1 and m2 separated by distance ‘r’ meters. Suppose mass of X is doubled and mass of Y is tripled also thedistance between them is doubled. Find the ratio of the gravitational force before and after changing the masses and distance. (3)

15. A circular track of radius 100 m is banked at an angle of 30°. If the coefficient of friction between the wheels of a car and the road is 0.5, then

what is the (i) optimum speed of the car to avoid wear and tear on its tires, and (ii) maximum permissible speed to avoid slipping? (3)

16. What are thermal radiations? Give some of its basic characteristics. (3)

17. State the law of equipartition of energy. Show that the ratio of specific heat at constant pressure to specific heat at constant volume is 7/5 for a rigid diatomic molecule. (3)

18. If the r.m.s speed of oxygen at NTP is x m/s. If the gas is heated at constant pressure till its volume is four fold, what will be its final temperature and r.m.s speed? (3)

19. State and prove Work-energy theorem. (3)20. A car drives along the straight level frictionless road by an engine delivering

constant power. How is velocity of the car related to time elapsed? (3)

21.Prove that the impulse received during an impact is equal to the total change in momentum produced during the impact. (3)

22. While approaching a planet circling a distant star, a space traveller determines the planet’s radius to be half that of the Earth. After landing on the surface he finds the acceleration due to gravity to be twice that on the surface of the Earth. Find the ratio of the mass of the planet to that of the Earth. (3)

23. Manu went to railway station to see off his uncle. At platform, he saw that an old coolie was carrying heavy load on his head. Suddenly the coolie tripped and a baggage fell off his head. The owner of the bag started shouting at the old man. Manu couldn’t tolerate this. He went to the old man, helped him in picking up the baggage and offered to carry some load for him.

(i) What does this tell you about the nature of Manu?(ii) A man weighing 55kg supports a body of 20kg on his head. Calculate work done by him if he

moves a distance of 20 m (a) on horizontal road (b) upon a smooth incline plane of 1/5 (g = 10 m/s²).

(iii) When is the work done negative? (4)

24. 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). (5)

OR

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn? (5)

25. A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.

(5)

OR

Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

(b) The instantaneous speed of the point of contact during rolling is zero.

(c) The instantaneous acceleration of the point of contact during rolling is zero.

(d) For perfect rolling motion, work done against friction is zero.

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion. (5)

26. (a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s). (5)

OR

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air. (5)

PHYSICS –XI

Q. No Value Points Marks

Ans1. The alloy is least affected by temperatures variations. It is non- corrosive and so does not wear out easily. (1)

Ans2. Coefficient of restitution is defined as the ratio of relative velocity of separation of the two bodies after collision to the relative velocity of approach before collision. It is denoted by ‘e’. (1)

Ans3. No. Radius of gyration depends on axis of rotation and distribution of mass. (1)

Ans4. In an adiabatic process work done by gas results in decrease in its internal energy. (1)

Ans5. Internal energy and volume (1/2 + 1/2)

Ans6. The average length of the rod is: 12.99/4 = 3.247 = 3.25 (Rounding off to two decimal points)

(1)

Hence, the absolute possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities. (1)

Ans7. Sea star is a polar satellite.Polar satellites are used for

(1/2)

1. Getting cloud images,2. Atmospheric data,3. To detect the ozone hole. (1/2 for each point)

Ans8. Free body diagram of block A

Forces on block A are:

(ii) WA = Weight of the block, acting downward(iii) nA = Normal reaction force from surface, acting vertically upward.(iv) T = Tension of chord(v) Fapplied = Applied force(vi) f = Frictional force, acting against the applied force (1)

Free body diagram of Forces on block B are:

T = Tension in chord, acting vertically upwardWB = Weight of the body, acting vertically downward (1)

Ans9.

Ans10.

Ans11. We know that horizontal range and maximum heights are independent of the mass of bodies. ThereforeHorizontal range of body A,

(1/2)And for body B,

(1/2)That is

(1/2)Now maximum height for body A,

Ans12. We are given here, Mass of the gun=M

Mass of the bullet=m Velocity of the bullet=v Recoil velocity of the gun=V

As initially both the bullet and gun are at rest, then applying principle of conservation of momentum, we get (1)

Total momentum of gun + bullet before firing= Total momentum of gun+ bullet after firing (1)

0= MV + mv

(1)

Ans13. Consider two masses and .Let the particle is moving with initial speed and be at rest.

In case of an perfectly inelastic collision in one dimension,Applying law of conservation of momentum, we get

where is the final velocity of the combined mass ( + ).

The loss in kinetic energy on collision is

(1)

(1)

Derivation (1) This is a positive quantity.

Ans14.

Ans15. Radius -R = 100 m, = 30o, = 0.5At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. (1)

So, the optimum speed is given by

The maximum permissible speed is given by

(1)

(1)

Ans16 Thermal radiations are radiations emitted by every body (that has temperature above 0 Kelvin) on account of its temperature. Thermal radiations are also called infra red radiations as their wavelength ranges from 8 x 10 -7 m to 4x 10-4 m. (1/2)

Basic characteristics of thermal radiations are

i. Thermal radiation requires no medium to propagate. They can travel through vacuum.

ii. Thermal radiations travel in straight lines with the speed of light.

iii. They do not heat the intervening medium through which they pass.

iv. Their intensity varies inversely as the square of the distance from the source.

v. Thermal radiations show the phenomena of interference, diffraction, reflection, refraction and polarization like light radiations.

(1/2 for each point)

Ans17.In equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to 1/2kT. This is the law of equipartition of energy. (1)

A diatomic molecule that can be taken as a rigid rotator with 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy the total internal energy of a mole of such a gas is given by

U 5

kT N 5

RT2 2

Then molar specific heat at constant volume CV

5

R2

(1)

and molar specific heat at constant pressure CP 7

R2

Ans18

HenceCP

7 (1) CV 5

dt

2

Ans19.Work-energy theorem states that the work done on a particle by a resultant force is equal to the change in its kinetic energy. (1)

W F.ds F. ds .dtdt

F.v.dt dK .dt

dK W K2 K1

(1)

(1)

Ans20. Power P Fv1

2

m dv

v1

dt 2

or vdv P

dt1

m 2

Integrating both sides, we get,1

2

v P

2 m t constant

or v2 t1

2

Ans21. We know,dP F

dt

or Fdt dP1

2

If the impact lasts for a small time dt and the momentum of thebody changes from P1toP2

then,

t P2 1 Fdt dP P2 P1 20 P1

t 1

or 0 Fdt P2 P1 2

F varies with time and does not remain constant. t 1

Fdt is a measure of the impulse of the force. 2 0

Let Fav be the constant force during the impact, then

r 2

r 2

e

P

t t 1

F dt 0 Fav dt

2 t Fav dt Fav t

0

F t P P1

av 2 1 2Thus, the impulse received during an impact is equal to the total change in momentum produced during the impact.

Ans 22. In case of the Earth, G M em

mg1

e 2

In case of the planet, G M p m

mg .1

p 2

Dividing these two equations, we get, M r 2 g 1 P e P ;M r 2 g 2 e P e

and r re

P 2

M P

M e 4 2 2Thus the ratio of the mass of the planet to the mass of the Earth is1 2.

Ans23. (i). Manu is helpful and kind heart. 2 (ii). W=fscosө, f=mg 1 (iii). When cosө is 180° 1Ans24.

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.

0

1212

2 1 1

(1)

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 (1)

(1)

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.

(1)

(1) OR

Speed of the cyclist,

Radius of the circular turn, r = 80 m

Centripetal acceleration is given as:

(1)

The situation is shown in the given figure:

(1)

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by0.5 m/s2.

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist. (1)

Since the angle between is 90°, the resultant acceleration a is given by:

Ans25. Disc - Radii of the ring and the disc, r = 10 cm = 0.1 m Initial

angular speed, ω0 =10 π rad s–1

Coefficient of kinetic friction, μk = 0.2

Initial velocity of both the objects, u = 0

Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma

μk mg= ma

Where, a = Acceleration produced in the objects, m = Mass

As per the first equation of motion, the final velocity of the objects can be obtained as:v = u + at

= 0 + μk gt

= μk gt … (ii)

The torque applied by the frictional force will act in perpendicularly outward

direction and cause reduction in the initial angular speed.

Torque, τ= –I α

α = Angular acceleration

μx mgr = –I α

(1)

Using the first equation of rotational motion to obtain the final angular speed:

Rolling starts when linear velocity, v = r ω

(1)

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

Since td > tr, the disc will start rolling before the ring. (1)

OR

(a) False; Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of thecentre of mass is backward. Hence, frictional force acts in the forward direction. (1)

(b) True; Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero. (1)

(c) False; when a body is rolling, its instantaneous acceleration is not equal to zero. It has some value. (1)

(d) True; when perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero. (1)

(e) True; the rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight. (1)

26. (a) Radius of the artery, r = 2 × 10–3 m

Diameter of the artery, d = 2 × 2 × 10–3 m = 4 × 10– 3 m (1/2)

Viscosity of blood,

Density of blood, ρ = 1.06 × 103 kg/m3

Reynolds’ number for laminar flow, NR = 2000 (1/2)

The largest average velocity of blood is given by the relation:

(1/2)

(1)

Therefore, the largest average velocity of blood is 0.983 m/s.

(b) Flow rate is given by the relation:

R = π r2 Vavg (1)

Therefore, the corresponding flow rate is . (1/2)

OR

Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10–10 N

Radius of the given uncharged drop, r = 2.0 × 10–5 m

Density of the uncharged drop, ρ = 1.2 × 103 kg m–3

Viscosity of air,

Density of air 0 can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g = 9.8 m/s2 (1)

Terminal velocity (v) is given by the relation:

Hence, the terminal speed of the drop is 5.8 cm s–1. (1)

The viscous force on the drop is given by:

(1)

Hence, the viscous force on the drop is 3.9 × 10–10 N. (1)