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Grade XIPhysics

Mock Test

CLASS - XI PHYSICS

Mock Test

Time allowed: 3 hrs Max. marks: 70

General Instructions:

(i) All questions are compulsory. There are 37 questions in all.

(ii) This question paper has four sections: Section A, Section B, Section C and Section D.

(iii) Section A contains twenty questions of one mark each, Section B contains seven questions of two marks each, Section C contains seven questions of three marks each, and Section D contains three questions of five marks each.

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

(v) You may use the following values of physical constants where ever necessary.

c = 3 × 108 ms-1

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

TmA-1π μo = 4 × 10−7

C2N-1m-2.85εo = 8 × 10−12

Nm2C-21

4πεo= 9 × 109

me= 9.1 × 10-31 kg

mp= 1.673 × 10-27 kg

mn= 1.675 × 10-27 kg

Avogadro’s Number = 6.023 × 1023 gram per mole

Kb = 1.38 × 10-23 JK-1

Section A

Directions (Q1-Q10) Select the most appropriate option from those given below each question

Q1. Conservation of the energy implies that A. The total energy of an isolated system is conserved. B. The total mechanical energy of an isolated system is conserved. C. Heat energy will not be generated in the system. D. The total mass and energy of a system are conservation. 1

Q2. The time period of a satellite is independent of A. Mass of the satellite. B. Radius of satellite’s orbit. C. Mass of the planet. D. None of the above. 1

Q3. When two objects of same mass have elastic collision, their velocities A. Becomes zero. B. Are interchanged. C. Are such that, one object stops and other moves with double speed. D. None of the above. 1

Q4. Mechanical advantage in a lever is given by A. The ratio of load arm and the effort arm length. B. The ratio of effort arm and the load arm length. C. The ratio effort to the load applied. D. None of these. 1

Q5. Which of the following is not a type of systematic errors? A. Random error. B. Instrumental error. C. Error in the procedure. D. Personal error. 1

Q6. In which of the following case, stress will be tensile stress? A. Book getting deformed by applying a force parallel to the surface of cover. B. An air filled balloon dipped inside water. C. A stretched spring. D. A twisted wire. 1

Q7. Which of the following is the odd one, from the perspective of surface tension? A. A bubble inside water of radius r. B. A water droplet of radius r. C. A soap bubble of radius r. D. Can not differentiated. 1

Q8. Which of the following is incorrect? A. An ideal spring mass system executes SHM about the normal position in

horizontal direction. B. A pendulum always executes SHM motion even for large angular displacement. C. A floating body executes SHM if a very small displacement is made about

equilibrium position. D. All are correct. 1

Q9. Average velocity and instantaneous velocity for which for which of the following would be the same.

A. A particle moving in uniform circular motion. B. A particle moving on a zig zag path. C. A particle oscillating about a mean position. D. A particle moving with zero acceleration. 1

Q10. The angle at which the range of the projectile is maximum is equal to A. 90o

B. 60o

C. 45o D. 30o 1

Directions (Q11 –Q15) Fill in the blanks with appropriate answer.

Q11. In CGS, FPS and MKS systems the unit of __________ is same. 1

Q12. In __________ the temperature of different points in a heat conductor does not change. 1

Q13. In a refrigerator the heat transfer is __________ at low temperature and it is __________ at high temp. 1

Q14. At the open end of the flute, there is __________ of the air molecule’s displacement. 1

Q15. Two perpendicular vectors have their __________ product as zero.

OR

Two parallel vectors have their __________ product as zero. 1

Directions (Q16 –Q20) Answer the following

Q16. Find the vector product of the given vectors.

1

Q17. Discuss in brief whether the function is periodic or not. e −wt 1

Q18. Under what conditions can a real gas be approximated to an ideal gas? 1

Q19. Draw a velocity−time graph illustrating a uniformly accelerated motion. 1

Q20. A ping-pong ball and a cricket ball are moving in air with the same velocity. Which ball is easier to catch?

OR In a collision of two objects, is momentum always conserved? Validate your answer with an example. 1

Section B

Q21. Show position-time graph of an object if the object is moving with 1. zero acceleration 2. positive acceleration 3. negative acceleration 2

Q22. Verify that , obtained from kinetic theory of gases, is dimensionally correct. 2

Q23. What are thermodynamic state variables? Give an example of thermodynamic state variable. Why is that quantity called thermodynamic state variable? 2

Q24. Establish and state the relationship between angular momentum and torque. 2

Q25. A simple pendulum of length 0.5 m has bob of mass 150 g. It is released from an angle of 15o with the vertical. (A) Find the frequency of vibration assuming simple harmonic motion. (B) Find the speed of bob when it passes through the lowest point of oscillation. (C) Find the total energy stored in this oscillation. 2

Q26. The cone of a loudspeaker of a music player vibrates in SHM at a frequency of 300 Hz where the amplitude at the centre of the cone is 6 × 10-4 m and x = A at t = 0. 2

(A) Find the equation that describes the motion of the centre of the cone. (B) Find the position of the cone at t = 2.4 ms.

OR

An object experiences a simple harmonic oscillation along the x = axis and its position varies with time according to the equation

Determine the amplitude, frequency and period of the motion.

Q27. Derive an expression for final velocities of two point objects colliding elastically with each other.

OR

A 15 g bullet is fired into a stationary block of wood (m = 5 kg). The relative velocity of bullet inside the block is zero. The speed of the wood after collision is 0.5 m/s. Find the original speed of the bullet.

2

Section C

Q28. The given figure shows the strain-stress curve for a given material.

Calculate (i) Young’s modulus of the material (ii) Ultimate strength for the material (iii) stress corresponding to the fracture point 3

Q29. What is the total amount of heat energy required to convert 5 kg of ice at −20°C into 3

steam at 100°C? The pressure stays constant throughout. [Specific heat of ice = 2100 J/kg/K−1, Latent heat of fusion of ice = 3.36 × 105 J/kg, Specific heat of water = 4200 J/kg/K, Latent heat of vaporization of water = 2.25 × 106 J/kg]

Q30. (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 × 105 Pa. Calculate the velocity with which water will come out from the small hole. 3

Q31. A large number of spherical raindrops of similar size are falling with a terminal velocity of 0.13 m/s. They combine together and form a large spherical drop. The terminal velocity of the large drop is 1.17 m/s. Calculate the number of small raindrops. 3

Q32. The mass and radius of a planet X are half of the mass and radius of earth. Obtain the ratio of the escape speed from the surface of the planet to that from the surface of the earth.

OR

An artificial satellite of mass 500 kg orbits at a height of 750 km above the Earth’s surface. If the radius of the earth is 6.4 × 106 m and the acceleration due to gravity at its surface is 9.81 m/s2, then find a) the weight of a 5 kg object at the height of the satellite b) the centripetal acceleration experienced by the satellite 3

Q33. A 2 kg block is initially at rest on a horizontal frictionless table and a horizontal force in the positive direction of x-axis is applied to the block. The force is given by

, where x is in metre and the initial position of the block is x = 0. What is the kinetic energy of the block when it passes through x = 2 m? 3

Q34. Two express trains A and B, each 400 m long, are running on straight parallel tracks heading in the same direction. B travelling with 40km/hr, which is just behind A, overtakes A which is travelling with a speed of 30km/hr. Calculate the time taken by B to overtake A and the distance covered by A in this time. 3

Section D

Q35. A stone of mass m is whirling in a vertical plane with the help of an attached string of length l. Velocity of the stone at the lowermost point is u. Obtain general expressions for the velocity and tension in the string at any point. Also, find the minimum velocity at the lowermost point to perform vertical circle. 5

OR

(i) Derive an expression for the maximum velocity with which a vehicle can take a turn on a circular banked track. (ii) A car is about to take a turn on a circular banked track of radius 3 m. The track is banked at a 45° angle. The coefficient of static friction between the track and the tyres is 0.11. What should be the maximum velocity of the car in order to avoid slip?

Q36. What is Carnot theorem? Describe Carnot cycle and derive an expression for efficiency of the Carnot engine.

OR

(a) A scientist claims to have constructed an engine that has an efficiency of 55% when operated between the boiling and freezing points of water. Can his claim be true? Give reasons. (b) Prove that the efficiency of all reversible heat engines operating

between two reservoirs is the same? 5

Q37. (a) State and prove the theorem of parallel axis. (b) What is the moment of inertia of uniform circular disc about an axis passing through it circumference and perpendicular to its plane? The mass of the disc is 250 gm and its radius is 25 cm.

OR

A cylinder measures 100 cm in length and has a radius of 20 cm. A sphere of radius 20 cm is cut out from the cylinder, such that the centre of the sphere lies at a distance of 30 cm from the central axis of the cylinder (as shown). By what distance does the centre of mass of the cylinder shift, after cutting out the sphere?

5

Mock Test Physics

Solutions

Section A

1. Although now we know that the conservation of mass-energy is true. The conservation of the energy implies that the total energy of an isolated system is conserved. Hence, the correct answer is option A.

2. The formula for the velocity of the satellite is where M is the mass of the planet. v = √ rGM

So, velocity is independent from the mass of the satellite. Hence, the correct answer is option A.

3. In case of two equal mass objects colliding elastically the momentum as well as the kinetic energy of the system is conserved. So the velocities are interchanged. Hence, the correct answer is option B.

4. Mechanical advantage in a lever is defined as the ratio of load and the effort which is equal to the ratio of length of the effort arm and the load arm. Hence, the correct answer is option B.

5. There are three types of systematic error. Instrumental error, error in the procedure and personal error. Random error is not a systematic error. Hence, the correct answer is option A.

6. In case of stretched spring there is tension in the body and the stress is called tensile stress. Hence, the correct answer is option C.

7. A soap bubble is the odd one out as it has 2 surfaces which accounts for the surface tension. All others have only one surface. Hence, the correct answer is option C.

8. A pendulum executes SHM only for small angular displacement. It will not have SHM, for example, if you release it from a horizontal position. So, statement B is incorrect. Hence, the correct answer is option B.

9. Average velocity and instantaneous velocity for a particle moving with zero acceleration will be the same as the velocity is not changing at all. Hence, the correct answer is option D.

10. The formula for the range in projectile motion is which is maximum at .R = gu sin2θ2 θ = 45o

Hence, the correct answer is option C.

11. In CGS, FPS and MKS systems the unit of time is same.

12. In steady state the temperature of different points in a heat conductor does not change.

13. In a refrigerator the heat transfer is positive at low temperature and it is negative at high temp.

14. At the open end of the flute, there is antinode of the air molecule’s displacement.

15. Two perpendicular vectors have their dot product as zero.

OR

Two parallel vectors have their cross product as zero.

16. Vector product is defined as

17. The function is not periodic. With increasing time, the function decreases monotonically and tends to zero as time tends to infinity. Hence, the function never repeats.

18. A real gas can be approximated to an ideal gas when its density is very low. This can be brought at very low pressure and very high temperature.

19.

20. The mass of a ping-pong ball is less than a cricket ball. Now, we know that the momentum transferred directly depends on mass and velocity of the body and inversely to the time in which the body is brought to rest. Since, mass of ping-pong ball is more than cricket ball, so it will be relatively easier to catch a ping-pong ball than a cricket ball.

OR No, momentum is not always conserved. For example, momentum is not conserved when external forces (along with internal forces) act on colliding objects.

Section B

21. (1) An object moving with zero acceleration possesses uniform motion i.e., moves with a constant velocity. Its position-time graph is a straight line as represented in the given figure.

(2) An object moving with a positive acceleration is said to have accelerating motion. Its position-time graph is a curve with positive curvature as represented in the given figure.

(3) An object moving with a negative acceleration is said to have decelerating motion. Its position-time graph is a curve with negative curvature as represented in the given figure.

22. According to the kinetic theory of gases,

L.H.S:-

R.H.S:-

Hence, we see that is dimensionally correct.

23. Macroscopic variables (such as pressure, volume, temperature, mass) used for describing the equilibrium state of a thermodynamic system completely are called state variables. The values of state variables depend only on the given state of a system. Internal energy is one of the examples of the thermodynamic state variables because it describes the state of a system in terms of the total kinetic and potential energy of individual molecules forming the system. It does not depend upon the path taken by the system to bring about a change in the internal energy. Any change in internal energy depends only on the initial and final states and not on the path taken by the system to bring about a change.

24. For a rigid body, the angular momentum is given as

Now, by differentiating the above equation with respect to 't', we get

Here, angular acceleration Now, we know that the torque is given as

By substituting this value in equation (2), we can arrive at a relation between angular momentum and torque.

Thus, So, the torque acting on a (rigid) body is defined as the rate of change of angular momentum exhibited by the body.

25.

(A) Frequency of the pendulum, f =

=> f = => f = 0.7046 Hz (B) Let us assume the lowest point as the zero level of gravitational potential energy. According to the law of conservation of energy, we have energy at top = energy at bottom

KEtop + PEtop = KEbottom + PEbottom

0 + mg ( L − L cos ) =

= 0.5778 m/s (C) Total energy at the bottom is also the total energy throughout the oscillation.

So,

Etotal = 0.025 J

26. (A) We have ω = 2πf = 2 × 3.14 × 300 rad/s = 1884 rod/s The motion is described as X = A cos (ωt) => X = (6 × 10-4) cos (1884t) (B) At t = 2.4 × 10-3s, the above equation gives x = (6 × 10-4 m) cos [(1884) × 2.4 × 10-3] = (6 × 10-4m) cos (4.5216 rad) = (6 × 10-4m) (−0.1896) x = −1.138 × 10-4m

OR The standard equation of motion for an oscillatory motion is

So, comparing this equation with the given equation, we get amplitude

Angular frequency, w = 3π rad/s

So, frequency, And time period is

27.

We know that linear momentum and kinetic energy are conserved in an elastic collision, m1v1 + m2v2 = m1u1 + m2u2 m2 (v2 − u2) = m1 (u1 − v1) ... (1) Also,

On dividing equation (2) by (1),

Now, on solving, we obtain:

Similarly,

OR

Mass of bullet mb = 15g = 0.015 kg Mass of wood m = 5 kg Initial speed of bullet = u Initial speed of wood vw= 0 Final speed of bullet-plus-wood combination = v = 0.5 m/s Applying liner conservation principle of momentum, we get, mb u + m × 0 = (mb + m)v ⇒ 0.015 × u = (0.015 + 5) × (0.5)

Section C

28. (i) It is evident from the given graph that for stress 8 × 107 N/m2, strain is 4 × 10−3.

∴Young’s Modulus, Hence, Young’s modulus for the given material is 2 ×1010 N/m2. (ii) Ultimate strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is evident from the given graph that this material has the ultimate strength equal to 16 × 107N/m2 or 1.6 × 108 N/m2. (iii) The fracture point exists beyond the yield strength. It can be observed in the graph that stress for the fracture point is approximately equal to 14 × 107 N/m2 or 1.4 × 108 N/m2.

29. Now, the following process takes place

So, the total heat energy required will be Qt = Q1 + Q2 + Q3 + Q4

= mciΔTi + mLi + mcwΔTw + mLw Here, m = 5 kg ΔTi = 0°C − (−20°C) = 20°C ΔTw = 100°C − 0°C = 100°C ci → specific heat of ice cw → specific heat of water Li → latent heat of fusion Lw → latent heat of vaporization So, Qt = (5 × 2100 × 20) + (5 × 3.36 × 105) + (5 × 4200 × 100) + (5 × 2.25 × 106) = 2.1 × 105 + 16.8 × 105 + 21 × 105 + 112.5 × 105

Thus, the total heat energy required will be Qt = 1.524 × 107 J

30. (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 = 103 kg/m3

Air pressure inside the tank above the water column, P = 1.05 × 105 Pa Atmospheric pressure, Pa = 1.01 × 105 Pa Acceleration due to gravity, g = 9.8 m/s2

Speed of efflux, v, is given as

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

31. Number of small rain drops = n Radius of each rain drop = r Radius of the larger rain drop = R Terminal velocity of small drop, v = 0.13 m/s

Terminal velocity of the larger drop, v’ = 1.17 m/s Volume of the larger rain drop is equal to the volume of n small raindrops.

Terminal velocity is given by the relation,

Hence, the ratio can be written as

Equation (1) reduces to

Hence, the number of raindrops is 27.

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

Let ME and RE be the mass and radius of the earth respectively. Hence, acceleration due to gravity for the earth is given as

It is given that mass and radius of a planet X is half of the mass and radius of earth. Hence, the expression for acceleration due to gravity for planet X is given as

Escape velocity from the surface of the earth,

Escape velocity from the surface of planet X,

Hence, the required ratio is 1:1.

OR

a) The variation of g due to height (h) is given as

Thus, g' = 8.78 m/s2

Now, the weight would be given as W = mg' So W = 5 × 8.78 or W = 43.9 N b) The centripetal acceleration (a) can be calculated using the following formula for centripetal force

So

Thus The centripetal acceleration experienced by the satellite will be a = 7.86 m/s2

33. Using work energy theorem W (all force) = kf − k i

⇒ W (T) + W (N) + W (Mg) + W (f) = kf − k i ……(1) W (N) = 0 [Since displacement is perpendicular to N and Mg]

W (Mg) = 0 ki = 0 [Initially at rest] kf = ?

Given that, Fx = (3 − x 2) N

Plugging these values in equation (1), we get, 0 + 0 + 3.33 = kf− = 0 ⇒ kf = 3.33 J

34. The relative velocity of train B with respect to train A is given as VBA = VB − VA VBA = 40 −(30) Thus, VBA = 10 km/hr Now, the total time taken by B to overtake

Here, the total distance travelled would be the sum of the lengths of the two trains which would be 400 m + 400m = 800 m = 0.8 km So, we have

Thus, t = 0.08 hr So, during this time, the total distance covered by A is d = (VA × t) So, by using appropriate values, we get d = (30 × 0.08) d = 2.4 km Thus, total distance travelled d = 2400 m

Section D

35. The given situation can be represented as

M and P are the lowermost and topmost points respectively. TP is the tension in the string when the stone is at point P. From the figure, it can be written that MQ = OM − OQ = l − l cosθ= l(1 − cosθ)

Velocity at point M = u Velocity at point N = vN vN can be obtained by using the relation,

Equation (1) represents the velocity of the stone at any point N on the vertical circle. At point N at equilibrium,

Equation (2) represents the tension in the string when the stone is at any point N on the vertical circle. Tension in the string should be non-zero at the topmost point P to perform a complete circle.

Hence, in order to perform a complete vertical circle, the minimum velocity required at the

lowermost point is .

OR (i) Free body diagram of a car taking a turn on a circular track is represented in the given figure.

θ = Angle of bank N = Normal reaction on the car f = Frictional force between the tyres and road m = Mass of the car g = Acceleration due to gravity At equilibrium, the net vertical force on the car is zero. Again, the net horizontal force on the car is also zero.

Where, v = Velocity of the car R = Radius of the circular track

= Centripetal force required for the circular motion The relation for frictional force is

Equation (1) reduces to

Substituting the values of N and f in equation (2), we obtain

(ii) It is given that, R = 3 m θ = 45°

= 0.11 g = 9.8 m/s2

Maximum allowed speed of the car is given by the relation,

36. 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 (T1) and cold (T2) 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 P1,V1 and T1 . From the curve it is observed that the gas first expands isothermally to the state having pressure, volume and temperature to be P2 ,V2 and T1. From this state it compressed isothermally to the state having pressure,volume and temperature to be P3 ,V3 and T2.From this state it is compressed adiabatically to the state having pressure,volume and temperature to be P1 ,V1 and T1 . The efficiency of the carnot engine can be found by calculating the work done in individual cycle. Isothermal expansion of gas Heat absorbed (Q1) by the gas from the reservoir is the work done. (W1 → 2) by the gas,

● Step 2 → 2

Adiabatic expansion of gas

Work done by the gas,

● Step 3 → 4

Isothermal compression

Heat released (Q2) by the gas to the reservoir is the work done (W3 → 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

OR a. The efficiency of a real engine must be less than the efficiency of a carnot engine, operating between two temperatures. Thus, Efficiency, E of a carnot engine operating between 0°C (273 K) and 100°C (373 K)

Hence, the claimed efficiency of 55% for this real engine is impossible. b. Let us consider two reversible heat engines operating between the same reservoirs.

Let us assume that heat engine 1 is more efficient than heat engine 2. Now, each engine is supplied with the same amount of heat QH. The work produced by 1 is W1, while by 2 is W2. We assume that the efficiency of engine 1, η1 is greater than the efficiency of engine 2, η2 i.e., η1 > η2. Now, let reversible engine 2 be made to operate as a refrigerator. This refrigerator receives an input work of W2 and rejects heat QH to the hot reservoir. We can say that this heat is being received by engine 1, since it receives QH amount of heat from the reservoir, thereby eliminating any net heat exchange with the hot reservoir. Thus, we have a combined heat engine that produces work while exchanging heat with a single reservoir. This is in direct contradiction to the Kelvin-Planck statement of the second law of thermodynamics, proving that the assumption η1 > η2 was incorrect.

37. (a)

Statement: The moment of inertia of a body about an axis, parallel to the axis passing through its center of mass, is equal to the sum of the moments of inertia of the body about the axis passing through its centre of mass and the product of the mass and the square of distance between the two parallel axes. Proof: Consider a particle of mass m at P. Let ‘d’ be the perpendicular distance between

parallel axes YY and . Let GP = x M.I of the particle about YY = m (x + d)2

M.I of the whole of lamina about YY,

However, Σ mx2 = Ig and Σ md2 = (Σ m)d2 = Md2

Where M (= Σ m) is the mass of the lamina. Also, Σ 2mxd = 2d Σ mx ΣI = Ig + Md2 + 2d Σmx The lamina will balance itself about its centre of gravity. Therefore, the algebraic sum of the moments of the 'weight of consistent particles' about the centre of gravity G should zero.

Σ mg × x = 0 g Σ mx = 0 Σ mx = 0 [ g ≠ 0] From (equation (i), I = Ig + Md2

(b) According to theorem of parallel axis, the moment of inertia will be I' = I + MR2 .....(1) for a circular disc, I = MR2

So, the moment of inertia of a circular disc about the said axis will be I' = 2MR2

Here M = mass of the disc = 250 g = 0.25 kg R = radius of the disc = 25 cm = 0.25 m So, I' = 2[0.25] × [0.25]2

Thus, the moment of inertial will be I' = 3.125 × 10–2 kg. m2

OR

Let the mass of the cylinder be ‘M’.

Density of the cylinder Mass of the sphere = Volume × Density

Mass of the remaining part of the cylinder is . Let the x−axis be the longitudinal axis of the cylinder. The origin is at the centre of the mass of the cylinder. Hence, the original centre of the mass of the cylinder lies at (0, 0). Let x1 be the position of the centre of the mass of the sphere.

Thus, x1 = −30 Let x2 be the coordinate of the centre of the mass of the remaining cylinder.

Thus, we have: Hence, the position of the centre of the mass shifts 10.9 cm to the right of its original position.