Category: Class 11 Physics Objective Questions

Physics Multiple Choice Questions on “Thermodynamics – Heat Engines”.

1. In a heat engine heat is converted to work and the process need not be cyclic. True or False?
a) True
b) False

Answer: b
Clarification: In heat engines heat is to be converted into work. The process is cyclic as we need the conversion of some heat into work for continuously carrying out our desired outcome.

2. If an engine takes 5J of heat energy from a reservoir and rejects 3J to the surroundings, what is its efficiency?
a) 2/5
b) 3/5
c) 1/5
d) 0

Answer: a
Clarification: Efficiency is defined as work done/heat input.
Work done = 5-3=2J.
Heat input = 5J.
Thus, efficiency = 2/5.

3. An engine works between 2 reservoirs of temperature 500K and 300K. A company claims that this engine of theirs takes 200J of energy from the hot reservoir and rejects 110J of energy. Is this True or False?
a) True
b) False

Answer: b
Clarification: The maximum possible efficiency is given by:
1 – T_cold/T_hot = 1 – 300/500 = 0.4.
The company claims that their engine has an efficiency of (200-110)/200 = 0.45.
This is not possible as maximum efficiency is 0.4.

Physics Multiple Choice Questions on “Oscillations – Damped Simple Harmonic Motion”.

1. Damping force on a spring mass system is proportional to which of the following quantities?
a) Velocity
b) Acceleration
c) Displacement from mean position
d) (velocity)²
Answer: a
Clarification: We know from Stoke’s law that damping force is proportional to velocity of a body. It also depends on the surface area in contact with the source of particles causing damping, like air.

2. If the restoring force on a body is given by: F = -kx -bv, then what is the expression for amplitude of motion? Let A be amplitude without damping forces
a) Ae^-at, where a is some constant
b) Ae^at, where a is some constant
c) A
d) 0
Answer: a
Clarification: The given force represents damping SHM. x(t) = Ae^-atcos(bt+c). The term Ae^-at is the amplitude which decreases with increase in time, until eventually the particle comes to a stop.

3. What happens to the energy of a particle, in SHM, with time in the presence of damping forces?
a) Stays constant
b) Decreases linearly
c) Decreases exponentially
d) Decreases cubically
Answer: c
Clarification: Energy in presence of damping forces is given by: 1/2kA²e^-bt/m. This shows that kinetic energy decreases exponentially with time.

4. Consider the damped SHM of a spring mass system. If the time taken for the amplitude to become half is ‘T’, what is the time taken for mechanical energy to become half?
a) T
b) T/2
c) 2T
d) T/4
Answer: b
Clarification: Amplitude = Ae^-bt/2m& Energy = 1/2kA²e^-bt/m. For amplitude to become half, e^-bt/2m = 1/2
Or bT/2m = ln(2)
Or T = 2m ln(2)/b For energy to become half, e^-bt/m = 1/2
Or t = m ln(2)/b = T/2.

Physics Multiple Choice Questions on “System of Particles – Motion of Centre of Mass”.

1. If forces are acting on a rigid body so that it has zero kinetic energy, then all forces will pass through the centre of mass and have a vector sum of zero. True or False?
a) True
b) False
Answer: b
Clarification: If a rigid body has zero kinetic energy means that vector sum of all forces will be zero and the torque on it will be zero. For torque to be zero the vector cross product of force and distance should be zero. It is not necessary for all forces to pass through the centre.

2. In a projectile motion, a particle breaks due to internal forces. We can say that its total momentum is conserved. True or False?
a) True
b) False
Answer: b
Clarification: For momentum to be conserved its value should remain constant, but in projectile motion the force of gravity is always changing its momentum. The internal forces will break the particle and the momentum of different particles can be conserved in a direction perpendicular to direction of gravity.

3. A ball of mass 1kg is thrown vertically upwards from a building with a speed of 10m/s, and another ball of mass 3kg is dropped from the same point towards the ground with the same speed. At what time will the centre of mass have the maximum height w.r.t the ground? Assume balls are left at t=0.
a) 0s
b) 1s
c) 2s
d) 2.8s
Answer: a
Clarification: When the balls are released in their respective directions, the ball going downwards is heavier and is covering more distance per unit time. This is because both the balls have the same initial speed and the heavier ball is gaining speed downwards while the other ball is losing speed as it moves upwards. Their mid-point will be below the point of release and centre of mass, will therefore, be even below that point, since heavier mass is moving downwards.

4. A boy of mass 50kg is standing on a frictionless surface. He throws a ball of mass 2kg away from him with a speed of 10m/s. Find the final speed of the centre of mass.
a) 0m/s
b) 20m/s
c) 10m/s
d) 0.4m/s
Answer: a
Clarification: Initially the boy was at rest and there is no friction, so the net force on the ball-boy system is zero. Hence, their momentum change will be zero and final velocity of centre will also be zero. Note that boy will move backwards with a speed of 2*10/50 = 0.4m/s.

5. A ball of mass 3kg is thrown at an angle of 30° with the horizontal & a speed of 10m/s. At the highest point the ball breaks into two parts, having mass ratio 2:1, due to internal forces. If the heavier part falls at a distance of 10m from the start, find the x coordinate of the second part.
a) 5.98m
b) 6.34m
c) 8.66m
d) 7.99m
Answer: a
Clarification: There is no net force on the particle in the x-direction, so the x-coordinate of centre of mass will be equal to the range of projectile motion.
The parts will have masses 2kg and 1kg.
The time of motion ‘t’ = 2u_y/g = 1s.
The range = u_xt = 5√3*1 = 8.66m.
The centre of mass will land at 8.66m.
So,
(m₁x₁ + m₂x₂)/(m₁ + m₂) = 8.66.
m₁ = 2kg, m₂ = 1kg, x₁ = 10m, we have to calculate x₂.
∴ m₂x₂ = (8.66*3) – 2*10
= 25.98 – 20 = 5.98.
∴ x₂= 5.98/1 = 5.98m.

Physics Multiple Choice Questions on “Gravitation – Earth Satellite – 1”.

1. The satellites orbiting the earth, eventually fall to the earth when they are left unsupervised or unattended because _____
a) their power supply runs out
b) of viscous forces causing the speed of the satellite and hence height to gradually decrease
c) the laws of gravitation predict such a trajectory
d) of collisions with other satellites

Answer: b
Clarification: Due to the presence of some gases of the atmosphere and other debris, the satellites speed is reduced due to viscous forces. This reduces the height of the satellite. Hence, it eventually falls to the earth.

2. The time period of the moon around the earth is 24 hours.
a) True
b) False

Answer: b
Clarification: The time period of the moon around the earth cannot be 24 hours because then it would be a geostationary satellite. However, we know it is not true. The time period of the moon around the earth is found to be approximately 27.3 earth days.

3. Consider 2 satellites A and B having time periods 16 hours and 1 hour, respectively. What is the ratio of the radius of their orbits?
a) 4:1
b) 8:1
c) 16:1
d) 64:1

Answer: d
Clarification: The time periods (T) of a satellite revolving around the earth of radius R and at height h is:
T = 2 x pi x [(R + h)³ / (G X M)]^1/2
Hence, the time period is directly proportional to (R + h)^3/2
The ratio of orbits is, therefore, 64:1.

4. A satellite is launched into a circular orbit of radius R while a second satellite is launched into an orbit of radius 1.02R. What is the percentage change in the time periods of the two satellites?
a) 0.7
b) 1
c) 1.5
d) 3

Answer: d
Clarification: The time periods (T) of a satellite revolving around the earth of radius R and at height h is:
T = 2 x pi x [(R + h)³ / (G X M)]^1/2
Hence, the time period is directly proportional to (R + h)^3/2
From this, we have the ratio of the time periods as 1.03.
1.03 x 100 = 103. Hence, there is a 3% change in the time period.

5. A satellite is revolving very close to a planet of density D. What is the time period of that satellite?
a) [3/(D*G)]^1/2
b) [3/(D*G)]^3/2
c) [3/(2*D*G)]^1/2
d) [(3*G)/D]^1/2

Answer: c
Clarification: The time period of a satellite flying very close to the surface of the earth is;
T = 2 x pi x [R³ / (G X M)]^1/2;  [Height is negligible compared to the radius of the earth].
Mass (M) = Density (D) x Volume
M = D x (4/3 x pi x R³)
Substituting the relation of mass into the time period, we get;
T = [3/(2 x D x G)]^1/2

6. A satellite orbits the earth at a height of R/5. What is its orbital speed?
a) [(2*G*M)/(R)]^1/2
b) [(G*M)/(R)]^1/2
c) [(G*M)/(7*R)]^1/2
d) [(5*G*M)/(6*R)]^1/2

Answer: d
Clarification: The orbital velocity of a satellite flying at a height “h” above the surface of the earth of radius “R” is:
v = [(G*M)/(R+h)]^1/2
Here, h = R/5
Therefore; v = [(5*G*M)/(6*R)]^1/2

7. The time period of a satellite is independent of the mass of the satellite.
a) True
b) False

Answer: a
Clarification: For a satellite of mass “m” revolving around a planet of mass “M” and radius
“R” at a height “h” above the surface, the satellite traverses a distance [2 x pi x (R + h)] in time “T” if the orbital velocity is “v”.
Hence, the time periods (T) = [2 x pi x (R + h)]/v
We know; orbital velocity (v) = [(G x M)/(R+h)]^1/2
Hence; T = [2 x pi x (R + h)^3/2/(G X M)^½], which is independent of the mass of the satellite.

8. The time period of a satellite of earth is 90 minutes. If the separation between the earth and the satellite is quadrupled, the new time period will be _____________
a) 90 minutes
b) 180 minutes
c) 560 minutes
d) 720 minutes

Answer: d
Clarification: The time period is directly proportional to the radius of orbit raised to the power of 1.5.
If the radius is quadrupled, the time period increases by 8 times.
Hence, the new time period = 90 x 8 = 720 minutes.