Category: Class 11 Physics Objective Questions

250+ TOP MCQs on Oscillations – Systems Executing Simple Harmonic Motion | Class 11 Physics

Physics Objective Questions and Answers for Class 11 on “Oscillations – Systems Executing Simple Harmonic Motion”.

1. Let T1 be the time period of a single spring mass system on a horizontal surface. Let T2 be the time period of the same system when hung from the ceiling. What is the expression for T1 & T2?
a) T1 = T2 = 2π √(k/m)
b) T1 = 2π √(m/k), T2 = 2π √(mg/k)
c) T1 = T2 = 2π √(m/k)
d) T1 = 2π √(m/k), T2 = 2π √(k/mg)
Answer: c
Clarification: For the spring mass system on the horizontal surface, F = -kx.
So, T1 = 2π/w = 2π √(m/k).
When the spring mass system is hung from the ceiling, at mean position mg = kx0.
Let spring be pulled further by x, extension = x+x0,
F = k(x+x0)-mg
= kx + kx0– mg = kx.
The force is proportional to x.
Thus, w = √(k/m) & T2 = 2π √(m/k).
Therefore, T1 = T2 = 2π √(m/k).

2. A spring block system is on a horizontal plane. If spring constant is 2N/m & mass of block is 1kg. By what amount should the block be extended from equilibrium position such that maximum velocity at mean position is 2m/s?
a) 2√2 m
b) 3√2 m
c) 4√2 m
d) 2 m
Answer: a
Clarification: Using, w = √ (k/m), we get: w = √ (1/2)s-1.
Maximum velocity = Aw.
2 = A/√2
∴ A = 2√2 m.

3. The springs in the given diagram are fixed to the walls. If the block is displaced by a distance x from the mean position, find the time period of oscillations in seconds. Assume that there is no friction.

a) 2π √(2m/3k)
b) 2π √(7m/3k)
c) 2π √(2m/7k)
d) 2π √(2m/5k)
Answer: c
Clarification: When the block is displaced by x in one direction, let’s say left side, all the springs will exert a force on the block towards right. The two springs will be compressed by length x and the right one will be extended by length x.
Net force on block = kx + 2kx + (k/2)x = 3.5kx.
Displacement is towards left and force is towards right, so we write F = -3.5kx.
F = ma = -3.5kx.
∴ a = -(3.5k/m)x.
We know that in SHM a = -w2x.
∴ w = √(3.5k/m).
∴ T = 2π/w = 2π √(m/3.5k) = 2π √(2m/7k) s.

4. If the block is displaced by ‘x’ from the mean position, what will be the time period of oscillations?

a) 2π √(2m/3k) s
b) 2π √(3m/k) s
c) 2π √(m/k) s
d) 2π √(4m/k) s
Answer: b
Clarification: Let spring with constant k be spring 1 and that with constant k/2 be spring 2.
For series connection, effective spring constant is given by (1/k + 2/k)-1 = k/3.
Thus, time period = 2π √ (3m/k) s.

5. Find the effective spring constant.

a) 8/5
b) 8/31
c) 5/8
d) 31/8
Answer: a
Clarification: The springs with spring constant 5 and 3 are parallel,
so their effective spring constant is 5+3 = 8.
Now, spring with spring constant 2 is in series with effective constant 8,
so net spring constant
= (1/8 + 1/2)-1 = 8/5.

6. A spring mass system is on a horizontal plane. Spring constant is 2N/m & mass of block is 2kg. The spring is compressed by 10cm and released. A small ball is released from a height of 2m above the block on the horizontal plane. At what time should the ball be left if the block is released from the extreme position at t=0, given that the ball has to fall on the block when the speed of block is maximum?
a) 1.57s
b) 0.94s
c) 0s
d) 0.63s
Answer: b
Clarification: w = √2/2 = 1. Time period of SHM = 2π/w = 2π s.
Speed of the block will be maximum at the mean position.
Time to reach mean position from extreme position = T/4
= 2π/4 = π/2 =1.57s.
Time for the small ball to fall down = √(2h/g) = √(4/9.8) = 0.63s.
∴ Time when it should be released
= 1.57 – 0.63 s
= 0.94s.

7. A spring mass system is hanging from the ceiling of a lift. It has been like this for quite some time. Suddenly the cables holding the lift are cut. What will be the amplitude of oscillation of the block when observed by a person standing (of mass M) in the lift?
a) mg/k
b) (M-m)g/k
c) The observer will not observe a SHM
d) 2mg/k
Answer: a
Clarification: When the system was stationary, the extension in the spring will be xo such that kxo= mg or x0 = mg/k.
When the cables are cut, the entire system will be in free fall.
The observer will add a pseudo force on the block upwards and equal to mg.
Now the spring has an extension of mg/k
and the restoring force by the spring will be corresponding to this extension,
so the amplitude of SHM for the observer is mg/k.

8. A spring block system is on a horizontal surface. The block is compressed by 10cm. What should be the coefficient of friction between the block and ground such that the block comes to a stop at the mean position? Mass of block = 0.5kg & spring constant = 3N/m.
a) 0.03
b) 0.06
c) 0.12
d) 0.05
Answer: b
Clarification: When the block is released, the spring force will act towards the mean position and friction force will act against velocity. The work done by these forces should be equal to potential energy of the system in the compressed state. ∴ Work done in dx displacement = (μmg – kx)dx.
On integrating this, with dx varying from 0 to 0.1, we get
Work = (μmgx – kx√/2)00.1= 0.5μ – 0.015.
This should be equal to initial potential energy of the system.
0.5μ – 0.015 = 1/2kA√
∴ 0.5μ – 0.015 = 1/2 * 3 * 0.01
∴ 0.5μ – 0.015 = 0.015
∴ 0.5μ = 0.03 OR μ = 0.06.

9. Two springs in horizontal spring mass systems have spring constants in the ratio 2:3. By what ratio should they be extended from their mean positions so that they have the same value of maximum speed? The masses of blocks are the same in both cases.
a) 3/2
b) √3 / √2
c) 9/4
d) 4/9
Answer: b
Clarification: w = √(k/m), Maximum speed = Aw.
According to question: A1w1 = A2w2.
Their extensions from mean positions will be their amplitudes.
A1/A2 = w2/w1
= √k2/√k1
= √3 / √2.

10. A spring is fixed to the ground as shown below. A block of mass m is fixed to the spring. The spring is compressed by 5cm. Imagine that we can remove the spring suddenly and the block will fly off with the speed it has at that instant. The spring has to be removed at such a position that the distance travelled upwards by the block relative to that position is maximum. Find that position.

a) Mean position
b) 2cm above mean position
c) 4cm above mean position
d) 2cm below mean position
Answer: a
Clarification: The given condition can be achieved when the spring is made to disappear at mean position because the block will have maximum speed.

11. Which of the following factors affects the time period of a simple pendulum?
a) Mass of bob
b) Shape of bob
c) Length of string
d) Angle of release from 0° to 20°
Answer: c
Clarification: We neglect air resistance, so the shape of the bob will not matter.
Time period is given by:
2π √(l/g).
Thus, it depends on the length of the pendulum and value of g.

12. A simple pendulum,hanging from the ceiling, is released from an angle . What is the restoring force?

a) mg
b) mgcosθ
c) mgsinθ
d) mgtanθ
Answer: c
Clarification: The component of mg which will cause change in speed is mgsinθ.
This force is always towards the mean position and is a function of θ.

13. What is the length of a seconds pendulum?
a) 1m
b) 2m
c) 9.8m
d) 0.5m
Answer: a
Clarification: Time period of seconds pendulum = 2s.
∴ 2 = 2π √(L/9.8)
OR L = 1m.

14. Two simple pendulums of the same length are released from 10° & 20° respectively. Select the correct statement.
a) The second pendulum have will have a double time period than that of the first
b) They both will reach the mean position at the same time
c) The first pendulum will have a double time period than that of the second
d) Their maximum speed will be the same
Answer: b
Clarification: Their time period will be the same as length is the same. So time to reach the mean position from their respective extremes will be the same. Therefore they will reach the mean position at the same time. They both have different potential energies at the start, so their maximum speeds will be different. Another way to look at this is that the speed of the 2nd pendulum should be more as it has to cover greater arc length in the same amount of time.

15. What will happen to the time period of a simple pendulum if temperature of the surroundings is significantly increased?
a) Increase
b) Decrease
c) Remain the same
d) The oscillations come to a stop
Answer: a
Clarification: The time period of a simple pendulum is given by: T = 2π √(l/g).
If the temperature increases,
length of string will increase and thus,
the time period will increase.

To practice Physics Objective Questions and Answers for Class 11,

250+ TOP MCQs on Motion in a Straight Line – Acceleration | Class 11 Physics

Physics Multiple Choice Questions on “Motion in a Straight Line – Acceleration”.

1. Which of the following is the correct formula for finding acceleration?
a) a = dx/dt
b) a = d2x/dt2
c) a = x/t
d) a = d3x/dt3
Answer: b
Clarification: Acceleration is the rate of change of velocity. Velocity is the rate of change of displacement. Therefore, acceleration is the rate of change of rate of change of displacement = d2x/dt2.

2. How does the displacement v/s time graph of a uniformly accelerated motion look like?
a) A straight line
b) A parabola
c) A hyperbola
d) An ellipse
Answer: b
Clarification: Here we use the second equation of motion. s = ut + (1/2)at2. In a uniformly accelerated motion, the acceleration remains constant. Therefore, the equation for displacement becomes the equation of a parabola.

3. The velocity of a body varies as, v = 2t+5t2. What is the acceleration at t = 10?
a) 102
b) 100
c) -100
d) 50
Answer: a
Clarification: Acceleration = dv/dt. To find the acceleration, we can differentiate the velocity equation with respect to time. a = 2 + 10t. On putting, t = 10, we get a = 102.

4. A ball falls from a building and covers 5m in 10s. What is the acceleration?
a) 0.1 m/s2
b) 0.2 m/s2
c) 9.81 m/s2
d) 10 m/s2
Answer: a
Clarification: Assuming the ball falls with zero initial velocity, then according to the second equation of motion, s = ut + (1/2)at2 = (1/2)at2. When we put t = 10s, we get a = 0.1 m/s2.

5. What is negative acceleration known as?
a) Deceleration
b) Acceleration
c) Escalation
d) Relaxation
Answer: a
Clarification: Negative acceleration is known is deceleration. Negative acceleration acts in the direction opposite to the direction of motion. It is also known as retardation.

6. A car is travelling in the north direction. To stop, it produces a deceleration of 60 m/s2. Which of the following is a correct representation for the deceleration?
a) 60 m/s2 Northwards
b) 60 m/s2 Southwards
c) 60 m/s2 Eastwards
d) 60 m/s2 Westwards
Answer: c
Clarification: Deceleration always acts in the direction opposite to the direction of motion. Magnitude of deceleration is 60 m/s2 and it acts in the direction opposite to North. Hence the answer is 60 m/s2 Southwards.

7. A uniformly accelerated body has ____
a) Constant speed
b) Constant velocity
c) Constant force
d) Constant momentum
Answer: c
Clarification: Since the body is accelerated, the speed and velocity will vary. Momentum depends on velocity; hence the momentum will also vary. The force remains constant as F = ma.

8. The velocity of a truck changes form 3 m/s to 5 m/s in 5s. What is the acceleration in m/s2?
a) 0.4
b) 0.5
c) 4.0
d) 5.0
Answer: b
Clarification: Acceleration is the rate of change of velocity. Here, the velocity changes form 3 m/s to 5 m/s in 5s. Hence, acceleration = (5-3)/5 = 0.4 m/s2.

9. The gradient of velocity v/s time graph is equal to____
a) Velocity
b) Acceleration
c) Distance
d) Momentum
Answer: a
Clarification: The gradient or slope of any graph tells us the value of its differential at that point. Since acceleration is the differential of velocity with respect to time, the gradient is equal to the acceleration.

10. The velocity of a ship varies with time as v = 5t3. What is the acceleration at t = 2?
a) 60
b) 56
c) 40
d) 100
Answer: a
Clarification: Acceleration of a body can be found out by differentiating the expression for velocity. Here v = 5t3. On differentiating, a = dv/dt = 15t2. On putting t = 2, we get a = 60 units.

11. If the velocity varies parabolically, how does the acceleration vary?
a) Linearly
b) Hyperbolically
c) Parabolically
d) Elliptically
Answer: a
Clarification: Velocity varies parabolically implies that velocity is of the form at + bt2. On differentiating the velocity, we get the equation for acceleration. Therefore, a = dv/dt = a+2bt, which is a linear equation. Hence the acceleration varies linearly.

12. The expression for displacement is x = sin(5t). The expression for acceleration is ______
a) 5sin(5t)
b) 25cos(5t)
c) -25sin(5t)
d) -5cost(5t)
Answer: a
Clarification: Acceleration = dv/dt = d2x/dt2. The velocity (v) can be expressed as, v = dx/dt = 5cos(5t). Hence acceleration (a) becomes, a = dv/dt = -25sin(5t).

250+ TOP MCQs on Laws of Motion – Conservation of Momentum | Class 11 Physics

Physics Question Bank on “Laws of Motion – Conservation of Momentum”.

1. Two bodies moving with constant velocities collide with each other. Which of the following quantities remain conserved?
a) Momentum
b) Speed
c) Force
d) Velocity
Answer: a
Clarification: Two bodies collide in the absence of any external force. In this case the momentum will remain constant because there is no force, hence, there is no change in momentum. The velocity and speed of the bodies might change after collision.

2. A ball is moving with a constant velocity. After some time, it collides with a wall. Which one of the following remains conserved except momentum?
a) Energy
b) Displacement
c) Power
d) Force
Answer: a
Clarification: Apart from the momentum, the energy also remains conserved during collision. As a matter of fact, the energy of a system often remains conserved.

3. A variable mass body with an initial mass of 5 kg is moving with a velocity of 5 m/s initially, after some time the velocity increases to 10 m/s, what is the new mass if the body is moving in the absence of any external force?
a) 2.5 Kg
b) 5 Kg
c) 7.5 Kg
d) 10 Kg
Answer: a
Clarification: The body is moving in the absence of any external force thus the momentum remains conserved. m1v1 = m2v2. v1 = 5 m/s, m1 = 5 Kg, v2 = 10 m/s. On solving, we get, m2 = 2.5 kg. Hence, the correct answer is 2.5 Kg.

4. Two bodies of masses 2 Kg and 7 Kg are moving with velocities of 2 m/s and 7 m/s respectively. What is the total momentum of the system in Kg-m/s?
a) 50
b) 53
c) 28
d) 0
Answer: b
Clarification: Both the bodies are moving with constant velocity, hence both of them have momentum. The momentum of the system is the sum of the momenta of the bodies. Hence total momentum = 2 x 2 + 7 x 7 = 53 Kg-m/s.

5. What is the plural of the word ‘momentum’?
a) Momenta
b) Momentums
c) Moments
d) Momentuma
Answer: a
Clarification: The plural for momentum is momenta and not momentums. Momenta, being a plural, is usually used in cases where we need to address a more than one momentum. This situation can possibly arise in systems having more than one moving body.

6. Two bodies are accelerating towards each other. After sometime they collide. Which one of the following statements is true about the system?
a) The total momentum remains conserved about the instant of collision
b) The total momentum remains conserved during collision
c) The total momentum does not remain conserved
d) The total momentum remains conserved
Answer: a
Clarification: The two bodies are accelerating and they collide. Since they are accelerating the total momentum does not remain conserved during the motion. But at the instant of collision, both the bodies have a constant velocity. Hence, just before and just after the collision, the total momentum will remain conserved.

7. When a body collides with a wall or the ground, what assumption do we make?
a) The mass of the body is not negligible
b) The body is stationary
c) The mass of the body is negligible as compared to the mass of the wall or the ground
d) The body is perfect
Answer: c
Clarification: When a body collides with a wall or the ground, we consider that the mass of the body is negligible as compared to the mass of the wall or the ground. We also make this assumption when something like a feather, which has a very small mass collides with a truck, which, comparatively, has a huge mass.

8. A variable mass body with an initial mass of 10 kg is moving with a velocity of 7 m/s initially, after some time the velocity increases to 14 m/s, what is the new mass if the body is moving in the absence of any external force?
a) 2.5 Kg
b) 5 Kg
c) 7.5 Kg
d) 10 Kg
Answer: b
Clarification: The body is moving in the absence of any external force thus the momentum remains conserved. m1v1 = m2v2. v1 = 7 m/s, m1 = 10 Kg, v2 = 14 m/s. On solving, we get, m2 = 5 kg. Hence, the correct answer is 5 Kg.

To practice Physics Question Bank,

250+ TOP MCQs on System of Particles – Centre of Mass | Class 11 Physics

Physics written test Questions & Answers on “System of Particles – Centre of Mass – 2”.

1. The centre of mass for an object always lies inside the object.
a) True
b) False

Answer: b
Clarification:The centre of mass for an object need not necessarily lie inside the object. A ring or a C-section element are classic examples for this.

2. For which of the following does the centre of mass lie outside the body?
a) Pen
b) Dice
c) Rectangular tile
d) Bangle

Answer: d
Clarification: The centre of mass of a bangle lies at its geometric centre which does not lie on or in the object. Hence the centre of mass of a bangle lies outside the body.

3. The centre of mass of the body always lies on the line of symmetry for an object with uniform density.
a) True
b) False

Answer: a
Clarification: The centre of mass of the body always lies on the line of symmetry for an object with uniform density if there exists a line of symmetry for that geometry. Because, by definition, symmetry implies an equivalence of mass on either side of the symmetrical axis if the density is uniform.

4. Numerous particles in a system are placed at a distance of “D” from the origin. The position of the centre of mass will definitely be _____
a) Equal to D
b) Lesser than or equal to D
c) Greater than D
d) Greater than or equal to D

Answer: b
Clarification: Since all objects are placed at the same distance, they are either placed in a spherical configuration, circular-arc configuration or a linear configuration. In all of the cases, the centre of mass of the system will lie within the geometry of the shape. Hence the centre of mass of the system is at a position lesser than or equal to D.

5. An asteroid enters the atmosphere of the earth and breaks into two pieces. One of the pieces is larger than the other. Which of the following is true considering the centre of mass of both the pieces together?
a) The centre of mass shifts horizontally towards the larger piece
b) The centre of mass shifts horizontally towards the smaller piece
c) The centre of mass remains in the same trajectory as before breaking apart of the asteroid
d) Depends on the velocity of the asteroid

Answer: c
Clarification: Since internal forces do not change the position of the centre of mass of a system as their net force is zero, the centre of mass remains in the same trajectory as before breaking apart of the asteroid. However, it is an influential factor for the trajectories of the two pieces after breaking apart.

6. Consider a system on which there are external forces acting. If the vector sum of all these external forces is zero, then the centre of mass _____
a) must not move
b) may move
c) may accelerate
d) must accelerate

Answer: b
Clarification: Since the vector addition of all external forces is zero, we know that the acceleration vector is zero for the resultant force vector as mass cannot be zero. Zero acceleration can imply zero or non-zero constant velocity. Thus, the centre of mass may move.

7. For an object with the centre of mass at the origin, the x-coordinates of particles of the object _____
a) may be all positive
b) may be all negative
c) may be all non-negative
d) cannot be predicted

Answer: c
Clarification: For an object to have the x-coordinate of its centre of mass, there should be particles on both the negative and positive coordinates of the plane for the sum to be zero. However, for 1-dimensional particles that all lie on the y-axis symmetric about the x-axis, the x-coordinate is zero – which is neither positive nor negative, i.e., non-negative.

8. Two bodies of masses 5 kg and 3 kg are moving towards each other at velocities of 3 m/s and 5 m/s, respectively. What is the velocity of their centre of mass?
a) 0.25 m/s towards 3 kg
b) Upredictable
c) The centre of mass is stationary
d) 0.5 m/s towards 5 kg

Answer: c
Clarification: The velocity of the centre of mass = (m1*v1 + m2*v2)/(m1 + m2)
m1 = 5kg
m2 =3 kg
v1 = 3 m/s
v2 = -5 m/s [opposite direction)
Velocity of centre of mass = (5*3 – 3*5)/(5+3)
= 0.

9. The velocities of three particles of masses 10 kg, 20 kg and 30 kg are 10i, 10j and 10k m/s, respectively. What is the velocity of their centre of mass?
a) (i + 2j + 3k) m/s
b) 10 (i + j + k) m/s
c) (10i + 20j + 30k) m/s
d) (3i + 2j + k) m/s

Answer: a
Clarification: The velocity of the centre of mass = (m1*v1 + m2*v2 + m3*v3)/(m1 + m2 + m3)
= (10*10i + 20*10j + 30*10k) / (10 + 20 + 30)
= (100i + 200j + 300k) / 100
= (i + 2j + 3k) m/s.

10. A system consists of 2 particles of the same mass. Let one particle be at rest and another particle have an acceleration of “2a”. What would be the acceleration of the centre of mass of the system?
a) a/4
b) a/2
c) a
d) 2a

Answer: c
Clarification:The acceleration of the centre of mass = (m1*a1 + m2*a2)/(m1 + m2)
m1 = m2 = m
a1 = 0
a2 = 2a
The acceleration of the centre of mass = (m*0 + m*2a) / (m + m)
= 2ma/2m
= a.

11. Two particles A and B, initially at rest, start moving towards each other under a mutual force of attraction. When the speed of A is “10 m/s” and the speed of B is “800 m/s”, what is the speed of their centre of mass?
a) 0
b) 0.8v
c) 80v
d) 8v

Answer: a
Clarification: Since both A and B are moving towards each other due to mutual force of attraction, we can infer that the particles of the system are in motion due to internal forces and hence the state of centre of mass will remain unchanged, i.e., at rest.

13. A uniform free rectangular steel plate is heated from 273 to 373 kelvin. The initial area of the plate is 15 sq. cm. What is the shift of the centre of mass?
a)0 cm
b) 1 cm
c) 2 cm
d) 3 cm

Answer: a
Clarification: Since the plate is free, the net force on the plate is zero. So, there will be no shift in the centre of mass of the plate, i.e., the shift in the centre of mass of the plate will be 0 cm.

250+ TOP MCQs on Gravitation – Escape Speed | Class 11 Physics

Physics Multiple Choice Questions on “Gravitation – Escape Speed”.

1. The ratio of the radius of a planet A to that of a planet B is “r”. The ratio of acceleration due to gravity on the planets is “p”. The ratio of escape velocities of the two planets is _____
a) (pr)1/2
b) (p/r)1/2
c) (pr)
d) (p/r)
Answer: a
Clarification: Escape velocity (v) = (2*g*R)1/2
g = Acceleration due to gravity
R = Radius of planet
Given;
Ra/Rb = r
ga/gb = p
va/vb = (2*ga*Ra])1/2/(2*gb*Rb) 1/2
= (p x r)1/2.

2. What is the minimum velocity required for an object of mass “m” to escape the gravitational pull of a planet of mass “M” and radius “R” from its surface?
a) [(G*M)/R]1/2
b) [(2*G*M)/R] 2
c) [(G*M*m)/R]1/2
d) [(2*G*M)/R]1/2
Answer: d
Clarification: If the object’s initial velocity is “v”, from the law of conservation of energy, we have;
(m x v2)/2 = (G x M x m)/R
v2 = (2 x G x M)/R
Therefore; v = [(2 x G x M)/R]1/2.

3. The escape velocity of an object from the surface of the planet does not depend on the mass of the object.
a) True
b) False
Answer: a
Clarification: The escape velocity of an object depends only on the mass of the planet from which it is escaping and not the mass of the object. The relation is as given below;
Escape velocity (v) = [(2*G*M)/R]1/2
R = Radius of planet
M = Mass of planet.

4. An object is projected vertically upwards with a velocity (gr) 1/2. What is the maximum height reached by the object if “R” is the radius of the earth and “g” is the acceleration due to gravity?
a) R/2
b) R
c) 2R
d) 4R
Answer: b
Clarification: From the law of conservation of energy, we have;
(m x v2)/2 – (G x M x m)/R = -(G x M x m)/(R+h)
m = Mass of object
M = Mass of earth
h = Maximum height of the object
v = Velocity of projection
(m x v2)/2 – (g x m x R) = -(g x m x R2)/(R+h)
(g x R)/2 = (g x R x h)/(R+h)
Therefore; h = R.

5. The escape velocity of the earth is “v”. If an object is thrown vertically upwards with a velocity “kv”, what is the speed of the object at infinity?
a) v/(k2-1)1/2
b) v(k2-1)1/2
c) v2/(k2-1)
d) v2 (k2-1)
Answer: b
Clarification: From the law of conservation of energy, we have;
(m x (kv) 2)/2 – (G x M x m)/R = (m x p2)/2
m = Mass of object
M = Mass of earth
R = Radius of the earth
p = Velocity of the object at infinity
(kv)2/2 –(g x R) = p2/2
(kv)2/2 – v2/2 = p2/2; v = (2 x g x R) ½
(k2-1)v2 = p2
Therefore; p = v(k2-1) 1/2.

6. The earth is able to retain its atmosphere because of _____
a) the mean velocities of atmospheric gas molecules which is less than that of the earth’s escape velocity
b) the mean velocities of atmospheric gas molecules which is greater than that of the earth’s escape velocity
c) the gravitational effect of the moon
d) the earth’s magnetic field
Answer: a
Clarification:The escape velocity of earth is approximately 11.2km/s. This is greater than the root-mean-square velocities of the atmospheric gases. Hence, the gases are unable to escape into the vacuum of space.

7. The escape velocity at the event horizon of a black hole is 3×108 m/s, i.e., the velocity of light. What is the mass of the black hole if the distance from its centre to the event horizon is 18km?
a) 12.15 x 10-30 kg
b) 12.15 x 1030 kg
c) Infinity
d) Cannot be determined
Answer: b
Clarification: Escape velocity (v) = [(2 x G x M)/R] 1/2
M = Mass of the black hole
R = Radius of the black hole
3 x 108 = [(2 x 6.67 x 10-11 x M)/18000]1/2
9 x 1016 = (2 x 6.67 x 10-11 x M)/18000
M = 12.15 x 1030kg.

8. A black hole is called so because it is a hypothetical object.
a) True
b) False
Answer: b
Clarification:A black hole is called so because of its high escape velocity. A black holes escape velocity is equal to the velocity of light at the event horizon, and hence, it absorbs any light hitting it without any reflection which essentially renders the object “black”.

9. The radius of the moon is approximately 3.7 times smaller than the radius of the earth. Assuming that their densities are same, what is the escape velocity of the moon compared to earth?
a) 0.07 times
b) 0.7 times
c) 7 times
d) 3.7 times
Answer:a
Clarification: Escape velocity (v) = [(2*G*M)/R]1/2
Mass of moon = (Density) x (Volume)
= (Density) x (Volume of the earth)/50.65;
[Volume is directly proportional to the cube of the radius]
Vmoon/vearth = [(2*G*Mmoon)/Rmoon1/2/[(2*G*Mearth)/Rearth]1/2
= [1/(50.65 x 3.7)] ½
= 0.07
Therefore; Vmoon = 0.07 (vearth).

10. The escape velocity of a planet depends on the radius of its orbit around the parent star.
a)True
b)False
Answer: b
Clarification:The escape velocity of a planet is independent of the radius of orbit of its parent star. It only depends on the radius of the planet and the mass of the planet.

11. The escape velocity for an object to leave the surface of the earth does not depend on its size, shape, mass or direction of projection of the body.
a) True
b) False
Answer: a
Clarification: The escape velocity for any object on earth is approximately constant at all points on the earth’s surface, i.e., the escape velocity of earth is approximately equal to 11.2 km/s and is not determined by any parameter of the projectile.

12. A planet as a radius “R” and density “P”. The escape velocity of this planet is _____
a) directly proportional to P
b) inversely proportional to P
c) directly proportional to P1/2
d) inversely proportional to P1/2
Answer: c
Clarification: Escape velocity (v) = [(2*G*M)/R] 1/2
Density (P) = Mass (M) x Volume
M = P / [(4/3) x pi x R3] Therefore;
v = [(8 x pi x P x G x R2) / 3]1/2.

13. A particle is kept at a distance R above the earth’s surface. What is the minimum speed with which it should be projected so that it does not return?
a) [(G*M)/(4*R)]1/2
b) [(G*M)/(2*R)]1/2
c) [(G*M)/R] 1/2
d) [(2*G*M)/R]1/2
Answer: c
Clarification:If the object’s initial velocity is “v”, from the law of conservation of energy, we have;
(m x v2)/2 = (G x M x m)/(R+R)
v2 = (2 x G x M)/(2x R)
Therefore; v = [(G x M)/R]1/2.

14. The radius of Jupiter is approximately 11 times larger than the earth. Jupiter has a mass 316 times that of earth. What is the approximate escape velocity of Jupiter compared to earth?
a) 100 times greater
b) 12 times greater
c) 5 times greater
d) Equal to that of the earth
Answer: c
Clarification:M = Mass of the earth
R = Radius of the earth
v = Escape velocity of the earth
We know;
v = [(2 x G x M)/R] 1/2
M’ = Mass of the Jupiter
R’ = Radius of the Jupiter
v’ = Escape velocity of the Jupiter
M’ = 316M
R’ = 11R
v’ = [(2 x G x M’)/R’] ½
= [(2 x G x 316M)/11R] ½
= 5.6 x [(2 x G x M)/R] 1/2
= 5.6 x v.

15. If the earth lost 99% of its mass, the escape velocity would _____
a) increase 10 times
b) decrease 10 times
c) decrease 90 times
d) increase 90 times
Answer: b
Clarification: New mass of the earth (M’) = 0.01 x Original mass of the earth (M)
Therefore;
M’ = M/100
Escape velocity (v) is directly proportional to the square root of the mass
“v” is directly proportional to the square root of “M’”.
Therefore;
New escape velocity (v’) = v/10.

250+ TOP MCQs on Fluids Mechanical Properties – Surface Tension | Class 11 Physics

Physics Multiple Choice Questions on “Fluids Mechanical Properties – Surface Tension”.

1. Surface energy is ________________
a) kinetic energy of the surface molecules
b) the force per unit length acting on surface particles
c) the energy of the molecules inside the beaker
d) the extra energy that the molecules at the surface have relative to molecules inside the liquid

Answer: d
Clarification: Molecules much below the surface are attracted by equal forces in all directions, while molecules on the surface have a net force acting on them as they are exposed to air from one side and water on another. This net force acting on the surface molecules has a corresponding energy called surface energy.

2. The unit of surface tension is same as that of _________
a) surface energy per unit volume
b) force per unit area
c) surface energy per unit area
d) surface energy per unit length

Answer: c
Clarification: Surface tension is defined as force per unit length or surface energy per unit area.
The unit is therefore N/m.
Another unit people use is J/m2 (where J refers to joule, the unit of energy).
But obviously it is equal to Nm/m2 = N/m.

3. What will be the change in surface energy when a drop of liquid (S = 0.08N/m) is divided into 10 equal droplets? Radius of initial single drop (R)= 5cm.
a) 0.1259 J
b) 0.1141 J
c) 0.2356 J
d) 0.1765 J

Answer: b
Clarification: Volume of drop = 4/3πR2.
Let the radius of droplets be ‘r’. Volume of big drop will be equal to volume of 10 droplets.
10*4/3πr3 = 4/3πR3 ⇒ r = 101/3 *R.
Surface area of big drop = 4πR2 & surface area of 10 droplets = 10 * 4πr2 = 40πR2(102/3).
Surface energy of big drop = 4πR2 *S = 4π(0.0025)*0.08 = 0.0025
Surface energy of droplets = 40πR2(102/3) *S = 40π(0.0025)(102/3)*0.08 = 0.1166.
Change in surface energy = 0.1166 – 0.0025 = 0.1141 J.

4. Consider a spherical drop of liquid having radius ‘R’. Assume a diametric cross-section which divides the drop into two hemispheres. The hemispheres pull each other due to surface tension along the circumference of the diametrical cross-section. What is the value of this force?
a) 2πRS
b) πR2 S
c) 2RS
d) 0

Answer: a
Clarification: The diametrical cross section of a sphere will be a circle of radius R. One hemisphere will exert a force of Sdl (dl is a small element of length along periphery) on dl length.
Total length = perimeter of circle = 2&7pi;R. Therefore, total force = 2πRS.

5. What is the value of pressure inside the bubble of radius 1cm, if the bubble is in water at a depth (H) of 1m? Assume atmospheric pressure to be 1atm, surface tension = 0.075N/m & density of water = 1000kg/m3.
a) 110015 Pa
b) 109985 Pa
c) 26 atm
d) 4 atm

Answer: a
Clarification: The pressure in the bubble will be more than the pressure just outside it by an amount 2S/R.
The pressure just outside the bubble is P0 + ρgh.
Therefore, pressure inside bubble = P0 + ρgh + 2S/R
= 105 + (1000*10*1) + (2*0.075/0.01)
= 110015 Pa.

6. A soap bubble contains two surface layers. What is the excess pressure just inside the first layer when seen from outside the bubble?
a) 2S/R
b) 4S/R
c) -2S/R
d) -4S/R

Answer: a
Clarification: Let the pressure outside the bubble be P1, between the two layers be P2 and pressure inside the bubble be P3.
P2 – P1 = 2S/R
P3 – P2 = 2S/R
∴ P3 – P1 = 4S/R
This shows that excess pressure just inside the first layer is 2S/R while excess pressure completely inside the bubble is 4S/R.

7. The contact angle determines whether liquid will rise or get depressed along a solid surface. Select the correct statement regarding the same.
a) If contact angle is found to be less than 90°, the liquid will have raise along the solid surface
b) If contact angle is found to be greater than 90°, then liquid will have depressed along the solid surface
c) Liquid always rises along solid surface, irrespective of contact angle
d) Liquid always gets depressed along solid surface, irrespective of contact angle

Answer: a

8. Consider a capillary tube in which water has risen. The contact angle is θ. The radius of the capillary tube is ‘r’. The surface tension is ‘S’. And the density of water is ‘ρ’. What is the expression for the height of water risen in the tube? Assume the radius of meniscus to be ‘R’. Let the height of water risen in the tube be ‘h’.
a) 2Ssinθ / Rρg
b) 2S / Rρg
c) 2Scosθ / Rρg
d) 2S / rρg

Answer: b

9. If a glass tube is of insufficient length, water rises and overflows.
a) True
b) False

Answer: b
Clarification: If the tube is of insufficient length then the contact angle changes and doesn’t let water overflow. The surface tension force balances the weight of water risen.
2πrcosθ*S = πr2 hρg.
Here, theta will be different from the known value of contact angle for water and glass tube.