Rotational Motion MCQ Quiz - Objective Question with Answer for Rotational Motion - Download Free PDF

Last updated on Jul 7, 2025

Latest Rotational Motion MCQ Objective Questions

Rotational Motion Question 1:

Statement 1: A person jumping freely from an airplane follows a parabolic path when observed from the ground, assuming no air resistance.

Statement 2: The rotation of the Earth causes the trajectory of the person to follow a parabolic path, accounting for the Coriolis effect.

Choose the correct option:

  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
  2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
  3. Statement 1 is true, statement 2 is false.
  4.  Statement 1 is false, statement 2 is true.

Answer (Detailed Solution Below)

Option 2 : Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Rotational Motion Question 1 Detailed Solution

Calculation:

Statement 1 is true: When a person jumps from an airplane, they are subject to the force of gravity and air resistance (neglecting air resistance in this case). The person follows a parabolic path relative to an observer on the ground, as the horizontal velocity remains constant, and the vertical velocity changes due to gravity.

Statement 2 is also true: The Earth's rotation does have an effect on the trajectory of moving objects. The Coriolis effect causes a deflection of the path of the person jumping from the airplane. However, this effect is very small at the altitude of an airplane and does not significantly alter the path's general parabolic shape.

Thus, the correct answer is (B).

Rotational Motion Question 2:

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

  1. \(\rm \frac{1}{2}Mr^2\)
  2. \(\rm \frac{3}{8}Mr^2\)
  3. \(\rm \frac{3}{2}Mr^2\)
  4. 2 Mr2

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{3}{8}Mr^2\)

Rotational Motion Question 2 Detailed Solution

Calculation:

Diameter is given as r.

qImage68637d8754fc443d2e95eb74

∴ Radius = r / 2

Using parallel axis theorem,

Moment of inertia about tangent, Itangent = 1/2 m(r/2)2 + m(r/2)2 = (3 / 2) × m × (r / 2)2 = (3 / 8) × m × r2

Rotational Motion Question 3:

qImage685d278221825cafe37b7757

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad/s2 . Moment of inertia of the wheel is ________ kg m2 .

(Acceleration due to gravity = 10 m/s2 )

Answer (Detailed Solution Below) 1

Rotational Motion Question 3 Detailed Solution

Calculation:

Given:

Force F = 10 N

Radius R = 0.2 m

Angular acceleration α = 2 rad/s2

Torque τ = F × R = 10 × 0.2 = 2 N·m

image-Photoroom (5)

Using the relation: τ = Iα

⇒ I = τ / α = 2 / 2 = 1 kg·m2

Hence, the correct option is (1).

Rotational Motion Question 4:

If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be \(\rm \frac{x}{7}\). The value of x is _______.

Answer (Detailed Solution Below) 5

Rotational Motion Question 4 Detailed Solution

Calculation:

Given the moment of inertia formulas:

I₁ = (2/5) m₁ R² + m₁ R² = m₁ R² (7/5)

qImage683569a139457df626dfcc30

So, I₁ = 7m₁ R²

Similarly, for I₂:

I₂ = (m₂ R² / 4) + m₂ R² = (5/4) m₂ R²

qImage683569a239457df626dfcc35

So, I₂ = 5m₂ R²

The ratio of I₂ to I₁ is:

I₂ / I₁ = (5m₂ R²) / (7m₁ R²) = 5 / 7

Final Answer: x = 5

Rotational Motion Question 5:

ICM is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IAB is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance \(\frac{2}{3}\)R from center. Where R is the radius of the disc. The ratio of IAB and ICM is x ∶ 9. The value of x is ______.

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Answer (Detailed Solution Below) 17

Rotational Motion Question 5 Detailed Solution

Calculation:

qImage682ee9cd1964ccc6b125fc5f

The moment of inertia of disc about center of mass is Icm = (mR2) / 2

The moment of inertia of disc about AB using parallel axis theorem is

IAB = (mR2) / 2 + m × (2R / 3)2 = (17 / 18) mR2

⇒ IAB / Icm = 17 / 9 ⇒ x = 17

Top Rotational Motion MCQ Objective Questions

The work done to increase the velocity of a 1500 kg car from 36 km/h to 72 km/h is

  1. 4.5 × 104 J
  2. 2.25 × 105 J
  3. 7.5 × 104 J
  4. 4.5 × 105 J

Answer (Detailed Solution Below)

Option 2 : 2.25 × 105 J

Rotational Motion Question 6 Detailed Solution

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The correct option is: 2

Concept Used:

  • Work-Energy Theorem: This theorem states that the net work done by the force on a body is equal to the change in its kinetic energy.
  • Mathematically, Work Done (W) = Final Kinetic Energy - Initial Kinetic Energy
  • Kinetic Energy (K.E) is given by the formula: K.E = (1/2) × m × v2
  • So, W = (1/2) × m × (v2 - u2), where:
    • m = mass of the object
    • v = final velocity
    • u = initial velocity

Calculation:

  • Given:
    • Mass of the car (m) = 1500 kg
    • Initial velocity (u) = 36 km/h = 36 × (1000 ÷ 3600) = 10 m/s
    • Final velocity (v) = 72 km/h = 72 × (1000 ÷ 3600) = 20 m/s
  • Using the work-energy theorem:
    W = (1/2) × m × (v2 - u2)
    ⇒ W = (1/2) × 1500 × (202 - 102)
    ⇒ W = (1/2) × 1500 × (400 - 100)
    ⇒ W = (1/2) × 1500 × 300
    ⇒ W = 750 × 300
    ⇒ W = 225000 J
    ⇒ W = 2.25 × 105 J

Additional Information:

  • Work done is a scalar quantity and is expressed in joules (J) in the SI system.
  • This type of question is commonly asked in mechanics under the topic of energy and work.
  • Always convert velocities from km/h to m/s before substituting in the kinetic energy formula: multiply by (1000 ÷ 3600) or simplify as 5/18.

Diagram/Visual Aid Suggestion:

  • A diagram showing a car moving initially at a lower speed and then at a higher speed with vectors labeled u and v.
  • Include a bar graph representing the initial and final kinetic energy for visual comparison.

In rotational motion, Power = Torque x ________.

  1. Angular velocity
  2. Angular momentum
  3. Angular displacement
  4. Angular acceleration

Answer (Detailed Solution Below)

Option 1 : Angular velocity

Rotational Motion Question 7 Detailed Solution

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CONCEPT:

  • Rotational motion: When a block is moving about a fixed axis on a circular path then this type of motion is called rotational motion.

Torque (τ): 

  • It is the twisting force that tends to cause rotation.
  • The point where the object rotates is known as the axis of rotation
  • Mathematically it is written as,

τ = rFsin θ 

EXPLANATION:

  • The power associated with torque is given by the product of torque and angular velocity of the body about an axis of rotation i.e.,

⇒ P = τω 

Where τ = torque and ω = angular velocity 

EXTRA POINTS:

In rotational dynamics

  • Moment of inertia is the analogue of mass
  • Angular velocity is analogue of linear velocity
  • Angular acceleration is analogue of linear acceleration

Thus, in linear motion mass x velocity = momentum

The analogues of the moment of inertia x angular velocity = angular momentum

 

Linear Motion

Rotational Motion

Position

x

θ

Velocity

v

ω

Acceleration

a

α

Motion equations

x = v̅ t

θ = ω̅t

 

v = v0 + at

ω = ω0 + αt

 

\(x = {v_0}t + \frac{1}{2}a{t^2}\)

\(\theta = {ω _0}t + \frac{1}{2}\alpha {t^2}\)

 

\({v^2} = v_0^2 + 2ax\)

\({ω ^2} = ω _0^2 + 2\alpha \theta\)

Mass (linear inertia)

M

I

Newton’s second law

F = ma

T = Iα

Momentum

p = mv

L = Iω

Work

Fd

T.θ

Kinetic energy

\(\frac{1}{2}m{v^2}\)

\(\frac{1}{2}I{ω ^2}\)

Power

Fv

Moment of inertia, rotational kinetic energy, and angular momentum of a body is I, E, and L respectively then :

  1. I = E/L
  2. L = EI
  3. E = 2IL
  4. L = √(2EI)

Answer (Detailed Solution Below)

Option 4 : L = √(2EI)

Rotational Motion Question 8 Detailed Solution

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The correct answer is option 4) i.e. L = √(2EI)

CONCEPT:

  • Angular momentum: The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity.
    • Angular momentum also obeys the law of conservation of momentum i.e. angular momentum before and after is conserved.

Angular momentum, L = I × ω 

Rotational kinetic energy: For a given fixed axis of rotation, the rotational kinetic energy is given by:

 \(KE = \frac{1}{2} Iω^2\)

Where I is the moment of inertia, ω is the angular velocity.

CALCULATION:

Angular momentum, L = I × ω      ----(1) 

Roational kinetic energy = \(E = \frac{1}{2} Iω^2\) ⇒ ω = \(\sqrt{\frac{2E}{I}}\)      ----(2)

Substituting (2) in (1) we get

L = I × \(\sqrt{\frac{2E}{I}}\)√(2EI)

If the moment of inertia of a rotating body is increased then what will be the effect on the angular velocity?

  1. It will increase
  2. It will decrease
  3. There will be no effect
  4. First increase and then decrease

Answer (Detailed Solution Below)

Option 2 : It will decrease

Rotational Motion Question 9 Detailed Solution

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CONCEPT:

  • The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.
  • It is measured as the product of linear momentum and the perpendicular distance of its line of action from the axis of rotation.
  • The relation between the angular momentum and moment of inertia is given by

L = Iω

Where I = moment of inertia, L = angular momentum, and ω = angular velocity.

EXPLANATION:

  • If there is no external torque acting on system then initial angular momentum (Linitial) of system is equal to final momentum (Lfinal).
  • Hence, the angular momentum in a closed system is a conserved.

∴ Iω = constant

⇒ I ∝ 1/ω

i.e. Moment of inertia is inversely proportional to the angular velocity.

  • Hence if the moment of inertia of a rotating body is increased then the angular velocity decreases.

The correct relationship between Moment of Inertia, Torque, and Angular acceleration is ________.

  1. Angular acceleration = Torque x Moment of inertia
  2. Angular acceleration = Torque / Moment of inertia
  3. Moment of inertia = Angular acceleration x Torque
  4. Moment of inertia = Angular acceleration + Torque

Answer (Detailed Solution Below)

Option 2 : Angular acceleration = Torque / Moment of inertia

Rotational Motion Question 10 Detailed Solution

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CONCEPT:

  • Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
  • If Δω is the change in angular velocity time Δt, then average acceleration is

\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)

Moment of Inertia (I):

  • Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.
  • Moment of inertia of a particle is

I = mr2

where r = perpendicular distance of the particle from the rotational axis.

Torque (τ): 

  • It is the twisting force that tends to cause rotation.
  • The point where the object rotates is known as the axis of rotation
  • Mathematically it is written as,

τ = rFsin θ 

EXPLANATION:

The relationship between the angular acceleration (α), torque (τ) and moment of inertia (I) is given by

⇒ τ = α × I

\( \Rightarrow \alpha = \frac{\tau }{I}\)

  • Therefore, angular acceleration = Torque / Moment of inertia. Hence option 2 is correct.

The expression Iω2/2 represents Rotational ____________.

  1. Kinetic energy
  2. Angular momentum
  3. Torque
  4. Power

Answer (Detailed Solution Below)

Option 1 : Kinetic energy

Rotational Motion Question 11 Detailed Solution

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CONCEPT:

  • Rotational kinetic energy: The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
  • A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
  • Mathematically rotational kinetic energy can be written as -

\(KE = \frac{1}{2}I{ω ^2}\)

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

  • From above it is clear that, \(\frac{1}{2}I{ω ^2}\) represents rotational kinetic energy.

quesImage364

Quantity

Expression

Torque

Iα 

Angular momentum

Iω  

Rotational kinetic energy

\(\frac{1}{2}I{ω ^2}\)

Work done

τθ  

Power

τω 

A man standing on a revolving platform spreading his hands outward. Then:

  1. angular velocity increase
  2. angular velocity decrease
  3. angular momentum increase
  4. angular momentum is zero

Answer (Detailed Solution Below)

Option 2 : angular velocity decrease

Rotational Motion Question 12 Detailed Solution

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CONCEPT:

Angular momentum (L):

  • The angular momentum of a rigid body is defined as the product of the moment of inertia and the angular velocity i.e.,

\(⇒ L = Iω \)

Where I = moment of inertia and ω = angular velocity

Law of conservation of angular momentum:

  • When the net external torque acting on a body about a given axis is zero, the total angular momentum of the body about that axis remains constant i.e.,

⇒ I1ω1 = I2ω2

EXPLANATION:

F2 J.K 28.5.20 Pallavi D5

  • When a man standing on a revolving platform and spreading his hands suddenly as he stretches his hand that he is increasing his moment of inertia thereby decreasing his angular velocity using the principle of conservation of angular momentum. Therefore option 2 is correct.

The product of the moment of inertia and the angular acceleration is:

  1. force
  2. torque
  3. work
  4. angular momentum

Answer (Detailed Solution Below)

Option 2 : torque

Rotational Motion Question 13 Detailed Solution

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CONCEPT:

Torque (τ): 

  • It is the twisting force that tends to cause rotation.
  • The point where the object rotates is known as the axis of rotation
  • Mathematically it is written as,

τ = rFsin θ 

  • Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
  • If Δω is the change in angular velocity time Δt, then average acceleration is

\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)

EXPLANATION:

  • Torque is the measure of the amount of force acting on an object that can cause it to rotate.
  • The torque that is needed to produce angular acceleration depends on the mass distribution of the object which is described by the moment of inertia.
  • Therefore, torque (\(\tau \)) is the product of the angular acceleration (a) and the moment of inertia (I) of an object. Therefore option 2 is correct.

           \(\tau = I \times a\)

A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their centre of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true?

  1. The ring has higher kinetic energy
  2. The disc has higher kinetic energy
  3. The ring and the disc have the same kinetic energy
  4. Kinetic energies of both the bodies are zero since they are not in linear motion

Answer (Detailed Solution Below)

Option 1 : The ring has higher kinetic energy

Rotational Motion Question 14 Detailed Solution

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CONCEPT:

Moment of inertia:

  • The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
  • The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

  • Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

Rotational kinetic energy: 

  • The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
  • A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
  • Mathematically rotational kinetic energy can be written as -

⇒ KE \( = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

  • The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by

⇒ Iring = MR2

  • Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{disc}} = \frac{1}{2}M{R^2}\)

  • As we know that mathematically rotational kinetic energy can be written as

\(⇒ KE = \frac{1}{2}I{\omega ^2}\)

  • According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
  • Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
  • So, from the equation, it is clear that,

⇒ Iring > Idisc

∴ Kring > Kdisc

  • The ring has higher kinetic energy.

quesImage483

                      Body    

Axis of Rotation

Moment of inertia

Uniform circular ring of radius R

perpendicular to its plane and through the center

MR2

Uniform circular ring of radius R

diameter

\(\frac{MR^2}{2}\)

Uniform circular disc of radius R perpendicular to its plane and through the center \(\frac{MR^2}{2}\)
Uniform circular disc of radius R diameter \(\frac{MR^2}{4}\)
A hollow cylinder of radius R Axis of cylinder MR2

A wheel has angular acceleration of 3 rad/s2 and an initial angular speed of 2 rad/s. In a time of two second it has rotated through an angle (in radians) of

  1. 10
  2. 12
  3. 4
  4. 6

Answer (Detailed Solution Below)

Option 1 : 10

Rotational Motion Question 15 Detailed Solution

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Concept:

The equations which are applied in linear motion can also be applied in angular motion.

 

Linear Motion

Rotational Motion

 

Position

x

θ

Angular position

Velocity

v

ω

Angular velocity

Acceleration

a

α

Angular acceleration

Motion equations

x = v̅ t

θ = ω̅t

Motion equations

 

v = v0 + at

ω = ω0 + αt

 

 

\(x = {v_0}t + \frac{1}{2}a{t^2}\)

\(θ = {ω _0}t + \frac{1}{2}α {t^2}\)

 

 

\({v^2} = v_0^2 + 2ax\)

\({ω ^2} = ω _0^2 + 2α θ\)

 

Mass (linear inertia)

M

I

Moment of inertia

Newton’s second law

F = ma

τ = Iα

Newton’s second law

Momentum

p = mv

L = Iω

Angular momentum

Work

Fd

τθ

Work

Kinetic energy

\(\frac{1}{2}m{v^2}\)

\(\frac{1}{2}I{ω ^2}\)

Kinetic energy

Power

Fv

τω

Power

Calculation:

Given:

α = 3 rad/s2, ωo = 2 rad/s, t = 2 s

We know that,

\(θ = {ω _0}t + \frac{1}{2}α {t^2}\)

\(θ = ({2}\times 2) + \frac{1}{2}\times3\times2^2\)

θ = 4 + 6 = 10 radians

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