Circular motion MCQ Quiz - Objective Question with Answer for Circular motion - Download Free PDF

Last updated on May 28, 2025

Latest Circular motion MCQ Objective Questions

Circular motion Question 1:

Consider a disc rotating in the horizontal plane with constant angular speed ω about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y–z plane and is the same for both pebbles with respect to the disc.

Assume that:

  1. They land back on the disc before the disc has completed 1/8 rotation.
  2. Their range is less than half the disc radius.
  3. ω is constant throughout.

Then:

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  1. P lands in the shaded region and Q in the unshaded region.
  2. P lands in the unshaded region and Q in the shaded region.
  3. Both P and Q land in the unshaded region
  4. Both P and Q land in the shaded region

Answer (Detailed Solution Below)

Option 3 : Both P and Q land in the unshaded region

Circular motion Question 1 Detailed Solution

Solution:

Let the radius of the disc be r₀ and the time period of rotation be T = 2π / ω.

Let the pebble Q be at the point (0, y₀) at the time of projection. For projectile motion, let r be the range and t be the time of flight. It is given that t < T / 8 and r < r₀ / 2.

The rotation of the disc provides additional initial velocity to the pebbles given by:

ΔūQ = −y₀ω î and ΔūP = r₀ω î

These additional velocities are perpendicular to the plane of projection. The time of flight t is not affected by these velocities. However, they introduce additional displacements in the x-direction:

Δr̄Q = −y₀ωt î and Δr̄P = r₀ωt î

So the new landing points Q₁ and P₁ shift by these respective distances.

From triangle OQ₁Q′₁:

θ = tan⁻¹ (y₀ωt / (y₀ + r)) ≤ tan⁻¹ (ωt) ≤ ωt

So the angle θ is less than the angle through which the disc rotates in time t. Hence, landing point Q′₁ lies in the unshaded region.

qImage681c84394e5bbe68330a1ffd

For pebble P, the horizontal displacement is r₀ωt, and the subtended angle is also ωt. But from the geometry of the disc, the angle of point P′₁ is greater than ωt. Therefore, P′₁ also lies in the unshaded region.

Answer: (C) Both P and Q land in the unshaded region.

Circular motion Question 2:

A car is performing circular motion with angular velocity 2π rad/s. Its time period will be :

  1. 0.5 sec
  2. π sec
  3. 0.5 π sec
  4. 1 sec

Answer (Detailed Solution Below)

Option 4 : 1 sec

Circular motion Question 2 Detailed Solution

CONCEPT:

Time Period in Circular Motion

  • The time period (T) of a circular motion is the time taken to complete one full revolution.
  • It is related to the angular velocity (ω) by the formula:

    T = /ω

EXPLANATION:

  • Angular velocity (ω) = 2π rad/s
  • Using the formula:
    • T = /ω
    • T = /
    • T = 1 second

Therefore, the time period (T) of the car's circular motion is 1 second.

Circular motion Question 3:

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while the angular position of point Q is α  with respect to point P. Which of the following graphs represents the correct relation between A and α when the ball goes from Q to R?

F9 Madhuri Engineering 29.09.2022 D1

  1. F9 Madhuri Engineering 29.09.2022 D2
  2. F9 Madhuri Engineering 29.09.2022 D3
  3. F9 Madhuri Engineering 29.09.2022 D4
  4. F9 Madhuri Engineering 29.09.2022 D5
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : F9 Madhuri Engineering 29.09.2022 D4

Circular motion Question 3 Detailed Solution

CONCEPT:

  • Centripetal acceleration is defined as the ratio of the square of the velocity and radius and it is written as,

           ac = \(\frac{v^2}{R}\) 

           here we have ac as the centripetal acceleration, v is the velocity and R is the radius.

CALCULATION:

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure below,

F9 Madhuri Engineering 29.09.2022 D6
Using the trigonometry identity to find h we have;

\(sin\alpha = \frac{h}{R}\)

⇒ h = R sin\(\alpha\)

and the first law of motion we have;

v2 - u2 = 2gh

Here is the initial velocity, u = 0 m/s we have;

v2 = 2gh

⇒ v = \(\sqrt{2gh}\) 

⇒ v = \(\sqrt{2gRsin\alpha}\) -----(1)

The centripetal acceleration is written as;

ac = \(\frac{v^2}{R}\) 

Now, on putting the value of equation (1) above we have,

ac = \(\frac{2gRsin\alpha}{R}\) 

⇒ ac = 2g sin\(\alpha\) -----(2)

At point Q we can see that normal N and force F which mg sin\(\alpha\) we have;

N -  mg sin α  = mac

⇒ N = mac + mg sinα ----(3)

Now, on putting the value of equation (2) in equation (3) we have;

N = m × 2 g sin α + mg sin α 

N = 3 × mg sin α 

The ratio of the centripetal force and normal reaction is written as;

\(\frac{ma_c}{N} = \frac{ 2mgsin\alpha}{3mg sin\alpha}\)

⇒ \(\frac{ma_c}{N} = \frac{2}{3}\) = constant

Therefore, the ratio of the centripetal force and normal reaction is constant.

Hence, option (3) is the correct answer.

Circular motion Question 4:

A car is negotiating a curved road of radius R. The road is banked at an angle θ. the coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:

  1. \(\sqrt{g R^{2} \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)
  2. \(\sqrt{g R \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)
  3. \(\sqrt{\frac{g}{R} \frac{\mu_{s}+\tan \theta}{1-\mu_{2} \tan \theta}}\)
  4. \(\sqrt{\frac{g}{R^{2}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}\)

Answer (Detailed Solution Below)

Option 2 : \(\sqrt{g R \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)

Circular motion Question 4 Detailed Solution

Calculation

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From the above diagram, we have

N = mg cosθ + (mv² / R) sinθ                        (1)

fmax = μN ⇒ fmax = μs mg cosθ + (μs mv² / R) sinθ

mg sinθ + fmax = (mv² / R) cosθ                     (2)

Putting the value

mg sinθ + μs mg cosθ + (μs mv² / R) sinθ = (mv² / R) cosθ

g sinθ + μs g cosθ = (v² / R)(cosθ − μs sinθ)

gR [ (tanθ + μs) / (1 − μs tanθ) ] = v²

v = √[ gR (tanθ + μs) / (1 − μs tanθ) ]

Circular motion Question 5:

A bead of mass ‘m’ slides without friction on the wall of a vertical circular hoop of radius ‘R’ as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ‘R’. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ‘R’, would be (spring constant is ‘k’, g is acceleration due to gravity) 

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  1. \(2 \sqrt{\mathrm{gR}+\frac{\mathrm{kR}^{2}}{\mathrm{~m}}}\)
  2. \(\sqrt{2 \mathrm{Rg}+\frac{4 \mathrm{kR}^{2}}{\mathrm{~m}}}\)
  3. \(\sqrt{2 \mathrm{Rg}+\frac{\mathrm{kR}^{2}}{\mathrm{~m}}}\)
  4. \(\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^{2}}{\mathrm{~m}}}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^{2}}{\mathrm{~m}}}\)

Circular motion Question 5 Detailed Solution

Calculation:

qImage67b717eb817b41a3ece8e43a

Work energy theorem 

\(\mathrm{Mg}(\mathrm{R}+\mathrm{R} \cos 60)+\frac{1}{2} \mathrm{k}\left(\mathrm{R}^{2}-0^{2}\right)=\frac{1}{2} \mathrm{mv}^{2}\)

\(\mathrm{Mg} \frac{3 \mathrm{R}}{2}+\frac{\mathrm{KR}^{2}}{2}=\frac{1}{2} \mathrm{mv}^{2}\)

\(V=\sqrt{3 g R+\frac{K R^{2}}{m}}\)

Top Circular motion MCQ Objective Questions

A body of mass 'm' is moving uniformly along a circle having radius 'r'. The centripetal force on the body is:

  1. \(\dfrac{mv^2}{r}\)
  2. \(\dfrac{mv^2}{r^2}\)
  3. \(\dfrac{mv}{r^2}\)
  4. \(\dfrac{mv}{r}\)

Answer (Detailed Solution Below)

Option 1 : \(\dfrac{mv^2}{r}\)

Circular motion Question 6 Detailed Solution

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Explanation:

Centripetal Force: It is a force required to move a body uniformly in a circular motion. This force acts along the radius and towards the center of the circle.

  • When a body moves in a circle, its direction of motion at any instant is along the tangent of a circle. But according to Newton’s first law of motion, A body cannot change its direction of itself an external force is required for this purpose. This external force is the centripetal force

F1 J.S 6.6.20 Pallavi D1

\({\bf{Centripetal}}\;{\bf{Force}}\;\left( {\bf{F}} \right) = \frac{{m{v^2}}}{r}\;\left[ {{\rm{m}} = {\rm{mass}},{\rm{\;v}} = {\rm{velocity}},{\rm{\;r}} = {\rm{radius}}} \right]\)

F1 J.S 6.6.20 Pallavi D2

  • The centripetal force required for circular motion along the surface of the road, towards the center of the turn. The Static friction between tire and road provides the necessary centripetal force.

A car travels in a circular motion. Which of the following statements is true?

  1. Distance travelled by the car will always be less than displacement of the car.
  2. Distance travelled by the car will always be more than displacement of the car.
  3. Distance travelled by the car will always be equal to displacement of the car.
  4. Distance travelled by the car will be zero when displacement is zero.

Answer (Detailed Solution Below)

Option 2 : Distance travelled by the car will always be more than displacement of the car.

Circular motion Question 7 Detailed Solution

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Explanation:

Distance: 

  • It is defined as the length of the path traveled by a body. 
  • It is a scalar quantity.
  • It cannot be in negative value.
  • It is non-zero in circular motion

​Displacement

  • It is the shortest distance between the initial and final position of the particle.
  • It is a vector quantity.
  • It can be positive, negative and zero.
  • It is zero in circular motion.

F1 Pritesh.K 21-01-21 Savita D2

When an object moves in a straight line without changing the direction the distance and displacement will be equal in magnitude.

  • When an object changes its direction during the motion its path length will become more compared to the distance between the initial and the final position, so in this case, the magnitude of distance becomes more than the displacement.
  • So the distance will always be more than or equal to displacement.

In a circular motion

  1. direction changes
  2. velocity changes
  3. centripetal force acts
  4. all of the above

Answer (Detailed Solution Below)

Option 4 : all of the above

Circular motion Question 8 Detailed Solution

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The correct answer is option 4) i.e. all of the above

CONCEPT:

  • Circular motion: The motion of an object along a circular path is called as circular motion.
  • Centripetal force: It is a net force acting on an object undergoing a circular path of motion, such that the force acts in a direction towards the centre of curvature.

F1 J.K 2.6.20 Pallavi D1

EXPLANATION:

  • A circle is assumed to be a polygon with infinitely many sides such that each side approximates to a point.
  • So, an object moving on a circular path undergoes a change in direction at every point. 
  • Since direction changes at every pointvelocity changes at every point.
  • Also, a centripetal force always acts on an object under circular motion since it is the force that makes a body follow a curved path.

For an object under uniform circular motion, which of the following remains constant?

  1. Kinetic energy
  2. Speed
  3. Magnitude of acceleration
  4. All of the above

Answer (Detailed Solution Below)

Option 4 : All of the above

Circular motion Question 9 Detailed Solution

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The correct answer is option 4) i.e. All of the above

CONCEPT:

  • Uniform circular motion is where a moving object traces a circular path with constant speed.
    • A circle is assumed to be a polygon with infinitely many sides such that each side approximates to a point.
    • So, the object moving on a circular path undergoes a change in direction at every point.
    • Since direction changes and speed remains constantvelocity is varying.
  • Kinetic energy is the measure of energy possessed by a moving object.

It is given by the equation KE = \(\frac{1}{2}mv^2\)

F1 J.K Madhu 19.05.20 D1

EXPLANATION:

Given object is under uniform circular motion.

  • An object under uniform circular motion has a constant speed.
  • In a uniform circular motion, the magnitude of velocity is uniform but the direction keeps changing at every point. So, the magnitude of acceleration is constant.
  • Kinetic energy is proportional to the magnitude of the velocity of a moving object ⇒ KE ∝ v2. Hence, kinetic energy remains constant in a uniform circular motion.

Confusion Points

  • Only the magnitude of velocity is considered for kinetic energy. 
  • Kinetic energy is a scalar quantity. 

In a uniform circular motion :

  1. both acceleration and speed changes
  2. both acceleration and speed are constant
  3. both acceleration and velocity are constant
  4. both acceleration and velocity changes

Answer (Detailed Solution Below)

Option 4 : both acceleration and velocity changes

Circular motion Question 10 Detailed Solution

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The correct answer is option 4) i.e. both acceleration and velocity changes.

CONCEPT:

  • Uniform motion is the type of motion where a moving object traces equal distances in equal intervals of time.
    • Since the distance and time intervals are the same, speed is constant in uniform motion.
  • Uniform circular motion is where a moving object traces a circular path with constant speed.
    • A circle is assumed to be a polygon with infinitely many sides such that each side approximates to a point.
    • So, if the object moving on a circular path undergoes a change in direction at every point.
    • Since direction changes and speed remains constant, velocity is varying.

F1 J.K Madhu 19.05.20 D1

EXPLANATION:

  • Velocity is changing in a uniform circular motion as the direction of the object keeps changing at every point.
  • Acceleration is the rate of change of velocity. Since velocity keeps changing at every instant, acceleration also changes.

Therefore, both acceleration and velocity changes in a uniform circular motion.

Which force is required to maintain a body in Uniform circular Motion?

  1. The Centripetal
  2. The force of Gravitation
  3. Muscular force
  4. Force of Friction

Answer (Detailed Solution Below)

Option 1 : The Centripetal

Circular motion Question 11 Detailed Solution

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Concept:

Circular Motion: The movement of an object along a circumference of a circle or rotation along a circular path is called circular motion.

  • Uniform circular motion: The circular motion in which the speed of the particle remains constant is called uniform circular motion. In a uniform circular motionforce supplies the centripetal acceleration.
    • The kinetic energy and the speed of the body remain constant.

 

Centripetal Force: 

It is a force required to move a body uniformly in a circle. This force acts along the radius and towards the centre of the circle.

\({\bf{Centripetal}}\;{\bf{Force}}\;\left( {\bf{F}} \right) = \frac{{m{v^2}}}{r}\)

F1 J.S 29.5.20 Pallavi D1

  • The centripetal force required for circular motion along the surface of the road, towards the centre of the turn. 

 

Explanation:

  • The Centripetal force is required to maintain a body in Uniform Circular Motion. So this is correct.
  • Centripetal Force acts when a body moves around the other body, the force that causes this acceleration and keeps the body moving along a circular path is acting towards the center

 

Which of the following is an example of uniform circular motion?

  1. The movement of a seconds hand of a watch
  2. The movement of a car on highway curves
  3. The movement of clothes in a dryer of a washing machine
  4. The movement of passenger cabin on a giant wheel

Answer (Detailed Solution Below)

Option 1 : The movement of a seconds hand of a watch

Circular motion Question 12 Detailed Solution

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CONCEPT:

  • Circular Motion: Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path.
    •  The force acts continuously at right angles to the velocity of the particle.
  • Uniform circular motion: The circular motion in which the speed of the particle remains constant is called uniform circular motion. In a uniform circular motionforce supplies the centripetal acceleration.
  • In a uniform circular motion, a body is experiencing a force in the inward direction.
  • For example: the motion of the hands of a clock, motion of the moon revolving around the earth.

F1 J.K Madhu 19.05.20 D1

EXPLANATION:

  • From the above, it is clear that the movement of the second hand of a watch is an example of uniform circular motion. Therefore option 1 is correct.
  • The motion of a giant wheel is a rotatory motion because all the objects within the wheel move in a circular manner in the same direction around a fixed axis. Therefore option 4 is incorrect.

In case of Uniform Circular Motion, the work done by the centripetal force is ________. 

  1. Equal to the displacement of the body 
  2. Equal to the force applied on the body 
  3. Zero 
  4. Equal to the impulse created on the body 

Answer (Detailed Solution Below)

Option 3 : Zero 

Circular motion Question 13 Detailed Solution

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CONCEPT:

  • Circular Motion:  Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path.
  •  The force acts continuously at right angles to the velocity of the particle.
  • Uniform circular motion: The circular motion in which the speed of the particle remains constant is called uniform circular motion.
  • In a uniform circular motionforce supplies the centripetal acceleration.

ac = v2/r, where ac is centripetal acceleration, v is velocity, r is the radius.

  • The speed and kinetic energy of the particle remains constant.

F1 J.K Madhu 19.05.20 D1

EXPLANATION:

  • If the Force F acts on a body and it gets displaced by a distance of S, the work done in that case is given by:

⇒ W = FS Cos θ

  • In the case of uniform circular motion, Cos 90= 0 and that is why W = 0 (The force and displacement are perpendicular to each other).  So option 3 is correct.

If the radii of circular track of two particles are in the ratio of 4:9 then, in order to have same centripetal acceleration, their speeds should be in the ratio of

  1. 4:9
  2. 2:3
  3. 3:2
  4. 9:4

Answer (Detailed Solution Below)

Option 2 : 2:3

Circular motion Question 14 Detailed Solution

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Correct option-2

Concept:

Circular Motion

  • An object is said to perform a circular motion when it moves on a circular path
  • While describing circular motion we often use the term circular motion of point object or particle.
  • This is because we learn circular motion with the brach of the translation motion of a body on a circular path and disregard any rotation.
  • Therefore, we represent the body as a particle.

Types of Circular Motion
Circular motion is classified into two categories
1.Uniform circular motion:

  • If a particle moves with a constant speed in a circle, then the motion is called uniform circular motion. 
  • In uniform-circular motion, particle possesses only centripetal acceleration (ar).

F1 Savita Defence 25-10-22 D2

2.Non-uniform circular motion:

  • If the particle moves with variable speed then the motion is called non-uniform motion.
  • In uniform-circular motion, particle possesses two kinds of acceleration.

F1 Savita Defence 25-10-22 D3

  • (i) Centripetal acceleration (ar&
  • (ii) Tangential acceleration (at)

 

Calculation:

Given:-

\(\frac{r_{1}}{r_{2}}=\frac{4}{9}\) &

\(a_{r_{1}}=a_{r_{2}}\)

In a circular motion, the centripetal acceleration ar is given by:-

\(a_{r}=\frac{v^{2}}{r}\)

where v is linear speed and r is the radius.

As \(a_{r_{1}}=a_{r_{2}}\)

\(\Rightarrow \frac{v_{1}^{2}}{r_{1}}=\frac{v_{2}^{2}}{r_{2}}\)

\(\Rightarrow \frac{v_{1}}{v_{2}}=\sqrt{\frac{r_{1}}{r_{2}}}\)

\(\Rightarrow \frac{v_{1}}{v_{2}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)

Hence, option-2 is correct

For a body moving in a horizontal circular path with uniform speed, which of the following remains unchanged?

  1. velocity
  2. acceleration
  3. direction of velocity
  4. kinetic energy

Answer (Detailed Solution Below)

Option 4 : kinetic energy

Circular motion Question 15 Detailed Solution

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The correct answer is option 4) i.e. kinetic energy.

CONCEPT:

  • Uniform circular motion is where a moving object traces a circular path with constant speed.
    • A circle is assumed to be a polygon with infinitely many sides such that each side approximates to a point.
    • So, the object moving on a circular path undergoes a change in direction at every point.
    • Since direction changes and speed remains constantvelocity is varying.
  • Kinetic energy is the measure of energy possessed by a moving object.

It is given by the equation KE\(\frac{1}{2}mv^2\)

F1 J.K Madhu 19.05.20 D1

EXPLANATION:

Given that the object is moving in a circular path with uniform motion. Thus, the object is under uniform circular motion.

  • Velocity is changing in a uniform circular motion as the direction of the object keeps changing at every point.
  • Acceleration is the rate of change of velocity. Since velocity keeps changing at every instant, acceleration also changes.
  • In a uniform circular motion, the magnitude of velocity is uniform but the direction keeps changing at every point. Kinetic energy is proportional to the magnitude of the velocity of a moving object.

Hence, kinetic energy remains unchanged in a uniform circular motion.

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