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Circular Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Circular Motion MCQs with step-by-step solutions covering Kinematics of Circular Motion, Dynamics of Circular Motion & Centripetal Force, Banking of Roads & Death Well, Vertical Circular Motion, Conical Pendulum. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Kinematics of Circular Motion · easy · theory
The SI unit of angular displacement is the:
A. Revolution
B. Metre
C. Degree
D. Radian ✓ Correct
Solution: Angular displacement is measured in radians (rad) in SI units; 1 revolution = 2π rad.
Q2 — Kinematics of Circular Motion · easy · theory
Angular velocity (ω) is defined as the rate of change of angular displacement, and its SI unit is:
A. rad/s²
B. rev/min
C. m/s
D. rad/s ✓ Correct
Solution: ω = dθ/dt, so its SI unit is radian per second (rad/s).
Q3 — Kinematics of Circular Motion · easy · theory
The linear (tangential) speed v of a particle in circular motion of radius r is related to its angular speed ω by:
A. v = ω/r
B. v = r/ω
C. v = ωr ✓ Correct
D. v = ω²r
Solution: The linear speed is v = ωr.
Q4 — Kinematics of Circular Motion · easy · theory
In uniform circular motion (UCM):
A. Both speed and velocity are constant
B. The speed changes but the velocity is constant
C. The speed is constant but the velocity changes continuously ✓ Correct
D. Both speed and velocity change
Solution: In UCM the magnitude of velocity (speed) is constant, but its direction changes continuously, so velocity is not constant.
Q5 — Kinematics of Circular Motion · easy · theory
The centripetal acceleration of a particle in circular motion is always directed:
A. Along the tangent
B. Away from the centre
C. Towards the centre of the circle ✓ Correct
D. Along the axis of rotation
Solution: Centripetal acceleration points radially inward, towards the centre of the circular path.
Q6 — Kinematics of Circular Motion · easy · theory
In uniform circular motion, the angular acceleration of the particle is:
A. Constant and non-zero
B. Equal to ω²r
C. Directed towards the centre
D. Zero ✓ Correct
Solution: Uniform circular motion means constant angular speed, so angular acceleration α = 0.
Q7 — Kinematics of Circular Motion · easy · theory
The number of radians in one complete revolution is:
A. π/2
B. π
C. 4π
D. 2π ✓ Correct
Solution: One complete revolution corresponds to an angular displacement of 2π radians (360°).
Q8 — Kinematics of Circular Motion · easy · numerical
A wheel rotates at 300 rpm. Its angular velocity in rad/s is (take π = 3.14):
A. 3.14 rad/s
B. 18.8 rad/s
C. 31.4 rad/s ✓ Correct
D. 5 rad/s
Solution: ω = 2πN/60 = 2π × 300/60 = 10π = 31.4 rad/s.
Q9 — Kinematics of Circular Motion · easy · numerical
The angular velocity corresponding to 60 rpm is:
A. 2π rad/s ✓ Correct
B. π rad/s
C. 60 rad/s
D. 4π rad/s
Solution: ω = 2πN/60 = 2π × 60/60 = 2π rad/s.
Q10 — Kinematics of Circular Motion · easy · numerical
A particle moves in a circle of radius 0.5 m with an angular velocity of 2 rad/s. Its linear speed is:
A. 4 m/s
B. 2.5 m/s
C. 0.25 m/s
D. 1 m/s ✓ Correct
Solution: v = ωr = 2 × 0.5 = 1 m/s.
Q11 — Kinematics of Circular Motion · easy · numerical
A car moves on a circular track of radius 5 m at a speed of 10 m/s. Its centripetal acceleration is:
A. 2 m/s²
B. 10 m/s²
C. 20 m/s² ✓ Correct
D. 50 m/s²
Solution: a_c = v²/r = 10²/5 = 100/5 = 20 m/s².
Q12 — Kinematics of Circular Motion · easy · numerical
A body moves in a circle of radius 2 m with angular velocity 4 rad/s. Its centripetal acceleration is:
A. 32 m/s² ✓ Correct
B. 8 m/s²
C. 64 m/s²
D. 16 m/s²
Solution: a_c = ω²r = 4² × 2 = 16 × 2 = 32 m/s².
Q13 — Kinematics of Circular Motion · easy · numerical
The period of revolution of a particle whose angular velocity is π rad/s is:
A. 1 s
B. 2 s ✓ Correct
C. π s
D. 0.5 s
Solution: T = 2π/ω = 2π/π = 2 s.
Q14 — Kinematics of Circular Motion · easy · numerical
A wheel starting from rest attains an angular velocity of 100 rad/s in 5 s. Its angular acceleration is:
A. 20 rad/s² ✓ Correct
B. 500 rad/s²
C. 10 rad/s²
D. 25 rad/s²
Solution: α = (ω − ω₀)/t = (100 − 0)/5 = 20 rad/s².
Q15 — Kinematics of Circular Motion · easy · numerical
A rotating body has an initial angular velocity of 10 rad/s and an angular acceleration of 2 rad/s². Its angular velocity after 5 s is:
A. 12 rad/s
B. 30 rad/s
C. 15 rad/s
D. 20 rad/s ✓ Correct
Solution: ω = ω₀ + αt = 10 + 2 × 5 = 20 rad/s.
Q16 — Kinematics of Circular Motion · easy · numerical
A particle moving in a circle of radius 2 m has an angular acceleration of 5 rad/s². Its tangential acceleration is:
A. 10 m/s² ✓ Correct
B. 2.5 m/s²
C. 20 m/s²
D. 7 m/s²
Solution: a_t = αr = 5 × 2 = 10 m/s².
Q17 — Kinematics of Circular Motion · easy · numerical
A disc rotates at 5 revolutions per second. Its angular velocity is (π = 3.14):
A. 5 rad/s
B. 15.7 rad/s
C. 10 rad/s
D. 31.4 rad/s ✓ Correct
Solution: ω = 2πn = 2π × 5 = 10π = 31.4 rad/s.
Q18 — Dynamics of Circular Motion & Centripetal Force · easy · theory
The centripetal force acting on a body in uniform circular motion is directed:
A. Towards the centre of the circle ✓ Correct
B. Along the tangent
C. Along the axis
D. Away from the centre
Solution: The net (centripetal) force required for circular motion always points towards the centre.
Q19 — Dynamics of Circular Motion & Centripetal Force · easy · theory
The magnitude of the centripetal force on a body of mass m moving with speed v (angular speed ω) in a circle of radius r is:
A. $\dfrac{mv}{r}$
B. $\dfrac{mr}{v^2}$
C. $\dfrac{mv^2}{r} = m\omega^2 r$ ✓ Correct
D. $m\omega r^2$
Solution: Centripetal force F = mv²/r = mω²r.
Q20 — Dynamics of Circular Motion & Centripetal Force · easy · theory
When a car takes a turn on a flat (unbanked) road, the centripetal force is provided by:
A. The normal reaction
B. The engine thrust
C. The weight of the car
D. The force of friction between the tyres and the road ✓ Correct
Solution: On a level road, static friction between the tyres and the road supplies the centripetal force.
Q21 — Dynamics of Circular Motion & Centripetal Force · easy · theory
A coin placed on a rotating turntable does not slip because:
A. The centrifugal force holds it in place
B. The normal reaction provides the centripetal force
C. The force of friction provides the necessary centripetal force ✓ Correct
D. Gravity provides the centripetal force
Solution: Static friction between the coin and the turntable provides the centripetal force mω²r that keeps the coin moving in a circle.
Q22 — Dynamics of Circular Motion & Centripetal Force · easy · theory
A passenger sitting in a car taking a sharp turn feels pushed outward. This apparent outward push is due to:
A. A real outward force from the road
B. The centripetal force
C. The (pseudo) centrifugal force in the car's rotating frame ✓ Correct
D. Increased gravity
Solution: In the non-inertial frame of the turning car, the passenger experiences a fictitious centrifugal force directed outward.
Q23 — Dynamics of Circular Motion & Centripetal Force · easy · theory
A stone tied to a string is whirled in a horizontal circle. The centripetal force on the stone is provided by:
A. The tension in the string ✓ Correct
B. Gravity
C. Friction
D. The normal reaction
Solution: The tension in the string acts along it towards the centre, providing the centripetal force.
Q24 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A body of mass 2 kg moves in a circle of radius 8 m with a speed of 4 m/s. The centripetal force acting on it is:
A. 8 N
B. 2 N
C. 4 N ✓ Correct
D. 16 N
Solution: F = mv²/r = 2 × 4²/8 = 2 × 16/8 = 4 N.
Q25 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A body of mass 0.5 kg moves in a circle of radius 2 m with angular velocity 4 rad/s. The centripetal force is:
A. 16 N ✓ Correct
B. 32 N
C. 8 N
D. 4 N
Solution: F = mω²r = 0.5 × 4² × 2 = 0.5 × 16 × 2 = 16 N.
Q26 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A 1000 kg car takes a turn of radius 50 m at 10 m/s. The centripetal force required is:
A. 1000 N
B. 5000 N
C. 200 N
D. 2000 N ✓ Correct
Solution: F = mv²/r = 1000 × 10²/50 = 1000 × 100/50 = 2000 N.
Q27 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A stone of mass 0.2 kg tied to a string is whirled in a horizontal circle of radius 1 m at 5 m/s. The tension in the string is:
A. 10 N
B. 5 N ✓ Correct
C. 25 N
D. 1 N
Solution: T = mv²/r = 0.2 × 5²/1 = 0.2 × 25 = 5 N.
Q28 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
The centrifugal force experienced by a body of mass 1 kg moving in a circle of radius 3 m at 6 m/s (in the rotating frame) has magnitude:
A. 12 N ✓ Correct
B. 6 N
C. 2 N
D. 18 N
Solution: Magnitude = mv²/r = 1 × 6²/3 = 36/3 = 12 N (directed outward in the rotating frame).
Q29 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A body of mass 2 kg is rotated in a circle of radius 0.5 m with angular velocity 10 rad/s. The centripetal force is:
A. 100 N ✓ Correct
B. 200 N
C. 10 N
D. 50 N
Solution: F = mrω² = 2 × 0.5 × 10² = 2 × 0.5 × 100 = 100 N.
Q30 — Dynamics of Circular Motion & Centripetal Force · easy · numerical
A body of mass 5 kg moving in a circle has a centripetal acceleration of 4 m/s². The centripetal force acting on it is:
A. 9 N
B. 1.25 N
C. 40 N
D. 20 N ✓ Correct
Solution: F = ma_c = 5 × 4 = 20 N.