Dynamics of Circular Motion & Centripetal Force — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Dynamics of Circular Motion & Centripetal Force MCQs with step-by-step solutions (30 questions). Part of Circular Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — Dynamics of Circular Motion & Centripetal Force · medium · theory
Centrifugal force is:
A. Always equal to the weight of the body
B. The reaction to gravity
C. A real force directed towards the centre
D. A pseudo (fictitious) force experienced in a rotating (non-inertial) frame, directed outward ✓ Correct
Solution: Centrifugal force is a pseudo force that appears only in a rotating (non-inertial) frame; it acts radially outward and has no real physical source.
Q3 — Dynamics of Circular Motion & Centripetal Force · medium · theory
The work done by the centripetal force on a particle in uniform circular motion is:
A. Zero ✓ Correct
B. Equal to the kinetic energy
C. Negative
D. Positive and constant
Solution: The centripetal force is always perpendicular to the velocity, so it does no work (W = 0).
Q4 — 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.
Q5 — 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.
Q6 — 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.
Q7 — Dynamics of Circular Motion & Centripetal Force · medium · theory
Which of the following is TRUE about centrifugal force?
A. It has no reaction force and appears only in a rotating frame of reference ✓ Correct
B. It obeys Newton's third law with the centripetal force
C. It is a real force acting on the body
D. It always acts towards the centre
Solution: Centrifugal force is fictitious — it arises only in a non-inertial (rotating) frame and has no reaction pair.
Q8 — Dynamics of Circular Motion & Centripetal Force · medium · theory
Centripetal force is:
A. Always provided only by friction
B. Always provided only by gravity
C. Not a new kind of force; it may be provided by tension, gravity, friction or normal reaction ✓ Correct
D. A fundamental force of nature
Solution: Centripetal force is the name for the net inward force; it can be supplied by any real force such as tension, gravitation, friction or the normal reaction.
Q9 — Dynamics of Circular Motion & Centripetal Force · medium · theory
If the string of a stone being whirled in a horizontal circle suddenly breaks, the stone will:
A. Move radially inward
B. Fly off along the tangent to the circle at that point ✓ Correct
C. Move radially outward
D. Continue in the same circle
Solution: With the centripetal force gone, the stone moves in a straight line along the tangent (Newton's first law).
Q10 — Dynamics of Circular Motion & Centripetal Force · medium · theory
The maximum safe speed of a vehicle on a flat unbanked road of radius r (coefficient of friction μ) is:
A. $\sqrt{\mu r g}$ ✓ Correct
B. $\sqrt{\dfrac{rg}{\mu}}$
C. $\sqrt{\dfrac{\mu g}{r}}$
D. $\mu r g$
Solution: Friction supplies the centripetal force: μmg = mv²/r ⇒ v_max = √(μrg), independent of the mass.
Q11 — 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.
Q12 — 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.
Q13 — 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.
Q14 — 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.
Q15 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
The maximum speed with which a car can take a turn of radius 20 m on a level road (μ = 0.5, g = 10 m/s²) is:
A. 10 m/s ✓ Correct
B. 100 m/s
C. 20 m/s
D. 5 m/s
Solution: v_max = √(μrg) = √(0.5 × 20 × 10) = √100 = 10 m/s.
Q16 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A coin is placed at 0.1 m from the centre of a turntable (μ = 0.4, g = 10 m/s²). The maximum angular speed at which it will not slip is:
A. ≈ 6.3 rad/s ✓ Correct
B. ≈ 2 rad/s
C. ≈ 40 rad/s
D. ≈ 4 rad/s
Solution: For no slipping, mω²r ≤ μmg ⇒ ω_max = √(μg/r) = √(0.4 × 10/0.1) = √40 ≈ 6.3 rad/s.
Q17 — 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.
Q18 — 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.
Q19 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
The maximum speed of a car on a level road of radius 45 m with μ = 0.2 (g = 10 m/s²) is:
A. ≈ 18 m/s
B. ≈ 90 m/s
C. ≈ 9.5 m/s ✓ Correct
D. ≈ 3 m/s
Solution: v_max = √(μrg) = √(0.2 × 45 × 10) = √90 ≈ 9.5 m/s.
Q20 — 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).
Q21 — 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.
Q22 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
If the speed of a body in circular motion is doubled (radius constant), the centripetal force becomes:
A. 2 times
B. Unchanged
C. Half
D. 4 times ✓ Correct
Solution: F = mv²/r ∝ v². Doubling v increases F by a factor of 4.
Q23 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A stone of mass 100 g is tied to a string of length 1 m. If the string can withstand a maximum tension of 40 N, the maximum speed of the stone in a horizontal circle is:
A. 4 m/s
B. 40 m/s
C. 20 m/s ✓ Correct
D. 2 m/s
Solution: T = mv²/r ⇒ v² = Tr/m = 40 × 1/0.1 = 400 ⇒ v = 20 m/s.
Q24 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A coin is placed 0.08 m from the centre of a turntable rotating at 5 rad/s (g = 10 m/s²). The minimum coefficient of friction needed so that it does not slip is:
A. 0.2 ✓ Correct
B. 0.1
C. 0.4
D. 0.5
Solution: μ_min = ω²r/g = 5² × 0.08/10 = 25 × 0.08/10 = 2/10 = 0.2.
Q25 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A body of mass 1 kg moves in a circle of radius 0.5 m making 2 revolutions per second. The centripetal force is (π² ≈ 9.87):
A. ≈ 158 N
B. ≈ 40 N
C. ≈ 79 N ✓ Correct
D. ≈ 20 N
Solution: F = 4π²mn²r = 4π² × 1 × 2² × 0.5 = 8π² ≈ 79 N.
Q26 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A body of mass 0.5 kg moves in a circle of radius 2 m with a period of 2 s. The centripetal force is (π² ≈ 9.87):
A. ≈ 2 N
B. ≈ 4.93 N
C. ≈ 9.87 N ✓ Correct
D. ≈ 19.7 N
Solution: ω = 2π/T = 2π/2 = π rad/s. F = mω²r = 0.5 × π² × 2 = π² ≈ 9.87 N.
Q27 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A car takes a turn of radius 100 m at a speed of 20 m/s (g = 10 m/s²). The minimum coefficient of friction between the tyres and the road required is:
A. 0.1
B. 0.8
C. 0.4 ✓ Correct
D. 0.2
Solution: μ_min = v²/(rg) = 20²/(100 × 10) = 400/1000 = 0.4.
Q28 — 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.
Q29 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A coin is placed 0.25 m from the centre of a turntable (μ = 0.5, g = 10 m/s²). The maximum angular speed before it slips is:
A. ≈ 8.9 rad/s
B. ≈ 20 rad/s
C. ≈ 4.47 rad/s ✓ Correct
D. ≈ 2 rad/s
Solution: ω_max = √(μg/r) = √(0.5 × 10/0.25) = √20 ≈ 4.47 rad/s.
Q30 — Dynamics of Circular Motion & Centripetal Force · medium · numerical
A 2 kg body moves in a horizontal circle of radius 0.5 m at 3 revolutions per second. The centripetal force is (π² ≈ 9.87):
A. ≈ 118 N
B. ≈ 710 N
C. ≈ 355 N ✓ Correct
D. ≈ 36 N
Solution: ω = 2πn = 6π rad/s. F = mω²r = 2 × (6π)² × 0.5 = 2 × 36π² × 0.5 = 36π² ≈ 355 N.