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Circular Motion — NEET Physics MCQs with Solutions
Free NEET Physics Circular Motion MCQs with step-by-step solutions covering Kinematics of Circular Motion, Uniform Circular Motion & Centripetal Acceleration, Centripetal & Centrifugal Force, Motion on Flat Circular Tracks & Friction, Banking of Roads & Well of Death, Conical Pendulum & Rotating Systems. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Kinematics of Circular Motion · easy · theory
In non-uniform circular motion, the total acceleration of a particle is:
A. √(a_c² + a_t²), since the centripetal and tangential accelerations are perpendicular ✓ Correct
B. a_c − a_t
C. Only a_c
D. a_c + a_t
Solution: a_c (radial, v²/r) ⊥ a_t (tangential, αr); they add vectorially: a = √(a_c² + a_t²). Trap: forgetting a_t in non-uniform motion.
Q2 — Kinematics of Circular Motion · easy · numerical
A wheel turns at 240 rpm. Its angular velocity in rad/s is:
A. 8π ≈ 25.1 rad/s ✓ Correct
B. 4π rad/s
C. 2π rad/s
D. 240 rad/s
Solution: ω = 2πN/60 = 2π×240/60 = 8π ≈ 25.1 rad/s.
Q3 — Kinematics of Circular Motion · easy · numerical
The angular speed of the minute hand of a clock is:
A. π/30 rad/s
B. 2π/60 rad/s
C. π/60 rad/s
D. π/1800 rad/s ✓ Correct
Solution: One revolution in 3600 s: ω = 2π/3600 = π/1800 rad/s.
Q4 — Uniform Circular Motion & Centripetal Acceleration · easy · theory
In uniform circular motion, which quantity remains constant?
A. Linear momentum
B. Velocity
C. Speed (and kinetic energy) ✓ Correct
D. Acceleration (as a vector)
Solution: Speed is fixed but the DIRECTIONS of v⃗, a⃗ and p⃗ rotate continuously — none of those vectors is constant.
Q5 — Uniform Circular Motion & Centripetal Acceleration · easy · numerical
A particle moves at 6 m/s on a circle of radius 3 m. Its centripetal acceleration is:
A. 2 m/s²
B. 6 m/s²
C. 12 m/s² ✓ Correct
D. 18 m/s²
Solution: a_c = v²/r = 36/3 = 12 m/s².
Q6 — Uniform Circular Motion & Centripetal Acceleration · easy · numerical
A cyclist rounds a circular track of radius 20 m at 10 m/s. The required centripetal acceleration is:
A. 2 m/s²
B. 10 m/s²
C. 0.5 m/s²
D. 5 m/s² ✓ Correct
Solution: a_c = v²/r = 100/20 = 5 m/s².
Q7 — Centripetal & Centrifugal Force · easy · theory
Centrifugal force is:
A. Always present in the ground frame
B. A pseudo force that appears only in the rotating (non-inertial) frame ✓ Correct
C. A real reaction to the centripetal force
D. The force that keeps a body in a circle
Solution: It exists only for observers in the rotating frame (magnitude mω²r, outward). In the inertial frame only the real centripetal force acts. Trap: it is NOT the Newton's-third-law pair of the centripetal force.
Q8 — Centripetal & Centrifugal Force · easy · numerical
A 0.5 kg stone moves in a horizontal circle of radius 2 m at 6 m/s. The centripetal force is:
A. 36 N
B. 3 N
C. 18 N
D. 9 N ✓ Correct
Solution: F = mv²/r = 0.5×36/2 = 9 N.
Q9 — Centripetal & Centrifugal Force · easy · numerical
A 1200 kg car takes a level turn of radius 60 m at 12 m/s. The centripetal force required is:
A. 1440 N
B. 2880 N ✓ Correct
C. 5760 N
D. 240 N
Solution: F = mv²/r = 1200×144/60 = 2880 N (provided by tyre friction).
Q10 — Motion on Flat Circular Tracks & Friction · easy · theory
On a flat (unbanked) road, the centripetal force for a turning car is provided by:
A. Static friction between the tyres and the road ✓ Correct
B. The car's weight
C. The normal reaction
D. The engine thrust directly
Solution: Friction is the only horizontal force available on level ground; v_max = √(μrg).
Q11 — Motion on Flat Circular Tracks & Friction · easy · numerical
A car turns on a flat road of radius 45 m with μ = 0.5. The maximum safe speed is:
A. 15 m/s ✓ Correct
B. 22.5 m/s
C. 9 m/s
D. 30 m/s
Solution: v = √(μrg) = √(0.5×45×10) = √225 = 15 m/s.
Q12 — Motion on Flat Circular Tracks & Friction · easy · numerical
With μ = 0.8 between tyres and road, the maximum speed on a level turn of radius 20 m is:
A. ≈ 16 m/s
B. ≈ 4 m/s
C. ≈ 12.6 m/s ✓ Correct
D. ≈ 8 m/s
Solution: v = √(μrg) = √(0.8×20×10) = √160 ≈ 12.6 m/s.
Q13 — Banking of Roads & Well of Death · easy · theory
Roads are banked at curves so that:
A. Cars can stop faster
B. A component of the normal reaction provides the centripetal force, reducing reliance on friction ✓ Correct
C. The weight of the car decreases
D. Friction increases
Solution: At the design speed v₀ = √(rg tanθ), N sinθ alone supplies mv²/r — no friction needed.
Q14 — Banking of Roads & Well of Death · easy · numerical
A curve of radius 40 m is banked at 45°. The optimum speed is:
A. 40 m/s
B. 14.1 m/s
C. 20 m/s ✓ Correct
D. 10 m/s
Solution: v₀ = √(rg tan45°) = √400 = 20 m/s.
Q15 — Banking of Roads & Well of Death · easy · numerical
A cyclist rounds a curve of radius 20 m at 10 m/s. The angle he must lean from the VERTICAL is:
A. tan⁻¹(0.2)
B. 30°
C. 45°
D. tan⁻¹(0.5) ≈ 26.6° ✓ Correct
Solution: tanθ = v²/rg = 100/200 = 0.5.
Q16 — Conical Pendulum & Rotating Systems · easy · theory
In a conical pendulum, the centripetal force on the bob is supplied by:
A. Friction
B. The weight of the bob
C. The vertical component of the tension
D. The horizontal component of the string tension (T sinθ) ✓ Correct
Solution: T cosθ = mg balances gravity; T sinθ = mω²r drives the circular motion.
Q17 — Conical Pendulum & Rotating Systems · easy · numerical
A conical pendulum's bob of mass 2 kg circles with the string at 60° to the vertical (cos60° = 0.5). The string tension is:
A. 34.6 N
B. 40 N ✓ Correct
C. 20 N
D. 10 N
Solution: T cosθ = mg ⇒ T = 20/0.5 = 40 N.
Q18 — Conical Pendulum & Rotating Systems · easy · numerical
A conical pendulum bob of mass 0.4 kg hangs at 45° to the vertical while circling. The tension is (√2 ≈ 1.41):
A. 4 N
B. ≈ 2.8 N
C. ≈ 5.7 N ✓ Correct
D. 8 N
Solution: T = mg/cos45° = 4/0.707 ≈ 5.7 N.
Q19 — VCM — Mass on a Light String · easy · theory
For a mass on a light STRING to complete a vertical circle of radius r, its minimum speed at the topmost point is:
A. √(gr) ✓ Correct
B. √(2gr)
C. √(5gr)
D. Zero
Solution: At the top the string can only pull; the critical case has T = 0, gravity alone centripetal: mg = mv²/r ⇒ v = √(gr).
Q20 — VCM — Mass on a Light String · easy · numerical
A stone on a string just loops a vertical circle of radius 1.6 m. Its speed at the top is:
A. 4 m/s ✓ Correct
B. 2 m/s
C. 8 m/s
D. 16 m/s
Solution: v_top = √(gr) = √16 = 4 m/s.
Q21 — VCM — Mass on a Light String · easy · numerical
The minimum speed at the bottom for a bucket of water to be whirled in a vertical circle of radius 0.8 m (without the water spilling at the top) is:
A. 8 m/s
B. √40 ≈ 6.3 m/s ✓ Correct
C. √8 ≈ 2.8 m/s
D. 4 m/s
Solution: Same critical condition as a string: v_b = √(5gr) = √40 ≈ 6.3 m/s.
Q22 — VCM — Rigid Rod & Spherical Shell · easy · theory
For a mass fixed to a light RIGID ROD moving in a vertical circle, the minimum speed at the topmost point is:
A. √(gr)
B. Zero ✓ Correct
C. √(2gr)
D. √(5gr)
Solution: A rod can PUSH as well as pull, so it can support the mass at the top; v_top can be zero.
Q23 — VCM — Rigid Rod & Spherical Shell · easy · numerical
A mass on a rigid rod just completes a vertical circle of radius 2.5 m. Its speed at the lowest point is:
A. 5 m/s
B. ≈ 11.2 m/s
C. √50 m/s
D. 10 m/s ✓ Correct
Solution: v_b = √(4gr) = √100 = 10 m/s.
Q24 — VCM — Rigid Rod & Spherical Shell · easy · numerical
A stone on a rigid rod (r = 0.9 m) needs what minimum bottom speed to just complete a vertical circle?
A. 3 m/s
B. ≈ 6.7 m/s
C. 9 m/s
D. 6 m/s ✓ Correct
Solution: v_b = 2√(gr) = 2×3 = 6 m/s.
Q25 — Variable Angular Acceleration & Non-Uniform Motion · easy · theory
In NON-uniform circular motion, the net acceleration is:
A. Purely tangential
B. Purely radial
C. Zero
D. Neither purely radial nor purely tangential — it has both components ✓ Correct
Solution: Speed changes (a_t ≠ 0) while direction changes (a_c ≠ 0): the total acceleration tilts away from the radius.
Q26 — Variable Angular Acceleration & Non-Uniform Motion · easy · numerical
A wheel's angular speed grows as ω = 4t (rad/s). Its angular acceleration and the angle swept in 2 s are:
A. 4 rad/s² and 8 rad ✓ Correct
B. 8 rad/s² and 8 rad
C. 2 rad/s² and 4 rad
D. 4 rad/s² and 16 rad
Solution: α = dω/dt = 4 rad/s²; θ = ∫ω dt = 2t² = 8 rad at t = 2 s.
Q27 — Variable Angular Acceleration & Non-Uniform Motion · easy · numerical
A wheel accelerates uniformly from 10 rad/s to 30 rad/s in 5 s. Its angular acceleration is:
A. 2 rad/s²
B. 4 rad/s² ✓ Correct
C. 8 rad/s²
D. 6 rad/s²
Solution: α = Δω/Δt = 20/5 = 4 rad/s².
Q28 — Energy & Work in Circular Paths · easy · theory
In UNIFORM circular motion, the kinetic energy of the particle:
A. Stays constant, since speed is constant ✓ Correct
B. Oscillates
C. Is zero
D. Increases steadily
Solution: KE = ½mv² depends only on speed. The centripetal force does no work, so KE cannot change.
Q29 — Energy & Work in Circular Paths · easy · numerical
A 2 kg stone moves in a horizontal circle of radius 2 m at 4 m/s. Its kinetic energy is:
A. 4 J
B. 8 J
C. 16 J ✓ Correct
D. 32 J
Solution: KE = ½×2×16 = 16 J (constant in UCM).
Q30 — Energy & Work in Circular Paths · easy · numerical
A cyclist (total mass 100 kg) rides a circular track at constant 10 m/s. The net work done on him per lap is:
A. Zero ✓ Correct
B. 10⁴ J
C. 2πr × 100 J
D. 5000 J
Solution: Constant speed ⇒ ΔKE = 0 ⇒ W_net = 0 per lap (work–energy theorem).