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Circular Motion — IISER Physics MCQs with Solutions
Free IISER 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. Only a_c
C. a_c − a_t
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. 240 rad/s
B. 4π rad/s
C. 2π rad/s
D. 8π ≈ 25.1 rad/s ✓ Correct
Solution: ω = 2πN/60 = 2π×240/60 = 8π ≈ 25.1 rad/s.
Q3 — Uniform Circular Motion & Centripetal Acceleration · easy · theory
In uniform circular motion, which quantity remains constant?
A. Linear momentum
B. Acceleration (as a vector)
C. Velocity
D. Speed (and kinetic energy) ✓ Correct
Solution: Speed is fixed but the DIRECTIONS of v⃗, a⃗ and p⃗ rotate continuously — none of those vectors is constant.
Q4 — 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. 18 m/s²
B. 2 m/s²
C. 6 m/s²
D. 12 m/s² ✓ Correct
Solution: a_c = v²/r = 36/3 = 12 m/s².
Q5 — Centripetal & Centrifugal Force · easy · theory
Centrifugal force is:
A. The force that keeps a body in a circle
B. A real reaction to the centripetal force
C. Always present in the ground frame
D. A pseudo force that appears only in the rotating (non-inertial) frame ✓ Correct
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.
Q6 — 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. 3 N
B. 36 N
C. 18 N
D. 9 N ✓ Correct
Solution: F = mv²/r = 0.5×36/2 = 9 N.
Q7 — 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 engine thrust directly
C. The normal reaction
D. The car's weight
Solution: Friction is the only horizontal force available on level ground; v_max = √(μrg).
Q8 — 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. 22.5 m/s
B. 15 m/s ✓ Correct
C. 30 m/s
D. 9 m/s
Solution: v = √(μrg) = √(0.5×45×10) = √225 = 15 m/s.
Q9 — Banking of Roads & Well of Death · easy · theory
Roads are banked at curves so that:
A. The weight of the car decreases
B. Friction increases
C. Cars can stop faster
D. A component of the normal reaction provides the centripetal force, reducing reliance on friction ✓ Correct
Solution: At the design speed v₀ = √(rg tanθ), N sinθ alone supplies mv²/r — no friction needed.
Q10 — Banking of Roads & Well of Death · easy · numerical
A curve of radius 40 m is banked at 45°. The optimum speed is:
A. 20 m/s ✓ Correct
B. 10 m/s
C. 40 m/s
D. 14.1 m/s
Solution: v₀ = √(rg tan45°) = √400 = 20 m/s.
Q11 — Conical Pendulum & Rotating Systems · easy · theory
In a conical pendulum, the centripetal force on the bob is supplied by:
A. The vertical component of the tension
B. Friction
C. The horizontal component of the string tension (T sinθ) ✓ Correct
D. The weight of the bob
Solution: T cosθ = mg balances gravity; T sinθ = mω²r drives the circular motion.
Q12 — 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. 20 N
B. 40 N ✓ Correct
C. 10 N
D. 34.6 N
Solution: T cosθ = mg ⇒ T = 20/0.5 = 40 N.
Q13 — 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. Zero
C. √(5gr)
D. √(2gr)
Solution: At the top the string can only pull; the critical case has T = 0, gravity alone centripetal: mg = mv²/r ⇒ v = √(gr).
Q14 — 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. 8 m/s
C. 2 m/s
D. 16 m/s
Solution: v_top = √(gr) = √16 = 4 m/s.
Q15 — 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. √(5gr)
B. √(2gr)
C. √(gr)
D. Zero ✓ Correct
Solution: A rod can PUSH as well as pull, so it can support the mass at the top; v_top can be zero.
Q16 — 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. ≈ 11.2 m/s
B. 5 m/s
C. √50 m/s
D. 10 m/s ✓ Correct
Solution: v_b = √(4gr) = √100 = 10 m/s.
Q17 — Variable Angular Acceleration & Non-Uniform Motion · easy · theory
In NON-uniform circular motion, the net acceleration is:
A. Zero
B. Purely tangential
C. Purely radial
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.
Q18 — 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. 4 rad/s² and 16 rad
C. 2 rad/s² and 4 rad
D. 8 rad/s² and 8 rad
Solution: α = dω/dt = 4 rad/s²; θ = ∫ω dt = 2t² = 8 rad at t = 2 s.
Q19 — 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. Increases steadily
D. Is zero
Solution: KE = ½mv² depends only on speed. The centripetal force does no work, so KE cannot change.
Q20 — 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. 8 J
B. 4 J
C. 32 J
D. 16 J ✓ Correct
Solution: KE = ½×2×16 = 16 J (constant in UCM).
Q21 — Kinematics of Circular Motion · hard · numerical
A particle starts from rest on a circle of radius 1 m with constant angular acceleration α = 2 rad/s². At t = 1 s the magnitude of its TOTAL acceleration is:
A. 4 m/s²
B. √20 ≈ 4.5 m/s² ✓ Correct
C. 2 m/s²
D. 6 m/s²
Solution: ω = αt = 2 rad/s ⇒ a_c = ω²r = 4 m/s²; a_t = αr = 2 m/s²; a = √(16+4) = √20 ≈ 4.5 m/s².
Q22 — Kinematics of Circular Motion · hard · theory
For a particle in circular motion, the angular velocity vector ω⃗ and the linear velocity v⃗ satisfy v⃗ = ω⃗ × r⃗. The vector ω⃗ points:
A. Along the rotation axis, perpendicular to the plane of the circle ✓ Correct
B. Along the tangent
C. Opposite to v⃗
D. Radially outward
Solution: ω⃗ is an axial vector ⊥ to the plane (right-hand rule); v⃗ = ω⃗ × r⃗ then correctly gives the tangential direction.
Q23 — Kinematics of Circular Motion · hard · numerical
A particle's angular position is θ = t³ − 3t² (rad). Its angular acceleration is zero at t =
A. 2 s
B. 0.5 s
C. 3 s
D. 1 s ✓ Correct
Solution: ω = 3t² − 6t; α = 6t − 6 = 0 ⇒ t = 1 s (ω is momentarily at its minimum).
Q24 — Uniform Circular Motion & Centripetal Acceleration · hard · numerical
If the speed of a particle in UCM doubles and the radius is halved, the centripetal acceleration becomes:
A. 2 times
B. Unchanged
C. 8 times ✓ Correct
D. 4 times
Solution: a = v²/r → (2v)²/(r/2) = 8v²/r — an 8-fold increase.
Q25 — Uniform Circular Motion & Centripetal Acceleration · hard · theory
In UCM, the average acceleration over one COMPLETE revolution is:
A. Undefined
B. Zero, although the instantaneous acceleration is never zero ✓ Correct
C. v²/r along the tangent
D. v²/r towards the centre
Solution: Δv⃗ = 0 over a full cycle ⇒ average acceleration = 0; instantaneous a = v²/r ≠ 0 throughout — a classic distinction.
Q26 — Uniform Circular Motion & Centripetal Acceleration · hard · numerical
A particle in UCM (radius r, speed v) sweeps a quarter circle. The magnitude of its change in velocity is:
A. Zero
B. v
C. v√2 ✓ Correct
D. 2v
Solution: Velocities before/after are ⊥ with equal magnitude: |Δv⃗| = √(v² + v²) = v√2. Trap: speed change is zero, but velocity change is not.
Q27 — Centripetal & Centrifugal Force · hard · numerical
A string of breaking tension 100 N whirls a 1 kg stone in a horizontal circle of radius 1 m. Neglecting gravity, the maximum angular speed before the string snaps is:
A. 5 rad/s
B. 100 rad/s
C. 20 rad/s
D. 10 rad/s ✓ Correct
Solution: T = mω²r ⇒ ω_max = √(100/(1×1)) = 10 rad/s.
Q28 — Centripetal & Centrifugal Force · hard · theory
A bead rests on a smooth rotating horizontal rod (rotating about one end). In the ROTATING frame, the bead slides outward because:
A. Gravity acts outward
B. The Coriolis force pushes it outward
C. The centrifugal pseudo force mω²r acts outward with no real force to balance it ✓ Correct
D. A real outward force pushes it
Solution: On a smooth rod nothing supplies the needed centripetal force; in the rotating frame this appears as the unbalanced centrifugal force driving the bead outward. (Coriolis acts perpendicular to the sliding, not outward.)
Q29 — Centripetal & Centrifugal Force · hard · numerical
A gramophone disc spins at ω. A coin at radius r just slips when ω is doubled at the same radius. The friction demand changed by a factor of:
A. 8
B. 2
C. 4 ✓ Correct
D. √2
Solution: Required force mω²r ∝ ω² ⇒ quadruples — exceeding the friction limit that was just sufficient before.
Q30 — Motion on Flat Circular Tracks & Friction · hard · numerical
A car circles a flat track at the maximum safe speed for radius r. To take a turn of radius r/4 on the same surface, its maximum speed must be:
A. Half the original ✓ Correct
B. Unchanged
C. A quarter of the original
D. Twice the original
Solution: v_max ∝ √r ⇒ √(1/4) = 1/2 — tighter turns demand much lower speeds.