Rotational Motion – Phase 1
Previous Year Objective Questions (IIT-JEE / JEE Main / AIEEE)
All questions below are previous exam–based or framed exactly in the same pattern. Each includes method + exam insight.
PYQ-1 (IIT-JEE)
A solid sphere rolls down an incline without slipping. Its acceleration is:
A) g sinθ
B) (5/7) g sinθ
C) (2/3) g sinθ
D) (1/2) g sinθ
Solution:
For rolling body:
a = g sinθ / (1 + I / MR²)
For solid sphere: I = 2/5 MR²
a = g sinθ / (1 + 2/5) = (5/7) g sinθ
Correct Answer: B
PYQ-2 (JEE Main)
A particle is moving in a circular path. Angular momentum about the centre is conserved because:
A) Velocity is constant
B) Force is zero
C) Torque about centre is zero
D) Energy is conserved
Solution:
Angular momentum changes only if torque exists.
Here, centripetal force passes through centre ⇒ torque = 0
Correct Answer: C
PYQ-3 (AIEEE)
Moment of inertia of a rod of length L and mass M about its centre is:
A) ML²
B) ML²/2
C) ML²/12
D) ML²/3
Solution:
Standard result:
Uniform rod about centre → I = ML²/12
Correct Answer: C
PYQ-4 (JEE Main)
A wheel rolling without slipping has velocity v at centre. Velocity of the lowest point is:
A) v
B) 2v
C) v/2
D) 0
Solution:
At point of contact:
v = v (translation) − v (rotation) = 0
Correct Answer: D
PYQ-5 (IIT-JEE)
If no external torque acts on a system, which quantity remains constant?
A) Angular velocity
B) Moment of inertia
C) Angular momentum
D) Angular acceleration
Solution:
From law:
τ = dL/dt
τ = 0 ⇒ L = constant
Correct Answer: C
PYQ-6 (Board Competitive)
Rotational kinetic energy of a body rotating with angular velocity ω is:
A) ½ mv²
B) Iω
C) ½ Iω²
D) τθ
Solution:
Direct formula:
Krot = ½ Iω²
Correct Answer: C
Examiner’s Observation (Very Important)
✔ Most mistakes happen in moment of inertia selection
✔ Always check rolling condition
✔ Conservation laws save time
✔ Memorize standard I values
Next: Phase 1 – Page 2 (Moderate IIT/JEE Numericals)
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