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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