JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Rotational Mechanics

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A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ______ and ______ respectively.

(A) $0.128 π Nm, 100$
(B) $0.0128 π Nm, 50$
(C) $0.128 π Nm, 50$
(D) $0.0128 π Nm, 100$ ✓ Correct

Answer: (D)

Solution

The initial spin is

ω_{i} = 1200 \times \frac{2 π}{60} = 40 π

rad/s, stopped in 10 s, so the angular deceleration has magnitude

α = \frac{40 π}{10} = 4 π

rad/s

^{2}

. For a solid sphere

I = \frac{2}{5} M R^{2} = \frac{2}{5} \times 5 \times (0.04)^{2} = 3.2 \times 1 0^{- 3}

kg·m

^{2}

, giving torque

τ = I α = 3.2 \times 1 0^{- 3} \times 4 π = 0.0128 π

N·m. The angle turned is

θ = \frac{ω _{i}^{2}}{2 α} = \frac{( 40 π ) ^{2}}{2 \times 4 π} = 200 π

rad, i.e.

\frac{200 π}{2 π} = 100

revolutions. The torque is

0.0128 π

N·m and the sphere makes 100 rotations.

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Rotational Mechanics

Solution

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