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.

Physics · Rotational Dynamics

JEE Main 2026 — 6 April, Morning Shift — Question 5

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

Option A:
$0.128\pi \mathrm{Nm}$ , 100
Option B:
$0.0128\pi \mathrm{Nm}$ , 50
Option C:
$0.128\pi \mathrm{Nm}$ , 50
Option D:
$0.0128\pi \mathrm{Nm}$ , 100
Correct

Answer: D

Step-by-step solution

$\omega_i = 1200 \times 2\pi/60 = 40\pi$ rad/s, $\alpha = (0-40\pi)/10 = -4\pi$ rad/s². $I = \frac25 MR^2 = \frac25 \times 5 \times (0.04)^2 = 0.0032$ kg m². Torque $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi$ Nm. Angular displacement $\theta = \omega_i^2/(2|\alpha|) = (1600\pi^2)/(8\pi) = 200\pi$ rad. Number of revolutions = $200\pi/(2\pi)=100$ .

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Torque, Equation of Motion and Toppling