Physics · Simple Harmonic Motion

JEE Main 2026 — 4 April, Evening Shift — Question 10

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is (g is acceleration due to gravity).

Option A:
$2\pi \sqrt{\frac{5R}{4g}}$
Correct
Option B:
$2\pi \sqrt{\frac{2R}{3g}}$
Option C:
$2\pi \sqrt{\frac{3R}{2g}}$
Option D:
$2\pi \sqrt{\frac{3R}{g}}$

Answer: A

Step-by-step solution

Moment of inertia about axis A: $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ ? The given solution uses $I = \frac{mR^2}{4} + mR^2 = \frac{5mR^2}{4}$ . Torque $\tau = -mgR\theta$ , $\alpha = \tau/I = -\frac{mgR}{5mR^2/4}\theta = -\frac{4g}{5R}\theta$ , so $\omega = \sqrt{4g/(5R)}$ , $T = 2\pi\sqrt{5R/(4g)}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Simple Pendulum and Angular SHM