JEE Main 2026 — Oscillations Question with Solution

From the given figure, the axis of rotation is a horizontal line passing through the point $A$ on the edge of the disc. The circular arrow around the axis indicates that the disc oscillates out of its plane, meaning the axis of rotation is a tangent to the disc in its own plane.

The moment of inertia of a uniform disc about its diameter is $I_{cm}=\frac{1}{4}M{R}^2$.

Using the parallel axis theorem, the moment of inertia of the disc about the given axis of rotation is:
$I=I_{cm}+M{R}^2=\frac{1}{4}M{R}^2+M{R}^2=\frac{5}{4}M{R}^2$

The distance from the center of mass of the disc to the axis of rotation is $d=R$.

The time period of a physical pendulum is given by:
$T=2\pi \sqrt{\frac{I}{Mgd}}$

Substituting the values of $I$ and $d$:
$T=2\pi \sqrt{\frac{\frac{5}{4}M{R}^2}{MgR}}=2\pi \sqrt{\frac{5R}{4g}}$

Answer: $2\pi \sqrt{\frac{5R}{4g}}$