JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Elasticity & SHM

The disc performs small oscillations as a physical pendulum about the horizontal axis $A$ tangent to its rim, with its centre of mass a distance $d=R$ below this axis. About a diameter through its centre a uniform disc has moment of inertia $\frac{1}{4}M{R}^2$, so by the parallel-axis theorem the moment of inertia about $A$ is For a physical pendulum $T=2\pi \sqrt{\frac{I}{Mgd}}=2\pi \sqrt{\frac{\frac{5}{4}M{R}^2}{MgR}}$. The time period of small oscillations is $2\pi \sqrt{\frac{5R}{4g}}$.