JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Rotational Mechanics

The moment of inertia about a diameter is $\frac{M{R}^2}{4}$. By the perpendicular axis theorem the moment of inertia about the central axis normal to the disc equals the sum of the two equal diameter values, $I_C=2\times \frac{M{R}^2}{4}=\frac{M{R}^2}{2}$. Shifting to a parallel axis through a point on the edge, a distance $R$ away, the parallel axis theorem gives $I=I_C+M{R}^2=\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}$. Comparing with $\frac{x}{2}M{R}^2$, the value of $x$ is $3$.