JEE Main · 2018 · 08 Apr 2018 · Offline · Rotational Mechanics

Treat the holed disc as the full disc minus the piece removed. The full disc of mass $9M$ has $I=\frac{1}{2}(9M)R^2=\frac{9M{R}^2}{2}$ about the central axis. The removed disc of radius $\frac{R}{3}$ has one ninth of the area and so mass $M$, and its centre sits $\frac{2R}{3}$ from the main axis, so by the parallel-axis theorem it accounts for $\frac{M}{2}{\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2=\frac{M{R}^2}{18}+\frac{4M{R}^2}{9}=\frac{M{R}^2}{2}$. Subtracting, the remaining disc has $\frac{9M{R}^2}{2}-\frac{M{R}^2}{2}=4M{R}^2$.