A circular hole of radius $\dfrac R4$ is made in a thin uniform disc having mass $M$ and radius $R$, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is:
Answer: (B) $\dfrac{237MR^2}{512}$
The removed piece has area (and mass) $\dfrac{1}{16}$ of the disc: $m = \dfrac{M}{16}$, radius $\dfrac R4$, centre at $\dfrac{3R}{4}$ from $O$. By the parallel axis theorem,
$$I_{\text{hole}} = \frac{M}{16}\left[\frac12\left(\frac R4\right)^2 + \left(\frac{3R}{4}\right)^2\right] = \frac{M}{16}\cdot\frac{19R^2}{32} = \frac{19MR^2}{512}$$
$$I = \frac{MR^2}{2} - \frac{19MR^2}{512} = \frac{256 - 19}{512}MR^2 = \frac{237MR^2}{512}$$
Solution by Sreeraj P, M.Sc Physics