A circular hole of radius $\dfrac R2$ is cut from a uniform disc of radius $R$ such that the edge of the hole passes through the centre $O$ of the disc. If the mass of the remaining part is $M$, its moment of inertia about the axis through $O$ perpendicular to the plane of the disc is
Answer: (D) $\dfrac{13}{24}MR^2$
Concept. Treat the remaining part as (full disc) $-$ (hole). Masses are in the ratio of areas. Use the parallel-axis theorem for the hole, whose centre is $\dfrac R2$ from $O$.
Step 1. Masses. The hole has $\dfrac14$ of the full disc's area. If the full disc has mass $M_0$, the remaining part has $\dfrac34M_0 = M$:
$$M_0 = \frac43M,\qquad m_{hole} = \frac13M$$
Step 2. Full disc about $O$.
$$I_1 = \frac12\cdot\frac43M\cdot R^2 = \frac23MR^2$$
Step 3. Hole about $O$.
$$I_2 = \frac12\cdot\frac M3\left(\frac R2\right)^2 + \frac M3\left(\frac R2\right)^2 = \frac{MR^2}{24} + \frac{MR^2}{12} = \frac{MR^2}{8}$$
Step 4. Remaining part.
$$I = \frac23MR^2 - \frac18MR^2 = \frac{13}{24}MR^2$$
Answer. $\dfrac{13}{24}MR^2$ (Option D).
Solution by Sreeraj P, M.Sc Physics