A circular hole of diameter $$R$$ is cut from a disc of mass $$M$$ and radius $$R$$; the circumference of the cut passes through the centre of the disc. The moment of inertia of the remaining portion of the disc about an axis perpendicular to the disc and passing through its centre is

Let the surface mass density of the original disc be $$\sigma$$. Since the disc is uniform, $$\sigma = \dfrac{M}{\pi R^{2}}$$.

Step 1 : Mass removed
Diameter of the hole = $$R$$, therefore its radius is $$a = \dfrac{R}{2}$$.
Area of the hole = $$\pi a^{2} = \pi \left(\dfrac{R}{2}\right)^{2} = \dfrac{\pi R^{2}}{4}$$.
Mass of the hole (to be removed) is therefore
$$m = \sigma \times \text{area} = \dfrac{M}{\pi R^{2}}\times\dfrac{\pi R^{2}}{4}= \dfrac{M}{4}$$.

Step 2 : Location of the hole
The circumference of the hole passes through the centre of the original disc, so the centre of the hole is at a distance
$$d = a = \dfrac{R}{2}$$
from the centre of the original disc.

Step 3 : Moment of inertia of the full disc
For a solid disc about an axis through its centre and perpendicular to the plane,
$$I_{\text{full}} = \dfrac{1}{2}\,M R^{2}.$$

Step 4 : Moment of inertia of the hole about the same axis
(a) About its own centre:
$$I_{\text{hole (cm)}} = \dfrac{1}{2}\,m a^{2} = \dfrac{1}{2}\left(\dfrac{M}{4}\right)\left(\dfrac{R}{2}\right)^{2} = \dfrac{M R^{2}}{32}.$$
(b) Shift this to the original disc’s centre using the parallel-axis theorem:
$$I_{\text{hole (origin)}} = I_{\text{hole (cm)}} + m d^{2}$$ $$= \dfrac{M R^{2}}{32} + \left(\dfrac{M}{4}\right)\left(\dfrac{R}{2}\right)^{2}$$ $$= \dfrac{M R^{2}}{32} + \dfrac{M R^{2}}{16}$$ $$= \dfrac{3 M R^{2}}{32}.$$

Step 5 : Moment of inertia of the remaining portion
Remove the hole’s contribution from the full disc:
$$I_{\text{remain}} = I_{\text{full}} - I_{\text{hole (origin)}}$$ $$= \dfrac{1}{2} M R^{2} - \dfrac{3 M R^{2}}{32}$$ $$= \dfrac{16 M R^{2}}{32} - \dfrac{3 M R^{2}}{32}$$ $$= \dfrac{13 M R^{2}}{32}.$$

Hence the required moment of inertia is $$\left(\dfrac{13}{32}\right)MR^{2}$$.

Option D which is: $$\left(\dfrac{13}{32}\right) MR^2$$