M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc. The moment of inertia of remaining part about an axis AB passing through the centre O and perpendicular to the plane of the disc is $$\frac{4}{x}MR^2$$. The value of x is _____.

$$I_{\text{remaining}} = I_{\text{total}} - I_{\text{removed}}$$

$$I = I_{\text{cm}} + md^2$$ (Parallel Axis Theorem)

$$\text{Mass of small disc } (m) = M \times \frac{\pi \left(\frac{R}{3}\right)^2}{\pi R^2} = \frac{M}{9}$$

$$\text{Distance between } O \text{ and } O' \ (d) = R - \frac{R}{3} = \frac{2R}{3}$$

$$I_{\text{total}} = \frac{1}{2}MR^2$$

$$I_{\text{removed}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 + md^2$$

$$I_{\text{removed}} = \frac{1}{2}\left(\frac{M}{9}\right)\left(\frac{R^2}{9}\right) + \left(\frac{M}{9}\right)\left(\frac{2R}{3}\right)^2$$

$$I_{\text{removed}} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{1}{18}MR^2$$

$$I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{1}{18}MR^2 = \left(\frac{9 - 1}{18}\right)MR^2 = \frac{8}{18}MR^2 = \frac{4}{9}MR^2$$

$$x = 9$$