A square shaped hole of side $$l = \frac{a}{2}$$ is carved out at a distance $$d = \frac{a}{2}$$ from the centre 'O' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is $$-\frac{a}{x}$$, value of X (to the nearest integer) is:

Mass of the original complete disk: $$M_1 = \sigma \cdot \pi a^2$$

Mass of the square portion removed: $$M_2 = \sigma \cdot l^2 = \sigma \left(\frac{a}{2}\right)^2 = \sigma \frac{a^2}{4}$$

Positions of the centres of mass relative to the origin O:

$$x_1 = 0$$

$$x_2 = d = \frac{a}{2}$$

Centre of mass of the remaining portion: $$x_{\text{cm}} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$$

$$x_{\text{cm}} = \frac{(\sigma \pi a^2)(0) - \left(\sigma \frac{a^2}{4}\right)\left(\frac{a}{2}\right)}{\sigma \pi a^2 - \sigma \frac{a^2}{4}} = \frac{-\frac{a^3}{8}}{\pi a^2 - \frac{a^2}{4}} = \frac{-\frac{a}{8}}{\pi - \frac{1}{4}}$$

$$x_{\text{cm}} = \frac{-a}{8\pi - 2} = -\frac{a}{8(3.14) - 2} = -\frac{a}{25.12 - 2} = -\frac{a}{23.12} \approx -\frac{a}{23}$$