A square shaped hole of side


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_____


Let $\sigma$ be the mass density of circular disc.

Original mass of the disc, $m_{0}=\pi a^{2} \sigma$

Removed mass, $m=\frac{a^{2}}{4} \sigma$

Remaining, mass, $m^{\prime}=\left(\pi a^{2}-\frac{a^{2}}{4}\right) \sigma$

$=a^{2}\left(\frac{4 \pi-1}{4}\right) \sigma$

New position of centre of mass

$X_{C M}=\frac{m_{0} x_{0}-m x}{m_{0}-m}=\frac{\pi a^{2} \times 0-\frac{a^{2}}{4} \times \frac{a}{2}}{\pi a^{2}-\frac{a^{2}}{4}}$

$=\frac{-a^{3} / 8}{\left(\pi-\frac{1}{4}\right) a^{2}}=\frac{-a}{2(4 \pi-1)}=\frac{-a}{8 \pi-2}=-\frac{a}{23}$

$\therefore x=23$

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