Q 11-06-027JEE MainMedium
A circular hole of radius $\dfrac{R}{2}$ is cut from a uniform disc of radius $R$. The edge of the hole touches the edge of the disc. The centre of mass of the remaining part is at a distance from the centre of the original disc of
Answer: (A) $\dfrac{R}{6}$
Treat the hole as negative mass. Areas are in the ratio $1 : \dfrac{1}{4}$, so take masses $4m$ (full disc) and $m$ (hole). The hole's centre is at $\dfrac{R}{2}$ from O.
$$x_{cm} = \frac{4m(0) - m\left(\frac{R}{2}\right)}{4m - m} = -\frac{R}{6}$$
The centre of mass shifts $\dfrac{R}{6}$ away from the hole.
Solution by Sreeraj P, M.Sc Physics