Q 11-06-150JEE MainJEE Main 2021 (24 Feb, Shift 2)Medium
A circular hole of radius $\left(\dfrac{a}{2}\right)$ is cut out of a circular disc of radius $a$ as shown in figure. The centroid of the remaining circular portion with respect to point $O$ will be:
Answer: (B) $\frac56 a$
Take $x$ from $O$. The full disc (area $\pi a^2$) has its centre at $x = a$; the hole (area $\pi a^2/4$) has its centre at $x = \frac{3a}{2}$.
$$x_{cm} = \frac{\pi a^2(a) - \frac{\pi a^2}{4}\cdot\frac{3a}{2}}{\pi a^2 - \frac{\pi a^2}{4}} = \frac{a - \frac{3a}{8}}{\frac34} = \frac{5a}{8}\cdot\frac43 = \frac56 a$$
Solution by Sreeraj P, M.Sc Physics