Q 11-06-189JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
A square shaped hole of side $l = \dfrac{a}{2}$ is carved out at a distance $d = \dfrac{a}{2}$ 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 $-\dfrac{a}{X}$, value of $X$ (to the nearest integer) is ______.
Numerical value type. Enter your answer.
Answer: 23
Take mass proportional to area. Full disc: area $\pi a^{2}$ at $O$. Hole: area $\dfrac{a^{2}}{4}$ with centre at $x = \dfrac a2$.
$$x_{cm} = \frac{0 - \dfrac{a^{2}}{4}\cdot\dfrac{a}{2}}{\pi a^{2} - \dfrac{a^{2}}{4}} = -\frac{a}{8\pi - 2} \approx -\frac{a}{23.1}$$
So $X \approx 23$.
Solution by Sreeraj P, M.Sc Physics