JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Conservation of Momentum

Treat the disc-with-hole using the idea of a negative mass superposed on the full disc.

Full disc radius $R=20\text{ cm}$, hole radius $r=5\text{ cm}$, and the hole's edge touches the disc's edge, so the hole's centre sits at $R-r=15\text{ cm}$ from the origin.

With surface mass density $\sigma$:

Mass of full disc: $m=\pi R^2\sigma$

Mass of removed piece: $m_{cut}=\pi r^2\sigma =\frac{m}{16}$

Taking moments about the origin for the remaining piece:

$x_{com}=\frac{m(0)-m_{cut}(15)}{m-m_{cut}}=\frac{-\frac{m}{16}(15)}{m-\frac{m}{16}}=\frac{-\frac{15m}{16}}{\frac{15m}{16}}=-1\text{ cm}$

The centre of mass of the remaining disc lies $1.0\text{ cm}$ from the origin, on the side opposite the hole.