Consider a circular disc of radius $20\ \text{cm}$ with centre located at the origin. A circular hole of radius $5\ \text{cm}$ is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of the centre of mass of the residual (remaining) disc from the origin will be
Answer: (C) $1.0\ \text{cm}$
Mass is proportional to area. Let the full disc have mass $16m$ (radius 20); the hole then has mass $m$ (radius 5, area $1/16$).
The hole touches the rim, so its centre is at $20 - 5 = 15\ \text{cm}$ from the origin.
Treat the remaining part as the full disc minus the hole:
$$x_{cm} = \frac{16m(0) - m(15)}{16m - m} = -\frac{15}{15} = -1\ \text{cm}$$
The centre of mass is $1.0\ \text{cm}$ from the origin, on the side away from the hole.
Solution by Sreeraj P, M.Sc Physics