Q 11-06-169JEE MainJEE Main 2020 (8 Jan, Shift 2)Hard
As shown in figure, a spherical cavity (centred at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$). The centre of mass of the remaining (shaded) part of the sphere is at $G$, i.e., on the surface of the cavity. $R$ can be determined by the equation:
Answer: (A) $(R^2+R+1)(2-R)=1$
The cavity touches the sphere from inside, so $CO = R - 1$. $G$ is on the cavity surface on the far side of $C$, so $OG = 1$ and
$$CG = 1 - (R-1) = 2 - R$$
Masses are proportional to volumes. Taking moments about $C$ (the full sphere has its centre of mass at $C$):
$$(R^3 - 1)\cdot CG = 1^3\cdot CO$$
$$(R-1)(R^2+R+1)(2-R) = R-1$$
$$(R^2+R+1)(2-R) = 1$$
Solution by Sreeraj P, M.Sc Physics