A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The smaller part having mass $M/8$ is converted into a sphere of radius $r$ and the larger part is converted into a circular disc of thickness $t$ and radius $2R$. If $I_1$ is moment of inertia of a sphere having radius $r$ about an axis through its centre and $I_2$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_2/I_1 =$ ______
Answer: (B) $70$
Same density, $\dfrac{1}{8}$ of the volume: $r = \dfrac{R}{2}$.
$I_1 = \dfrac{2}{5}\cdot\dfrac{M}{8}\cdot\dfrac{R^2}{4} = \dfrac{MR^2}{80}$.
Disc of mass $\dfrac{7M}{8}$, radius $2R$, about a diameter: $I_2 = \dfrac{1}{4}\cdot\dfrac{7M}{8}\cdot 4R^2 = \dfrac{7MR^2}{8}$.
$\dfrac{I_2}{I_1} = \dfrac{7}{8} \times 80 = 70$.
Solution by Sreeraj P, M.Sc Physics