Q 11-06-210JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\dfrac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio $I_1/I_2$ is given by:
Answer: (A) $140$
Disc: $I_1 = \tfrac12\cdot\dfrac{7M}{8}(2R)^2 = \dfrac{7MR^2}{4}$.
The new sphere has mass $\dfrac M8$, so its volume is $\tfrac18$ of the original and its radius is $\dfrac R2$:
$$I_2 = \frac25\cdot\frac M8\cdot\frac{R^2}{4} = \frac{MR^2}{80}$$
$$\frac{I_1}{I_2} = \frac74\times80 = 140$$
Solution by Sreeraj P, M.Sc Physics