JEE Main 2026 — Rotational Motion Question with Solution

Let the mass of the original solid sphere be $M$ and its radius be $R$.

Mass of the smaller sphere, $m_1=\frac{M}{8}$.
Since the material remains the same, the density is constant. The volume is directly proportional to the mass.
$\frac{4}{3}\pi r^3=\frac{1}{8}\left(\frac{4}{3}\pi R^3\right)\Rightarrow r^3=\frac{R^3}{8}\Rightarrow r=\frac{R}{2}$

Moment of inertia of the smaller sphere about an axis through its centre is:
$I_1=\frac{2}{5}{m}_1{r}^2=\frac{2}{5}\left(\frac{M}{8}\right){\left(\frac{R}{2}\right)}^2=\frac{2}{5}\times \frac{M}{8}\times \frac{R^2}{4}=\frac{M{R}^2}{80}$

Mass of the larger part (which is converted into a disc) is:
$m_2=M-\frac{M}{8}=\frac{7M}{8}$

The radius of the disc is given as $2R$. The moment of inertia of a disc about its diameter is:
$I_2=\frac{1}{4}{m}_2(\text{radius}{)}^2=\frac{1}{4}\left(\frac{7M}{8}\right)(2R{)}^2=\frac{1}{4}\times \frac{7M}{8}\times 4R^2=\frac{7M{R}^2}{8}$

The ratio of their moments of inertia is:
$\frac{I_2}{I_1}=\frac{\frac{7M{R}^2}{8}}{\frac{M{R}^2}{80}}=\frac{7}{8}\times 80=70$

Answer: $70$