JEE Main 2026 Jan 28 Shift 1, Physics Q23: Moment of Inertia

Moment of inertia of a solid sphere about a diameter: $\frac{2}{5}m{R}^2$. By parallel axis theorem about an axis at distance $d$ from the centre:

$I=\frac{2}{5}m{R}^2+m{d}^2=m{k}^2$, so $k^2=\frac{2}{5}{R}^2+d^2$.

Plug $R=10$ cm, $d=15$ cm:

$k^2=\frac{2}{5}(100)+225=40+225=265$ cm${}^2$.

$k=\sqrt{265}$ cm $\Rightarrow n=265$.