Moment of Inertia & Radius of Gyration question from NEET 2025

Answer: (a)

1. Let big sphere mass be $M$ (radius $2R$); cut sphere has $\frac{1}{8}$ the volume, so mass $\frac{M}{8}$, centre at distance $R$ from $Y$-axis.
2. $I_{small} = \frac{2}{5}\frac{M}{8}R^2 + \frac{M}{8}R^2 = \frac{7MR^2}{40}$
3. $I_{big} = \frac{2}{5}M(2R)^2 = \frac{64MR^2}{40}$, so $I_{rest} = \frac{57MR^2}{40}$
4. Ratio $= \frac{7}{57}$, so the answer is (a).