NEET 2025 — Physics solution

NEET2025Physics

A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

A
$\frac{7}{8}$
B
$\frac{7}{40}$
C
$\frac{7}{57}$
D
$\frac{7}{64}$

Solution

Big sphere mass $M$ , radius $2 R$ : $I_{bi g} = \frac{2}{5} M (2 R)^{2} = \frac{8 M R ^{2}}{5}$ .

Mass of small sphere (volume ratio $1/8$ ): $m = M /8$ . Its centre is at distance $R$ from the Y-axis.

Small sphere about Y-axis (parallel-axis): $I_{s ma l l} = \frac{2}{5} (\frac{M}{8}) R^{2} + \frac{M}{8} R^{2} = \frac{M R ^{2}}{20} + \frac{M R ^{2}}{8} = \frac{7 M R ^{2}}{40}$ .

Rest of sphere: $I_{r es t} = I_{bi g} - I_{s ma l l} = \frac{8 M R ^{2}}{5} - \frac{7 M R ^{2}}{40} = \frac{64 M R ^{2} - 7 M R ^{2}}{40} = \frac{57 M R ^{2}}{40}$ .

Ratio $\frac{I _{s ma l l}}{I _{r es t}} = \frac{7/40}{57/40} = \frac{7}{57}$ .

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