Physics · Rotational Dynamics

NEET (UG) 2025 — Question 36

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:

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

Answer: C

Step-by-step solution

$\mathrm{I}_{1}=\frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}$

$\mathrm{I}_{2}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{md}^{2}$

$=\frac{2}{5}\left(\frac{\mathrm{M}}{8}\right) \mathrm{R}^{2}+\frac{\mathrm{M}}{8}(\mathrm{R})^{2}$

$=\frac{7 \mathrm{MR}^{2}}{40}$

$\mathrm{I}=\mathrm{I}_{1}-\mathrm{I}_{2}=\frac{8 \mathrm{MR}^{2}}{5}-\frac{7 \mathrm{MR}^{2}}{40}=\frac{57}{40} \mathrm{MR}^{2}$

$\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{7}{57}$

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present…

Answer: C

Step-by-step solution

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