A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R / 2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

Physics · Rotational Dynamics

JEE Main 2025 — 22 January, Morning Shift — Question 53

A uniform circular disc of radius ' $R$ ' and mass ' $M$ ' is rotating about

an axis perpendicular to its plane and passing through its centre. A small

circular part of radius $\mathrm{R} / 2$ is removed from the original disc

as shown in the figure. Find the moment of inertia of the remaining part of the

original disc about the axis as given above.

Option A:
$\frac{7}{32} \mathrm{MR}^{2}$
Option B:
$\frac{9}{32} \mathrm{MR}^{2}$
Option C:
$\frac{17}{32} \mathrm{MR}^{2}$
Option D:
$\frac{13}{32} \mathrm{MR}^{2}$
Correct

Answer: D

Step-by-step solution

$I=\frac{M R^{2}}{2}-\left[\frac{\frac{M}{4}\left(\frac{R}{2}\right)^{2}}{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2}\right]$ $\mathrm{I}=\frac{13}{32} \mathrm{MR}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia

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Step-by-step solution

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