Physics · Rotational Dynamics

JEE Main 2025 — 7 April, Morning Shift — Question 68

A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure. The moment of inertia of the given system about PQ axis is $\frac{x}{15} I$ , where $I$ is the moment of inertia of the disc about its diameter. The value of $x$ is $\qquad$ .

Answer: 199

Numerical answer — enter this value.

Step-by-step solution

(disk) $I_{A}=\frac{m R^{2}}{4}$

$I=\frac{m R^{2}}{4}$ (solid sphere)

$I_{B}=\frac{7}{5} m R^{2}$ (Spherical shell)

$I_{C}=\frac{5}{3} m R^{2}$

$I_{P Q}=m R^{2}\left[\frac{1}{4}+\frac{7}{5}+\frac{5}{3}\right]=m R^{2}\left(\frac{199}{4}\right) \times \frac{1}{15}$

So, $\frac{x}{15} \times \frac{m R^{2}}{4}=\frac{m R^{2} \times 199}{4 \times 15}$

$\Rightarrow x=199$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia