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JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Physics
Chapter: Rotational Motion

JEE Main 2025 · 2 April · Evening Shift · Physics · Rotational Motion

The moment of inertia of a circular ring of mass M and diameter r...

The moment of inertia of a circular ring of mass

M

and diameter r about a tangential axis lying in the plane of the ring is :

Options

A $\frac{1}{2} \mathrm{Mr}^2$
B $\frac{3}{8} \mathrm{Mr}^2$
C $\frac{3}{2} \mathrm{Mr}^2$
D $2 \mathrm{Mr}^2$

Correct answer

B. $\frac{3}{8} \mathrm{Mr}^2$

Step-by-step solution

Diameter is given as R.

\begin{aligned}& \therefore \text{ Radius }=\mathrm{R} / 2 \\ & \mathrm{I}_{\text{tan gent }}=\frac{3}{2} \mathrm{ m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2\end{aligned}

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