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Physics · Rotational Dynamics

JEE Main 2025 — 2 April, Evening Shift — Question 49

The moment of inertia of a circular ring of mass MM and diameter rr about a tangential axis lying in the

plane of the ring is

  1. Option A:

    38Mr2\frac{3}{8} M r^{2}

    Correct
  2. Option B:

    2Mr22 M r^{2}

  3. Option C:

    32Mr2\frac{3}{2} M r^{2}

  4. Option D:

    12Mr2\frac{1}{2} M r^{2}

Answer: A

Step-by-step solution

I1=MR22I_{1}=\frac{M R^{2}}{2}

I′=I1+MR2=32MR2\begin{aligned} I^{\prime} & =I_{1}+M R^{2} & =\frac{3}{2} M R^{2} \end{aligned}

r=2Rr=2 R

∴I′=38Mr2\therefore \quad I^{\prime}=\frac{3}{8} M r^{2}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

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