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Physics · Rotational Dynamics

JEE Main 2025 — 8 April, Evening Shift — Question 47

A rod of linear mass density ' λ\lambda ' and length ' LL ' is bent to form a ring of radius ' RR '. Moment of inertia of ring about any of its diameter is

  1. Option A:

    λL34π2\frac{\lambda L^{3}}{4 \pi^{2}}

  2. Option B:

    λL38π2\frac{\lambda L^{3}}{8 \pi^{2}}

    Correct
  3. Option C:

    λL312\frac{\lambda L^{3}}{12}

  4. Option D:

    λL316π2\frac{\lambda L^{3}}{16 \pi^{2}}

Answer: B

Step-by-step solution

2πR=L2 \pi R=L ⇒M=λL\Rightarrow M=\lambda L

R=L2πR=\frac{L}{2 \pi}

Moment of Inertia about diameter

=MR22=M2(L2π)2=λL38π2=\frac{M R^{2}}{2}=\frac{M}{2}\left(\frac{L}{2 \pi}\right)^{2}=\frac{\lambda L^{3}}{8 \pi^{2}}

Answer key and solution verified before publishing.

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

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