Q141 / 150 · MHT CET 2025 (26 Apr Shift 1) · Physics

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' $\mathrm{I}_1$ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' $I_2$ '. Then $I_1 / I_2$ is

  • A
    $\frac{3 \pi^2}{2}$
  • Correct answer
    B
    $\frac{2 \pi^2}{3}$
  • C
    $\frac{\pi^2}{3}$
  • D
    $\frac{\pi^2}{9}$

Solution

The moment of inertia for a rod of mass $M$ and length $L$ about its center perpendicular axis is $I_1 = \frac{1}{12} M L^2$.

When bent into a ring, the circumference becomes $L = 2\pi R$, giving radius $R = \frac{L}{2\pi}$. The moment of inertia for a ring about its diameter is $I_2 = \frac{1}{2} M R^2 = \frac{1}{2} M \left(\frac{L}{2\pi}\right)^2 = \frac{M L^2}{8\pi^2}$.

The ratio is $\frac{I_1}{I_2} = \frac{\frac{1}{12} M L^2}{\frac{1}{8\pi^2} M L^2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$, matching option B.

Hence, the correct answer is option: 'B'.

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