MHT CET · Physics · Rotational Motion
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 \(\frac{3 \pi^2}{2}\)
- B \(\frac{2 \pi^2}{3}\)
- C \(\frac{\pi^2}{3}\)
- D \(\frac{\pi^2}{9}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \pi^2}{3}\)
Step-by-step Solution
Detailed explanation
\(I_1 = \frac{1}{12}ML^2\) \(L = 2\pi R \implies R = \frac{L}{2\pi}\)
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