MHT CET · Physics · Rotational Motion
Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is 'I'. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' \(\mathrm{I}_1{ }^{\prime}\). If \(I_1=x I\), then the value of ' \(x\) ' is
- A \(\frac{2 \pi^2}{3}\)
- B \(\frac{3}{2 \pi^2}\)
- C \(\frac{3 \pi^2}{4}\)
- D \(\frac{4}{3 \pi^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{2 \pi^2}\)
Step-by-step Solution
Detailed explanation
\(I = \frac{1}{12} ML^2\) \(L = 2 \pi R \implies R = \frac{L}{2 \pi}\)
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