MHT CET · Physics · Rotational Motion
A thin metal wire of length ' \(L\) ' and mass ' \(M\) ' is bent to form semicircular ring as shown. The moment of inertia about \(\mathrm{XX}^1\) is
- A \(\frac{M L^2}{4 \pi^2}\)
- B \(\frac{2 \mathrm{ML}^2}{\pi^2}\)
- C \(\frac{M L^2}{2 \pi^2}\)
- D \(\frac{\mathrm{ML}^2}{\pi^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{M L^2}{2 \pi^2}\)
Step-by-step Solution
Detailed explanation
\(R = \frac{L}{\pi}\) \(I = \frac{1}{2} M R^2\)
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