MHT CET · Physics · Rotational Motion
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\lambda\) ' is bent into a circular ring. The moment of inertia of ring about a tangential axis in its plane is
- A \(\frac{3 \lambda L^2}{8 \pi^2}\)
- B \(\frac{8 \pi^2}{3 \lambda L^3}\)
- C \(\frac{3 \lambda L^3}{8 \pi^2}\)
- D \(\frac{8 \pi^2}{3 \lambda L^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \lambda L^3}{8 \pi^2}\)
Step-by-step Solution
Detailed explanation
\( M = \lambda L \) \( R = \frac{L}{2 \pi} \)
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