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A thin wire of length L and uniform linear mass density λ is bent into a circular ring. The MI of the 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) \(\frac{8\pi^2}{3\lambda L^{2-}}\)
A thin wire of length L and uniform linear mass density λ is bent into a circular ring. The MI of the 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) \(\frac{8\pi^2}{3\lambda L^{2-}}\)
Correct Option is (C) \(\frac{3 L^3 \lambda}{8 \pi ^2}\)
Moment of inertia of the coil about the diameter
\(I = \frac{1}{2}MR^2\)
\(\because M = V \times \lambda\)
\(L = 2 \pi R\)
\(R = \frac{L}{2 \pi}\)
\(I = \frac{1}{2} L \lambda \frac{L^2}{4 \pi ^2}\)
\(I = \frac{1}{2} \times L \times \lambda \times \frac{L^2}{4 \pi ^2}\)
\(I = \frac{L^3 \lambda}{8 \pi ^2}\)
Using parallel axis theorem
\(I = I_{cm} + MR^2\)
\(I = \frac{L^3 \lambda}{8 \pi ^2} + L \lambda (\frac{L}{2 \pi})^2\)
\(I = \frac{L^3 \lambda}{8 \pi ^2} + \frac{L \lambda L^2}{4 \pi^2}\)
\(I = \frac{3 L^3 \lambda}{8 \pi ^2}\)
(C) \(\frac{3\lambda L^3}{8\pi^2}\)