1.

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) \(\frac{8\pi^2}{3\lambda L^{2-}}\)

Answer»

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) \(\frac{3\lambda L^3}{8\pi^2}\)


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