Question

A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX ' is

• A. $ \frac{ \tau L^3}{ 8 {\pi}^2}$
• B. $ \frac{ \tau L^3}{1116 {\pi}^2}$
• C. $ \frac{ 5 \tau L^3}{ 16 {\pi}^2}$
• D. $ \frac{ 3\tau L^3}{ 8 {\pi}^2}$

Answer 1

The Correct Option is D

Mass of the ring M = $\tau$L. Let R be the radius of the ring, then
L = 2n R or R = $\frac{L}{2 \pi }$
Moment of inertia about an axis passing through O and parallel to XX 'will be
$ \, \, \, \, \, \, \, \, \, \, \, I_0 = \frac{1}{2} MR^2$
Therefore, moment of inertia about XX' (from parallel axis theorem) will be given by
$ \, \, \, I_{XX ' } = \frac{ 1}{2} MR^2 +MR^2 = \frac{3}{2} MR^2$
Substituting values of M and R
$ \, \, \, \, \, I_ { XX ' } = \frac{ 3}{2} (\tau L) \bigg( \frac{ L^2}{{\pi}^2} \bigg) = \frac{ 3 \tau L^3}{ 8 {\pi}^2}$