A thin wire AB of length l and mass m is bent in the form of a semicircle as shown in fig. Its moment
of inertia about an axis joining its free ends will be

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${ml}^{2}$

${ml}^{2} / π^{2}$

${ml}^{2} / 2 π^{2}$

Zero

answer is D.

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Detailed Solution

Here, $πr = l or r = l / π$
The moment of inertia of a ring about its diameter is given
by $I = \frac{1}{2} {Mr}^{2}$
$∴$ Moment of inertia of semicircle
$= \frac{1}{2} [m {(\frac{l}{π})}^{2}] = \frac{{ml}^{2}}{2 π^{2}}$

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