Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is:

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$\frac{M L^{2}}{π^{2}}$

$\frac{1}{4} \frac{M L^{2}}{π^{2}}$

$\frac{1}{2} \frac{M L^{2}}{π^{2}}$

$\frac{2}{5} \frac{M L^{2}}{π^{2}}$

answer is A.

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Detailed Solution

Since all element of wire are same distance from axis, moment of Inertia is $I = M r^{2}$
Where $r = \frac{L}{π}$ is radius of semicircle $I = M {[\frac{L}{π}]}^{2} \Rightarrow I = \frac{M L^{2}}{π^{2}}$

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