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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Answer & Solution

Correct Answer

(C) \(\frac{ ML ^{2}}{\pi^{2}}\)

Step-by-step Solution

Detailed explanation

\(\pi L \Rightarrow r =\frac{ L }{\pi}\)

\(I = Mr ^{2}=\frac{ ML ^{2}}{\pi^{2}}\)

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