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Q.
A wire of length $l$ and mass $M$ in the form of a circular ring. The moment of inertia of the ring about its axis is
System of Particles and Rotational Motion
Solution:
Wire is bent into circular ring, then radius of wire
$l=2 \pi R$
$ \Rightarrow R=\frac{l}{2 \pi}$
$\therefore $ Moment of inertia of ring about its own axis
$\therefore I=M R^{2}=M\left(\frac{l}{2 \pi}\right)^{2}=\frac{M l^{2}}{4 \pi^{2}}$