A thin wire of length ' \(L\) ' and linear mass density ' \(m\) ' is bent into a circular ring (in \(x-y\) plane) with centre ' \(C^{\prime}\) as shown in figure. The moment of inertia of the ring about an axis \(y y^{\prime}\) will be :

(B) \(\frac{3 m L^3}{8 \pi^2}\)
Total mass of the ring, \(M=m L\).
Since the wire of length \(L\) is bent into a circular ring of radius \(R\), its circumference is \(L\).
\(2 \pi R=L \Rightarrow R=\frac{L}{2 \pi}\)
The axis \(y y^{\prime}\) is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is
\(I_d=\frac{1}{2} M R^2\)
Using the parallel axis theorem, the moment of inertia of the ring about the tangent \(y y^{\prime}\) is:
\(I_{y y^{\prime}}=I_d+M R^2\)
\(I_{y y^{\prime}}=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2\)
Substituting the values of \(M\) and \(R\) :
\(I_{y y^{\prime}}=\frac{3}{2}(m L)\left(\frac{L}{2 \pi}\right)^2\)
\(I_{y y^{\prime}}=\frac{3}{2} m L\left(\frac{L^2}{4 \pi^2}\right)\)
\(I_{y y^{\prime}}=\frac{3 m L^3}{8 \pi^2}\)