A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre O as shown in fig. The moment of inertia of the loop about the axis XX' is

The moment of inertia of the loop about the diameter = M R²/ 2. So, the moment of inertia of the loop about XX' is given by
${\mathrm{I}}_{\mathrm{XX}}=\frac{{\mathrm{MR}}^2}{2}+{\mathrm{MR}}^2=\frac{3}{2}{\mathrm{MR}}^2$
where M = mass of the loop and .R = radius of the loop Now
$\mathrm{M}=\mathrm{L\rho }\text{ and }\mathrm{R}=\mathrm{L}/2\mathrm{\pi }$
$\therefore {\mathrm{I}}_{{\mathrm{XX}}^{\mathrm{\prime }}}=\frac{3}{2}\left(\mathrm{L\rho }\right)(\mathrm{L}/2\mathrm{\pi }{)}^2=3{\mathrm{L}}^3\mathrm{\rho }/8{\mathrm{\pi }}^2$