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

Length of the wire = L

Density of the wire = $\rho$

Mass of the wire and hence of the loop is, $\mathrm{M}=\mathrm{\rho L}$ and radius of the loop is, $\mathrm{r}=\frac{\mathrm{L}}{2\mathrm{\pi }}$

Moment of inertia of a ring about an axis passing through its center and lying in its plane, ${\mathrm{I}}_1=\frac{{\mathrm{Mr}}^2}{2}$.

Using parallel axis theorem, we can say that

${\mathrm{I}}_2={\mathrm{I}}_1+{\mathrm{Mr}}^2$

$\Rightarrow {\mathrm{I}}_2=\frac{{\mathrm{Mr}}^2}{2}+{\mathrm{Mr}}^2$

$\Rightarrow {\mathrm{I}}_2=\frac{3{\mathrm{Mr}}^2}{2}$

$\Rightarrow {\mathrm{I}}_2=\frac{3}{2}\left(\mathrm{\rho L}\right){\left(\frac{\mathrm{L}}{2\mathrm{\pi }}\right)}^2$

$\Rightarrow {\mathrm{I}}_2=\frac{3{\mathrm{\rho L}}^3}{8{\mathrm{\pi }}^2}$