A thin wire of mass ‘m’ and length ‘L’ is bent in the form of a ring. The M.I. of the ring about a tangential axis parallel to any of its diameters is

Let the radius of the ring formed be r

thus,

$l=2\mathrm{\pi r}\Rightarrow \mathrm{r}=\frac{l}{2\mathrm{\pi }}\mathrm{\_\_\_\_\_\_}\left(1\right)$

Moment of inertia ring about axis AB

${\mathrm{I}}_1=\frac{1}{2}{\mathrm{Mr}}^2$

By parallel axis theorem:

Moment of inertia about tangential axis P

$\mathrm{I}=\frac{1}{2}{\mathrm{Mr}}^2+{\mathrm{Mr}}^2=\frac{3}{2}{\mathrm{Mr}}^2$

putting the value of r from equation 1,

$\mathrm{I}=\frac{3}{2}\mathrm{M}\frac{l^2}{4{\mathrm{\pi }}^2}=\frac{3\mathrm{M}{l}^2}{8{\mathrm{\pi }}^2}$

Hence the correct answer is $\frac{3\mathrm{M}{l}^2}{8{\mathrm{\pi }}^2}.$