A thin wire of length l having linear density p is bent into a circular loop with C as its centre, as shown in figure. The moment of inertia of the loop about the line AB is

Length of the wire = l

linear mass density = $\mathrm{\rho }$

We know that mass of the wire,

$\mathrm{M}=\mathrm{\rho L}$

$\mathrm{Also} \mathrm{from}, l=2\mathrm{\pi R}$

$\mathrm{R}=\frac{l}{2\mathrm{\pi }}$

The loop's moment of inertia about an axis that passes through C and is parallel to its plane is given by

${\mathrm{I}}_{\mathrm{c}}={\mathrm{MR}}^2$

Moment of inertia of loop about PQ ${\mathrm{I}}_{\mathrm{c}}={\mathrm{MR}}^2$

Moment of inertia of loop PQ = ${\mathrm{I}}_1=\frac{1}{2}{\mathrm{MR}}^2$

Let ${\mathrm{I}}_{\mathrm{AB}}$ be the MOI of the loop around axis AB.

Using parallel axis theorem,

${\mathrm{I}}_{\mathrm{AB}}={\mathrm{I}}_1+{\mathrm{MR}}^2=\frac{3}{2}{\mathrm{MR}}^2$

${\mathrm{I}}_{\mathrm{AB}}=\frac{3}{2}{\mathrm{MR}}^2$

${\mathrm{I}}_{\mathrm{AB}}=\frac{3}{2}\times \mathrm{\rho l}\times \frac{{\mathrm{l}}^2}{4{\mathrm{\pi }}^2}=\frac{3{\mathrm{\rho l}}^3}{8{\mathrm{\pi }}^2}$

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