A thin wire of length 'I' and mass m is bent in the form of semi circle . Its moment of inertia about an axis joining its free ends will be 

From the question,

length is equal to circumference of semicircle.

$l=\mathrm{\pi r}\Rightarrow \mathrm{r}= \frac{l}{\mathrm{\pi } }$

Moment of inertia for a half-ring 

$=\mathrm{m}\left(\frac{{\mathrm{r}}^2}{2}\right)=\mathrm{m}\left(\frac{l^2}{2{\mathrm{\pi }}^2}\right)$

Moment of inertia = $\frac{\mathrm{m}{l}^2}{2{\mathrm{\pi }}^2}.$