If a rod has resistance 4 $\mathrm{\Omega }$ and if rod is turned as half cycle then the resistance along diameter

Suppose the length of the rod = /. Given that its resistance is 4 $\mathrm{\Omega }$
$\therefore$ Resistance per unit length $=(4/\mathrm{l})\mathrm{\Omega }$
The half circle of turned rod is shown in fig.

From figure $\mathrm{\pi r}+\mathrm{r}+\mathrm{r}=\mathrm{l}$
or $\mathrm{r}=\frac{\mathrm{l}}{(2+\mathrm{\pi })}$
Its diameter $=2\mathrm{r}=\frac{2\mathrm{l}}{(2+\mathrm{\pi })}$
Now, resistance along diameter $=\frac{2\mathrm{l}}{(2+\mathrm{\pi })}\times \frac{4}{\mathrm{l}}$
$=\frac{8}{(2+3\cdot 14)}=1\cdot 56\mathrm{ohm}$