Answer
(e) )$0.80\Omega$
Work Step by Step
Since both are made from same material, it means resistivity($\rho$) is same for both. Therefore
$ R \propto \frac{l}{A}$
in first wire
$l_1 =l$ and $ r_1 =r $
$\implies A_1 = \pi r^2$
In second wire
$ l_2 =2l $ and $r_2 = \frac {1}{2}r$
$\implies A_2 = \pi (\frac{r}{2})^2 = \pi \frac{r^2}{4}$
Now $ \frac{R_1}{R_2} = \frac{l_1 }{A_1 }*\frac{A_2 }{L_2 }$
$\implies \frac{0.10\Omega }{ R_2} =\frac{l }{\pi r^2} * \frac{\pi r^2 }{4*2l }$
$\implies R_2 = 8 * 0.10\Omega\\ \implies R_2 = 0.80 \Omega$
