Explanation

(b) Give that,
$\frac{l_{1}}{l_{2}} = \frac{d_{1}}{d_{2}} = \frac{\rho_{1}}{\rho_{2}} = \frac{1}{2}$
$and\ \ \ \ R_{1} = 10\Omega$
We know that, the resistance of the wire
$R = \frac{\rho l}{A} = \frac{\rho l}{\pi\left( \frac{d}{2} \right)^{2}} = \frac{4\rho l}{\pi d^{2}}\ \ \ \ \left\lbrack \because A = \pi\left( \frac{d}{2} \right)^{2} \right\rbrack$
So, the resistance of first wire is
$R_{1} = \frac{4\rho_{1}l_{1}}{\pi d_{1}^{2}}\ \ \ \ \ \ \ \ ..(i)$
and the resistance of the second wire is
$R_{2} = \frac{4\rho_{2}l_{2}}{\pi d_{2}^{2}}\ \ \ \ \ \ \ \ \ \ \ ..(ii)$
On dividing Eq. (ii) by Eq (i)
$\ \ \ \ \ \ \frac{R_{2}}{R_{1}} = \frac{\rho_{2}}{\rho_{1}} \times \frac{l_{2}}{l_{1}} \times \frac{d_{1}^{2}}{d_{2}^{2}}$
$\Longrightarrow \frac{R_{2}}{10} = \frac{2}{1} \times \frac{2}{1} \times \left( \frac{1}{2} \right)^{2}$
$\Longrightarrow \frac{R_{2}}{10} = 1 \Longrightarrow R_{2} = 10\Omega$