Answer
$0.981$
Work Step by Step
We can find the required ratio of radii of silver and copper as follows:
The resistance for silver is given as $R_s=\rho_s(\frac{L_s}{\pi r_s^2})$.......eq(1)
The resistance for copper is given as $R_c=\rho_c(\frac{L_c}{\pi r_c^2})$......eq(2)
dividing eq(1) by eq(2), we obtain:
$\frac{R_s}{R_c}=\frac{\rho_s(\frac{L_s}{\pi r_s^2})}{\rho_c(\frac{L_c}{\pi r_c^2})}$
$\implies \frac{R_s}{R_c}=(\frac{\rho_s}{\rho_c})(\frac{L_s}{L_c})(\frac{r_c^2}{r_s^2}) $....eq(3)
Given that $R_s=R_c$ and $V_s=V_c$ $\implies \pi r_s^2L_s=\pi r_c^2L_c$
$\implies \frac{L_s}{L_c}=\frac{\pi r_c^2}{\pi r_s^2}=\frac{r_c^2}{r_s^2}$
Putting these values in eq(3), we obtain:
$\implies 1=(\frac{\rho_s}{\rho_c})(\frac{r_c^2}{r_s^2})(\frac{r_c^2}{r_s^2}) $
$1=(\frac{\rho_s}{\rho_c})(\frac{r_c}{r_s})^4$
$\implies (\frac{r_c}{r_s})^4=(\frac{\rho_c}{\rho_s})$
$\implies \frac{r_c}{r_s}=(\frac{\rho_c}{\rho_s})^{\frac{1}{4}}$
$\implies \frac{r_s}{r_c}=(\frac{\rho_s}{\rho_c})^{\frac{1}{4}}$
We plug in the known values to obtain:
$\implies \frac{r_s}{r_c}=(\frac{1.59\times 10^{-8}\Omega.m}{1.72\times 10^{-8}\Omega.m})^{\frac{1}{4}}=0.981$
