The masses of the three wires of copper are in the ratio 5 : 3 : 1 and their lengths are in the ratio 1 : 3 : 5. The ratio of their electrical resistance is:

1 : 5 : 125

Given, $l_1=1 k, l_2=3 k, l_3=5 k$ and $m_1=5 m, m_2=3 m, m_3=1 m$

Resistance, $R=\frac{\rho l}{A}$

Where, $\rho$ is resistivity of the material of conductor.

So, $R_1: R_2: R_3=\frac{l_1}{A_1}: \frac{l_2}{A_2}: \frac{l_3}{A_3}=\frac{l_1^2}{V_1}: \frac{l_2^2}{V_2}: \frac{l_3^2}{V_3}$

$=\frac{l_1^2}{m_1}: \frac{l_2^2}{m_2}: \frac{l_3^2}{m_3}=\frac{1}{5}: \frac{9}{3}: \frac{25}{1}$ = 1 : 15 : 125