Mass of three wires of copper are in the ratio $\text{1}:\text{3}:\text{5}$ and their lengths are in the ratio $\text{5}: \text{3}:\text{1}$ The ratio of their electrical resistances is

The relation between resistance of the wire with its mass and density is given by

$R=\frac{\rho l}{\text{a}}=\frac{\rho l^2}{\text{v}}=\frac{\rho l^{2}d}{m}$

$R\propto l^2/\text{m}$

From data $l_1:l_2:l_3=5:3:1,m_1:m_2:m_3=1:3:5$

$\Rightarrow {\text{R}}_1\text{:}{\text{R}}_2{\text{: R}}_{\text{3}}=\left(\frac{l_1^2}{m_1}\right):\frac{l_2^2}{m_2}:\frac{l_3^3}{m_3}$

$=\frac{25}{1}:\frac{9}{3}:\frac{1}{5}=125:15:1$