Question

The mass of the three wire of copper are in the ratio 1:3:5, and their lengths are in ratio 5:3:1. The ratio of their electrical resistance is

• A. 1:3:5
• B. 5:3:1
• C. 1:15:125
• D. 125:15:1

Answer 1

The correct option is D

125:15:1

$R=\frac{\rho l}{A}=\frac{\rho l^2}{Al}=\frac{\rho l^2}{V}=\frac{\rho l^2}{m/d}=\frac{\rho d{l}^2}{m}OrR\propto \frac{l^2}{m}{R}_1:R_2:R_3=\frac{l_1^2}{m_1}:\frac{l_2^2}{m_2}:\frac{l_3^2}{m_3}=\frac{25}{1}:\frac{9}{8}:\frac{1}{5}=125:15:1$