Discussion Forum

The masses of the three copper wires are in the ratio 2:3:5  and their  lengths are in the ratio 5:3:2  . Then , the ratio of their electrical resistance is 

Answer:

Explanation:

Using ,   $R=\rho \frac{l}{A}$

$R_{1 }:R_{2 }:R_{3 }= \frac{l_{1}}{A_{1}}:\frac{l_{2}}{A_{2}}:\frac{l_{3}}{A_{3}}$

$= \frac{l_{1}^{2}}{V_{1}}:\frac{l_{2}^{2}}{V_{2}}:\frac{l_{3}^{2}}{V_{3}}$

$= \frac{l_{1}^{2}}{(m_{1}d)}:\frac{l_{2}^{2}}{(m_{2}d)}:\frac{l_{3}^{2}}{(m_{3}d)}$

$= \frac{l_{1}^{2}}{(m_{1})}:\frac{l_{2}^{2}}{(m_{2})}:\frac{l_{3}^{2}}{(m_{3})}$

$= \frac{5^{2}}{2}:\frac{3^{2}}{3}:\frac{2^{2}}{5}=125:30:8$