## In This post you will get answer of Q.

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

Solution:

## Resistance $R=frac{s l}{A}$

$A=frac{volume}{l}=frac{mass}{ltimes Density}=frac{m}{l d}$

$R=frac{sl}{frac{m}{l d}}=frac{sdl^{2}}{m}$

$R alphafrac{l^{2}}{m}$

$R_{1}:R_{2}:R_{3}=frac{l_{1}^{2}}{m_{1}}:frac{l_{2}^{2}}{m_{2}}:frac{l_{3}^{2}}{m_{3}}$

$=frac{left(5 lright)^{2}}{m}:frac{left(3lright)^{2}}{3 m}:frac{l^{2}}{5m}$

$=25:3:frac{1}{5} ldots left(frac{l}{m} is,a, constantright)=125:15:1$