Question

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

• A. $25:1:125$
• B. $1:125:25$
• C. $125:1:25$
• D. $125:15:1$

Answer 1

The correct option is D $125:15:1$

Given, Ratio of masses $=1:3:5$
As density will remain same
Ratio of volumes $=1:3:5$ ......(i)
and Ratio of lengths $=5:3:1$ .....(ii)
For ratio of resistance
$R_1:R_2:R_3=\frac{\rho l_1}{A_1}:\frac{\rho l_2}{A_2}:\frac{\rho l_3}{A_3}$
Since, resistivity remian same
$=\frac{l_1}{A_1}:\frac{l_2}{A_2}:\frac{l_3}{A_3}=\frac{l_1^2}{A_1{l}_1}:\frac{l_2^2}{A_2{l}_2}:\frac{l_3^2}{A_3{l}_3}$
From equations (i) and (ii)
$R_1:R_2:R_3=\frac{{\left[5\right]}^2}{1}:\frac{{\left[3\right]}^2}{3}:\frac{{\left[1\right]}^2}{5}=25:3:\frac{1}{5}=125:15:1$