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. None of these

Answer 1

Answer: (D)

Given, Ratio of masses $=1:3:5$
As density will remain same
Ratio of volumes $=1:3:5...(i)$
and Ratio of length $5:3:1...(ii)$
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 remain 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 Eqs. (i) and (ii)
$R_1:R_2:R_3=\frac{[5{]}^2}{1}:\frac{[3{]}^2}{3}:\frac{[1{]}^2}{5}=25:3:\frac{1}{5}$
$=125:15:1$