Question

Masses of three wires of copper are in the ratio of $1 : 3 : 5$ and their lengths are in the ratio of $5 : 3 : 1$. The ratio of their electrical resistances is

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

Answer 1

Answer: (D)

125 : 15 : 1

Answer 2

Given, $m_{1}: m_{2}: m_{3}=1: 3: 5$
and $l_{1}: l_{2}: l_{3}=5: 3: 1$
We know that, electrical resistance,
$R=\rho \frac{1}{A}$
and $d=\frac{m}{V}=\frac{m}{A l}$
or $A=\frac{m}{d l}$
$\Rightarrow A \propto \frac{m}{l}$
($\because$ density is same for the three wires)
$\therefore R_{1}: R_{2}: R_{3}=\frac{I_{1}}{A_{1}}: \frac{I_{2}}{A_{2}}: \frac{I_{3}}{A_{3}}$
or $R_{1}: R_{2}: R_{3} =\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}$
$=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}$
$\Rightarrow R_{1}: R_{2}: R_{3} =125: 15: 1$