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)

$m=l \times$ area $\times$ density
Area $\propto \frac{m}{l}$
$R \propto \frac{l}{\text { Area }} \propto \frac{l^{2}}{m}$
$R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{2}^{2}}{m_{3}}$
$R_{1}: R_{2}: R_{3}=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}$
$=125: 15: 1$