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 $\propto \frac{m}{l}$
$R\propto \frac{l}{\text{ Area }}\propto \frac{l^2}{m}$
$R_1:R_2:R_3=\frac{l_1^2}{m_1}:\frac{l_2^2}{m_2}:\frac{l_2^2}{m_3}$
$R_1:R_2:R_3=\frac{25}{1}:\frac{9}{3}:\frac{1}{5}$
$=125:15:1$