Question

The masses of three copper wires are in the ratio 2 : 3 : 5 and their lengths are in the ratio 5 : 3 : 2. Then, the ratio of their electrical resistances is

• A. 1 : 9 : 15
• B. 2 : 3 : 5
• C. 5 : 3 : 2
• D. 125 : 30 : 8

Answer 1

Answer: (D)

125 : 30 : 8

Using, $R=\rho \frac{1}{A}$
$R_1:R_2:R_3=\frac{l_1}{A_1}:\frac{l_2}{A_2}:\frac{l_3}{A_3}$
$=\frac{l_1^2}{V_1}:\frac{l_2^2}{V_2}:\frac{l_3^2}{V_3}$
$=\frac{l_1^2}{\left(m_{1}d\right)}:\frac{l_2^2}{\left(m_{2}d\right)}:\frac{l_3^2}{\left(m_{3}d\right)}$
$=\frac{l_1^2}{m_1}:\frac{l_2^2}{m_2}:\frac{l_3^2}{m_3}$
$=\frac{5^2}{2}:\frac{3^2}{3}:\frac{2^2}{5}=125:30:8$