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

Answer 2

As we know that, $R = \rho \frac{1}{A}$
$R_{1} : R_{2} : R_{3} = \frac{1_{1} }{A_{1}} : \frac{1_{2}}{A_{2}} : \frac{1_{3}}{A_{3}}$
$= \frac{1_{1}^{2}}{V_{1}} : \frac{1^{2}_{2}}{V_{2}} : \frac{1^{2}_{3}}{V_{3}}$
$= \frac{1^{2}_{1}}{\left(m_{1}d\right)}: \frac{1^{2}_{2}}{\left(m_{2}d\right)} : \frac{1^{2}_{3}}{\left(m_{3}d\right) }$
$= \frac{1^{2}_{1}}{m_{1}} : \frac{1^{2}_{2}}{m_{2}} : \frac{1^{2}_{3}}{m_{3}}$
$= \frac{5^{2}}{2} : \frac{3^{2}}{3} : \frac{2^{2}}{5} = 125 : 30 : 8$