The masses of three wires of copper are in the ratio of 1:3:5 and their lengths are in the ratio 5:3:1. The ratio of their electrical resistance is :

1:3:5

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5:3:1

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1:15:125

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125:15:1

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Solution

The correct option is D
125:15:1

$m_{1} : m_{2} : m_{3} = 1 : 3 : 5 A_{1} l_{1} : A_{2} l_{2} : A_{3} l_{3} = 1 : 3 : 5 Also l_{1} : l_{2} : l_{3} = 5 : 3 : 1 l_{1}^{2} : l_{2}^{2} : l_{3}^{2} = 25 : 9 : 1 \Rightarrow \frac{A_{1} l_{1}}{l_{1}^{2}} : \frac{A_{2} l_{2}}{l_{1}^{2}} : \frac{A_{3} l_{3}}{l_{1}^{2}} = \frac{1}{25} : \frac{3}{9} : \frac{5}{1} Since they have same resistivity's R_{1} : R_{2} : R_{3} = \frac{l_{1}}{A_{1}} : \frac{l_{2}}{A_{2}} : \frac{l_{3}}{A_{3}} = 25 : 3 : \frac{1}{5} = 125 : 15 : 1$

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