Masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1, the ratio of their electrical resistance is

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125:15:1

5:3:1

1:3:5

11:15:125

answer is C.

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Detailed Solution

$\begin{array}{l} m = {dA}_{1} . 5 l \Rightarrow A_{1} = \frac{K}{5} K = \frac{m}{d . 5 l} \\ 3 m = {dA}_{2} \cdot 3 l \Rightarrow A_{2} = K \\ 5 m = {dA}_{3} . l \Rightarrow A_{3} = 5 K \\ R_{1} : R_{2} : R_{3} = \frac{5 l}{A_{1}} : \frac{3 l}{A_{2}} : \frac{l}{A_{3}} = \frac{5 l}{K 15} : \frac{3 l}{k} : \frac{l}{5 k} \end{array}$
=125:15:1

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