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Q.

Masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 10:6:1, then the ratio of electrical resistances is

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a

$225 : 15 : 10$

b

$10 : 5 : 125$

c

$500 : 60 : 1$

d

$1 : 100 : 500$

answer is C.

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Detailed Solution

$m = dv \begin{aligned} m = dA (l 0 l) \\ A_{1} = k / 10 \\ 3 m = {dA}_{2} 6 \\ A_{2} = \frac{3 k}{6} \\ 5 m = {dA}_{3} 1 \\ A_{3} = \frac{5 k}{1} \end{aligned} R_{1} : R_{2} : R_{3} = \frac{{ρl}_{1}}{A_{1}} : \frac{{ρl}_{2}}{A_{2}} : \frac{{ρl}_{3}}{A_{3}} = \frac{10 l}{k / 10} : \frac{6 l}{3 k / 6} : \frac{l}{5 k / 1} = \frac{10 l \times 10}{k} : \frac{6 l \times 6}{3 k} : \frac{l \times 1}{5 k} = 100 : 12 : 1 / 5 = 500 : 60 : 1$

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