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Q. 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 resistances is :-

NTA AbhyasNTA Abhyas 2020

Solution:

$R=\rho \frac{l}{\pi r^{2}}$ and $m=\pi r^{2}ld$
$\therefore R \propto \frac{\ell ^{2}}{ m}$
$\therefore R_{1}:R_{2}:R_{3}=\frac{25}{1}:\frac{9}{3}:\frac{1}{5}=125:15:1$