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Masses of three wires of copper are in the ratio of $1: 3: 5$ and their lengths are in the ratio of $5: 3: 1$. The ratio of their electrical resistances is
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The correct answer is:
$125: 15: 1$
Given, $\quad m_{1}: m_{2}: m_{3}=1: 3: 5$
and $\quad I_{1}: I_{2}: I_{3}=5: 3: 1$
We know that, electrical resistance,
$R=\rho \frac{l}{A}$
and $\quad d=\frac{m}{V}=\frac{m}{A l}$
or $\quad A=\frac{m}{d l}$
$\Rightarrow \quad A \propto \frac{m}{l}$
( $\because$ density is same for the three wires)
$\therefore \quad R_{1}: R_{2}: R_{3}=\frac{l_{1}}{A_{1}}: \frac{I_{2}}{A_{2}}: \frac{I_{3}}{A_{3}}$
or $R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}$
$=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}$
$\Rightarrow R_{1}: R_{2}: R_{3}=125: 15: 1$
and $\quad I_{1}: I_{2}: I_{3}=5: 3: 1$
We know that, electrical resistance,
$R=\rho \frac{l}{A}$
and $\quad d=\frac{m}{V}=\frac{m}{A l}$
or $\quad A=\frac{m}{d l}$
$\Rightarrow \quad A \propto \frac{m}{l}$
( $\because$ density is same for the three wires)
$\therefore \quad R_{1}: R_{2}: R_{3}=\frac{l_{1}}{A_{1}}: \frac{I_{2}}{A_{2}}: \frac{I_{3}}{A_{3}}$
or $R_{1}: R_{2}: R_{3}=\frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}$
$=\frac{25}{1}: \frac{9}{3}: \frac{1}{5}$
$\Rightarrow R_{1}: R_{2}: R_{3}=125: 15: 1$
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