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The resistance of a wire is $R$ ohm. If it is melted and stretched to $n$ times its original length, its new resistance will be
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Verified Answer
The correct answer is:
$n^{2} R$
Volume of material remain same after stretching. As volume remains same,
$$
\begin{aligned}
& & A_{1} l_{1} &=A_{2} l_{2} \\
& H e r e, \quad & l_{2} &=n l_{1} \\
\Rightarrow \quad & A_{2} &=\frac{A_{1} l_{1}}{l_{2}}=\frac{A_{1}}{n}
\end{aligned}
$$
Resistance of wire after stretching,
$$
R_{2}=\frac{\rho l_{2}}{A_{2}}=\rho \frac{n l_{1}}{A_{1} / n}=n^{2} \rho \frac{l_{1}}{R_{1}}=n^{2} R \quad\left(\because R=\rho \frac{l_{1}}{A_{1}}\right)
$$
$$
\begin{aligned}
& & A_{1} l_{1} &=A_{2} l_{2} \\
& H e r e, \quad & l_{2} &=n l_{1} \\
\Rightarrow \quad & A_{2} &=\frac{A_{1} l_{1}}{l_{2}}=\frac{A_{1}}{n}
\end{aligned}
$$
Resistance of wire after stretching,
$$
R_{2}=\frac{\rho l_{2}}{A_{2}}=\rho \frac{n l_{1}}{A_{1} / n}=n^{2} \rho \frac{l_{1}}{R_{1}}=n^{2} R \quad\left(\because R=\rho \frac{l_{1}}{A_{1}}\right)
$$
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