Question

If a uniform wire of resistance $R$ is uniformly stretched to $n$ times the original length, then new resistance of the wire becomes :

• A. $n R$
• B. $ n^{2}R $
• C. $ \frac{R}{n} $
• D. $ \frac{R}{n^{2}} $

Answer 1

Answer: (B)

Given: $l_{1} A_{1}=l_{2} A_{2}$
As the volume of wire does not change after stretching so,
$\Rightarrow \frac{A_{2}}{A_{1}}=\frac{l_{1}}{l_{2}}=\frac{l}{n l}=\frac{1}{n}$
The resistance of the wire is $R=\rho \frac{l}{A}$
$\Rightarrow R \propto \frac{l}{A}$
Hence, $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}}$
$=\frac{1}{n} \times \frac{1}{n}=\frac{1}{n^{2}}$
Therefore, $R_{2}=n^{2} R$
$\left(\because R_{1}=R\right)$