Question

The resistance of a wire is $R\Omega$. If it is melted and stretched to $n$ times its original length, its new resistance will be

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

Answer 1

The correct option is B $n^{2}R$

Resistance of wire is $R=\frac{\rho l}{A}$
$\Rightarrow R=\frac{\rho l.l}{A.l}=\frac{\rho l^2}{V}$
After stretching, volume of wire will be constant
$\Rightarrow R\propto l^2$
So, if length increases $n$ times
New resistance is
$R^1=n^{2}R$