Question

The resistance if a wire is '$R$' ohm. If it is melted and stretched to '$n$' times its original length, its new resistance will be:

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

Answer 1

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

$\text{Step 1: Resistance in terms of Volume and Length}$

We know that Resistance:

$R=\frac{\rho \ell }{A}$

$R=\frac{\rho \ell }{A}\times \frac{\ell }{\ell }$ $=\frac{\rho {\ell }^2}{V}$ $(\because V=A\ell )$

When the wire is melted, its volume remains constant.

Also, since the material is same, therefore resistivity $\rho$ is also constant.

Hence, $R\propto {\ell }^2$

$\text{Step 2: Ratio of initial and final Resistances}$

$\frac{R_1}{R_2}={\left(\frac{{\ell }_1}{{\ell }_2}\right)}^2$ $(\rho =\text{constant})$

$R_2={\left(\frac{{\ell }_2}{{\ell }_1}\right)}^2{R}_1={\left(\frac{n{\ell }_1}{{\ell }_1}\right)}^{2}R=R{n}^2$

Therefore, New Resistance $=n^{2}R$