Physics - Current Electricity Question with Solution | TestHub

The resistance of a wire $(R)$ is related to its resistivity $\left(\rho \right)$, length$(l)$ and cross-sectional area $\left(A\right)$ as:

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

As the length of wire increases, the cross-sectional area decreases. The cross-sectional area in terms of volume $(V)$ and length $(l)$ is

$A=\frac{V}{l}$

$\therefore R=\rho \frac{l^2}{V}$Taking the volume of wire constant,

$R\propto l^2$

So if $R^{{}^{\prime }}$ is the resistance of wire after getting redrawn and $R$ is the initial resistance of wire,

$$\begin{gathered} \therefore \frac{R^{{}^{\prime }}}{R}=\frac{(5l{)}^2}{l^2}=25 \\ \Rightarrow R^{{}^{\prime }}=25\times R=125\Omega \end{gathered}$$