Physics - Current Electricity Question with Solution | TestHub

The resistance of a wire $\left(R\right)$ is related to its resistivity $\left(\rho \right)$, length$\left(l\right)$  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 $\left(V\right)$ and length $\left(l\right)$ is

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

$\therefore R=\rho \frac{l^2}{V}$

Taking the volume of wire constant,

$R\propto l^2$

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

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