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:

Answer 1

Step 1: Given data

$\text{Original length}=l$

$\text{New length}=nl$

$\text{Original length}=R$

The value of the resistance is written as the original length of the wire. As the length of the wire is stretched thus, the area of the wire also changes.

Step 2: To find

We have to find a new resistance.

Step 3: Calculate new resistance

The resistance of a cylindrical section of a conductor is equivalent to the material's resistivity multiplied by the length divided by the area.

$\mathrm{R}=\mathrm{\rho }\frac{\mathrm{L}}{\mathrm{A}}$

Here,

$R$ is the resistance.

$\mathrm{\rho }$ is the resistivity of copper.

$\mathrm{L}$ is the length of wire.

$\mathrm{A}$ is the cross-sectional area.

In this case volume is constant. So

$$\begin{gathered} \frac{R\text{'}}{R}={\left(\frac{l\text{'}}{l}\right)}^2 \\ \frac{R\text{'}}{R}={\left(\frac{nl}{l}\right)}^2 \\ R\text{'}=n^{2}R \end{gathered}$$

Therefore, the new resistance is ${\mathrm{n}}^2\mathrm{R}$.