Question

A copper wire is stretched to double its length, keeping the volume same. If the original resistance of the wire is $4\Omega$, What is the final resistance?

Answer 1

Step 1:Given data and assumptions

The copper wire is stretched to double its length, keeping the volume same.

Original resistance $R=4\Omega$

Let the wire's length and cross-section area be $l$ and $A$, respectively.

Step 2:Finding final resistance
$Resis\tan ce,R=p\times \frac{l}{A}$

Since volume is kept constant, doubling the length reduces the area to half.

For $I\text{'}=2I$ and $A\text{'}=\frac{A}{2}$, resistance of the wire will be

$$\begin{gathered} R\text{'}=\frac{\rho \times I\text{'}}{A\text{'}} \\ =\frac{\rho \times 2I}{A/2} \\ R\text{'}=\frac{4\rho l}{A}=4R \\ R\text{'}=4\times 4\Omega \\ =16\Omega \end{gathered}$$

Hence, the final resistance is $16\Omega$.