Question

A wire of resistance $2\Omega$ is stretched to double its length such that its volume remains constant. What is the resistance of the stretched wire?

• A. 8

Answer 1

The correct option is A 8

Volume of the wire, $V=$ Length ($l$) $\times$ Area of cross-section ($A$)
i.e. $V=lA$

Let the new volume be $V^{\prime }$ and the new area be $A^{\prime }$

So, $V^{\prime }=2l\times A^{\prime }$
Since the volume remains constant, $l\times A=2l\times A^{\prime }$
$\Rightarrow A^{\prime }=\frac{A}{2}$

We know, $R=\rho \frac{l}{A}$
So, new resistance $R^{\prime }=\rho \times \frac{2l}{\frac{A}{2}}$
$\Rightarrow R^{\prime }=4\rho \frac{l}{A}=4R=4\times 2=8\Omega$
Hence, the resistance of the stretched wire is $8\Omega$.