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. $2\Omega$
• B. $8\Omega$
• C. $6\Omega$
• D. $10\Omega$

Answer 1

The correct option is B

$8\Omega$

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 resitance of the stretched wire is $8\Omega$