Question

If a wire of resistance R is stretched twice its original length keeping volume constant then what will be its new resistance?

• A. R/4
• B. 2R
• C. 4R
• D. R/2

Answer 1

The correct option is C 4R

The resistance of a wire is given by,
R=$\rho \frac{1}{A}$
The resistivity $\left(\rho \right)$ of the material does not change in this case. Given that its length $\left(l\right)$ is increased by 2 times, but keeping the volume constant.
Volume, V = Area (A) x Length (l)
For the constant volume, $A\propto \frac{1}{l}$
$⟹\frac{{\text{A}}_2}{{\text{A}}_1}=\frac{{\text{l}}_1}{{\text{l}}_2}$

So, the new resistance is given by ${\text{R}}_2$, as
$\frac{{\text{R}}_1}{{\text{R}}_2}=\frac{{\text{l}}_1}{{\text{l}}_2}\times \frac{{\text{A}}_2}{{\text{A}}_1}=\frac{l}{2l}\times \frac{l}{2l}=\frac{1}{4}$

$⟹{\text{R}}_2=4{\text{R}}_1=4\text{R}$