Question

A metallic wire of resistance R is melted and recast to half of its original length while maintaining the original volume. What will the new resistance of the wire be?

• A. $\frac{R}{4}$
  
• B. $\frac{R}{2}$
  
• C. $4R$
  
• D. $2R$

Answer 1

The correct option is A

$\frac{R}{4}$

Let $L$ and $A$ be the original length and area of cross-section respectively of the metallic wire.

Volume of the metallic wire $=L\times A$

Let $L^{\prime }$ and $A^{\prime }$ be the new length and new cross-sectional area of the recast metallic wire.
Since the volume of the wire is not changing, we can say $AL=A^{\prime }{L}^{\prime }$.

$\Rightarrow \frac{L^{\prime }}{L}=\frac{A}{A^{\prime }}$

$\frac{L^{\prime }}{L}=\frac{1}{2}\text{[Given]}\therefore \frac{A}{A^{\prime }}=\frac{1}{2}$

Now, $R=\frac{\rho L}{A}$ [Original resistance]

New resistance, $R^{\prime }=\frac{\rho L^{\prime }}{A^{\prime }}$ [Since the material of the wire is same, $\rho$ will be the same]
$Now\frac{R^{\prime }}{R}=\frac{\frac{\rho L^{\prime }}{A^{\prime }}}{\frac{\rho L}{A}}$
$=\left(\frac{L^{\prime }}{L}\right)\left(\frac{A}{A^{\prime }}\right)$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$
$\Rightarrow R^{\prime }=\frac{R}{4}$
Therefore, the new resistance will be one-fourth of the original resistance.