Question

If a metallic wire of resistance $R$ is melted and recast to half of its length, the new resistance of the wire will be?

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

Answer 1

The correct option is A

$\frac{R}{4}$

Volume of the metallic wire = $Length\times Areaofcrosssection$
If $l$ and $A$ are the original length and area of cross -section and $l^{\prime }$ and $A^{\prime }$ are their corresponding values on recasting.
$Al=A^{\prime }{l}^{\prime }or\frac{l^{\prime }}{l}=\frac{A}{A^{\prime }}\because \frac{l^{\prime }}{l}=\frac{1}{2}\left(Given\right)\therefore \frac{A}{A^{\prime }}=\frac{1}{2}$

Original resistance $R=\frac{\rho l}{A}$

New resistance, $R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}$ (Since the material is the same, its resistivity will remain the same)

$\therefore \frac{R^{\prime }}{R}=\frac{\frac{\rho l^{\prime }}{A^{\prime }}}{\frac{\rho l^{\prime }}{A}}$
$=\left(\frac{l^{\prime }}{l}\right)\left(\frac{A}{A^{\prime }}\right)$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$
or $R^{\prime }=\frac{R}{4}$