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 $\frac{R}{4}$

Given:
Resistance of wire before recast = R
Length of the wire after recast = half of length of wire before recast

Let:
Original length be $l$
Length after recasting be $l^{\prime }$
Original area of cross-section be $A$
Area of cross-section after recasting be $A^{\prime }$

Volume of the metallic wire is the product of length and area of the cross section and does not change.
$Al=A^{\prime }{l}^{\prime }$
$\frac{A}{A^{\prime }}=\frac{l^{\prime }}{l}$
$\therefore \frac{A}{A^{\prime }}=\frac{1}{2}$

Resistance is given by:
$R=\frac{\rho l}{A}$

Ratio of resistances are:
$\frac{R^{\prime }}{R}=\frac{\rho l^{\prime }/A^{\prime }}{\rho l/A}$
$=\left(\frac{l^{\prime }}{l}\right)\left(\frac{A}{A^{\prime }}\right)$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$
$R^{\prime }=\frac{R}{4}$