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

$\frac{R}{4}$

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

Let and A are the original length and area of cross-section respectively.

Let l' and A' be the corresponding values of length and cross sectional area on recasting.

$Al=A^{\prime }{l}^{\prime }or\frac{l^{\prime }}{l}=\frac{A}{A^{\prime }}$

$\frac{l^{\prime }}{l}=\frac{1}{2}\left(Given\right)\therefore \frac{A}{A^{\prime }}=\frac{1}{2}$

Now, $R=\frac{\rho l}{A}$ (Original resistance)

New resistance, $R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}$

$\therefore \frac{R^{\prime }}{R}=\frac{\frac{\rho l^{\prime }}{A^{\prime }}}{\frac{\rho l}{A}}$

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