Question

A conducting wire of length $l$, area of cross-section $A$ and electrical resistivity $\rho$ is connected between the terminals of a battery. A potential difference $V$ is developed between its ends, causing an electric current. If the length of the wire is doubled and the area of cross-section is halved, the resultant current would be:

• A. $\frac{1}{4}\frac{\rho l}{VA}$
• B. $\frac{3}{4}\frac{VA}{\rho l}$
• C. $4\frac{VA}{\rho l}$
• D. $\frac{1}{4}\frac{VA}{\rho l}$

Answer 1

The correct option is D $\frac{1}{4}\frac{VA}{\rho l}$

We know that
$R=\rho \frac{l}{A}$
Now, new length: $l^{\prime }=2l$
& new area of cross section, $A^{\prime }=A/2$

$\therefore$ New resistance, $R^{\prime }=\rho .\frac{2l}{\frac{A}{2}}$
$\Rightarrow R^{\prime }=4\frac{\rho l}{A}$
$\Rightarrow R^{\prime }=4R$

$\therefore$ Resultant current: $I=\frac{V}{4R}$
$I=\frac{1}{4}\frac{VA}{\rho l}$
Hence, option (d) is correct.