A conducting wire of length $l$, area of cross-section $A$ and electric 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 of the same material is doubled and the area of cross-section is halved, the resultant current would be:

The correct option is D

$\frac{1}{4}\frac{VA}{\rho l}$

Step 1. Given data:

Length of wire, $l$

Cross-section Area, $A$

Electrical Resistivity, $\rho$

Potential difference, $V$

New area of cross-section, $A\text{'}=\frac{A}{2}$

New length of wire, $l\text{'}=2l$

Step 2. Calculating new resistance $R\text{'}$ using formula:

As we know that, formula for resistance, $R$

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

Therefore, New resistance, $R\text{'}$

$$\begin{gathered} \Rightarrow R’=\frac{\rho l\text{'}}{A\text{'}} \\ \Rightarrow R\text{'}=\frac{\rho \times 2l}{{\displaystyle \frac{A}{2}}} \end{gathered}$$

$$\begin{gathered} \Rightarrow R\text{'}=4\frac{\rho l}{A} \\ \therefore R\text{'}=4R \end{gathered}$$ $\left[R=\frac{\rho l}{A}\right]$

Therefore, Resultant current, $I$:

$$\begin{gathered} I=\frac{V}{R\text{'}} \\ \Rightarrow I=\frac{V}{4R} \\ \therefore I=\frac{1}{4}\left(\frac{VA}{\rho l}\right) \end{gathered}$$

Hence, option (D) is correct.