1. Initial resistance of the wire

For a wire of length $l$, cross-sectional area $A$, and resistivity $\rho$,

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

2. Initial current

Using Ohm’s law,

$I = \frac{V}{R} = \frac{V}{\rho l/A} = \frac{VA}{\rho l}$

3. New dimensions of the wire

The wire is of the same material, so resistivity remains $\rho$.

• New length: $l' = 2l$
• New area of cross-section: $A' = \frac{A}{2}$

4. New resistance

$R' = \frac{\rho l'}{A'} = \frac{\rho (2l)}{A/2}$

Multiply by reciprocal:

$R' = \rho (2l) \cdot \frac{2}{A} = \frac{4\rho l}{A}$

So the new resistance is 4 times the original resistance.

5. Resultant current

With the same potential difference $V$,

$I' = \frac{V}{R'} = \frac{V}{4\rho l/A} = \frac{VA}{4\rho l}$

Thus,

$I' = \frac{1}{4}\frac{VA}{\rho l}$

6. Matching with the options

This matches:

Option C:

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