Two resistance wires are of same material having length, $l$ and $2l$ and area of cross-section, $4A$ and $A$ respectively. When they are connected in parallel, their equivalent resistance can be expressed as

$\frac{2}{9}(ρ\frac{l}{A})$

The correct answer is Option (1) → $\frac{2}{9}(ρ\frac{l}{A})$

Resistance of a wire is given by

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

For first wire:

$R_{1} = \rho \frac{l}{4A}$

For second wire:

$R_{2} = \rho \frac{2l}{A} = \frac{2\rho l}{A}$

When connected in parallel,

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{R_{eq}} = \frac{4A}{\rho l} + \frac{A}{2\rho l}$

$\frac{1}{R_{eq}} = \frac{8A + A}{2\rho l}$

$\frac{1}{R_{eq}} = \frac{9A}{2\rho l}$

$R_{eq} = \frac{2\rho l}{9A}$