Two wires with resistances $R_1$ and $R_2$ are stretched $n$ times their original length. After stretching, the two wires are connected in parallel. The equivalent resistance of the combination will be

$\frac{n^2 R_1 R_2}{R_1+R_2}$

The correct answer is Option (1) → $\frac{n^2 R_1 R_2}{R_1+R_2}$

Resistance of a wire: $R = \rho \frac{l}{A}$

If a wire is stretched $n$ times, its new length $l' = n l$, and volume remains constant: $A' l' = A l \text{ implies } A' = \frac{A}{n}$

New resistance after stretching:

$R' = \rho \frac{l'}{A'} = \rho \frac{n l}{A/n} = n^2 \rho \frac{l}{A} = n^2 R$

Thus, resistances become $R_1' = n^2 R_1$, $R_2' = n^2 R_2$

When connected in parallel:

$\frac{1}{R_\text{eq}} = \frac{1}{R_1'} + \frac{1}{R_2'} = \frac{1}{n^2 R_1} + \frac{1}{n^2 R_2} = \frac{1}{n^2} \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$

$R_\text{eq} = \frac{n^2}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{n^2 R_1 R_2}{R_1 + R_2}$

Final Answer: $R_\text{eq} = \frac{n^2 R_1 R_2}{R_1 + R_2}$