$(A)\;\frac{K 2 \lambda_1 \lambda_2}{R} \\ (B)\;K \frac{2 \lambda_1^2}{R} \\ (C)\; Zero \\ (D)\;none $

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Electric field due to wire (2)$= \large\frac{1}{4 \pi \in _0} \frac{2 \lambda_2}{R}$

Charge per unit length on wire $(1)=\lambda _2$

Force per unit length on wire $(1) =\lambda_1 \times E$

$\qquad= \lambda \times \large\frac{1}{4 \pi \in _0}\frac{2 \lambda_2}{R}$

$\qquad= \large\frac{K(2 \lambda _1 \lambda_2)}{R}$

Hence A is the correct answer.

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