$\begin {array} {1 1} (A)\;\alpha i^2 \large\frac{R}{\lambda} & \quad (B)\;\alpha i R \lambda \\ (C)\;\alpha^2i\large\frac{R}{\lambda} & \quad (D)\;\alpha i^2 \large\frac{R}{2\lambda} \end {array}$

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This question requires basic concepts from several chapters but can be solved simply by

realizing the principle of conservation of energy: power developed = power dissipated.

$ i^2R = \lambda ( T - T_0)$

or, $(T – T_0) = i^2 \large\frac{R}{ \lambda}$

This is also the increase in temperature of the resistance

$ \Delta l =l \alpha \Delta T$

Strain = $ \large\frac{ \Delta l}{l}$

Hence, the strain $= \alpha \times ( T-T_0)= \alpha i^2 \large\frac{R}{\lambda}$

Ans : (A)

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