V-I graph for a platinum wire at temperatures T_1 and T_2 is shown in…

The correct answer is Option (4) → \( \cot 2\theta \)

On a V-I graph the slope \( V/I \) is the resistance. The resistance of a metal wire rises with temperature as

    \( \displaystyle R = R_0\left[1 + \alpha\,(T - T_0)\right] \;\Rightarrow\; R_2 - R_1 = R_0\,\alpha\,(T_2 - T_1) \)

So \( (T_2 - T_1) \propto (R_2 - R_1) \).

Readings from the figure:

    \( T_1 \text{ line makes } \theta \text{ with the } I\text{-axis: } R_1 = \tan\theta \)
    \( T_2 \text{ line makes } \theta \text{ with the } V\text{-axis, so } (90^\circ - \theta) \text{ with the } I\text{-axis: } R_2 = \cot\theta \)

    \( \displaystyle R_2 - R_1 = \cot\theta - \tan\theta = \frac{\cos^{2}\theta - \sin^{2}\theta}{\sin\theta\cos\theta} \)

Using \( \cos^{2}\theta - \sin^{2}\theta = \cos 2\theta \) and \( \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta \):

    \( \displaystyle R_2 - R_1 = \frac{\cos 2\theta}{\frac{1}{2}\sin 2\theta} = 2\cot 2\theta \)

    \( \displaystyle T_2 - T_1 \propto \cot 2\theta \)