The current-voltage graphs for a given metallic wire at two different temperatures, $T_1$ and $T_2$ are shown in the figure. The correct relation of temperatures and resistances will be:

$R_{T_2}>R_{T_1},T_1>T_2$

The correct answer is Option (3) → $R_{T_2}>R_{T_1},T_1>T_2$

Slope of I–V = $\frac{I}{V}=\frac{1}{R}$ (conductance).

Line at $T_1$ is steeper than at $T_2$ ⇒ $\frac{1}{R_{T_1}}>\frac{1}{R_{T_2}}$ ⇒ $R_{T_1}<R_{T_2}$.

For a metallic wire $R$ increases with temperature ⇒ $T_1>T_2$.