The figure shows the voltage (V) versus the current (I) graphs for a wire at two temperatures \( T_1 \) and \( T_2 \). One can conclude that:
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In an I–V graph, a steeper slope means higher conductance (lower resistance). Since resistance increases with temperature, a steeper slope means a lower temperature.
The provided figure illustrates two I-V graphs for a wire at temperatures \( T_1 \) and \( T_2 \). The graph plots current \( I \) against voltage \( V \). The slope of the line represents conductance \( \left( \frac{I}{V} \right) \), which is the reciprocal of resistance: \[\text{slope} = \frac{I}{V} = \frac{1}{R}\]Observing the graph, the slope at \( T_1 \) is steeper than at \( T_2 \), indicating:\[\frac{1}{R_1}>\frac{1}{R_2} \Rightarrow R_1<R_2\]For typical metallic conductors, resistance escalates with increasing temperature. Therefore:\[R_1<R_2 \Rightarrow T_1<T_2\]
