The current (I) and voltage (V) graphs for a | vedclass

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(C) The slope of the $I-V$ graph is given by $	ext{Slope} = \frac{I}{V} = \frac{1}{R}$.From the graph,the slope of the line corresponding to $T_1$ is

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$T_1 < T_2$

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(C) The slope of the $I-V$ graph is given by $	ext{Slope} = \frac{I}{V} = \frac{1}{R}$.From the graph,the slope of the line corresponding to $T_1$ is greater than the slope of the line corresponding to $T_2$.Therefore,$\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 For a metallic wire,resistance increases with an increase in temperature.Since $R_1 < R_2$,it follows that $T_1 < T_2$.

The current $(I)$ and voltage $(V)$ graphs for a given metallic wire at two different temperatures $T_1$ and $T_2$ are shown in the figure. It is concluded that

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