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

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(B) From Ohm's law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ grap

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$T_1 < T_2$

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(B) From Ohm's law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ graph is given by $	ext{slope} = \frac{dI}{dV} = \frac{1}{R}$.From the given graph,the slope for temperature $T_1$ is greater than the slope for temperature $T_2$.Therefore,$\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 For a metallic wire,resistance increases with an increase in temperature $(R \propto T)$.Since $R_1 < R_2$,it follows that $T_1 < T_2$.

The current-voltage $(I-V)$ graphs for a given metallic wire at two different temperatures $T_1$ and $T_2$ are shown in the figure. The temperatures $T_1$ and $T_2$ are related as

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