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(B) Let the resistivity at a distance $x$ from the left end be $\rho = (\rho_0 + ax)$,where $\rho_0$ is the resistivity at $x = 0$ and $a$ is a positi

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(B) Let the resistivity at a distance $x$ from the left end be $\rho = (\rho_0 + ax)$,where $\rho_0$ is the resistivity at $x = 0$ and $a$ is a positive constant.According to Ohm's law,the current density $J$ is given by $J = \frac{i}{A} = \frac{E}{\rho}$,where $i$ is the constant current and $A$ is the uniform cross-sectional area.Therefore,the electric field intensity $E$ at a distance $x$ is $E = \frac{i \rho}{A} = \frac{i(\rho_0 + ax)}{A}$.This equation shows that $E$ is a linear function of $x$ of the form $E = mx + c$,where $m = \frac{ia}{A}$ and $c = \frac{i\rho_0}{A}$.Since $\rho_0 > 0$,the intercept $c$ is non-zero,and the slope $m$ is positive.Thus,the graph of $E$ versus $x$ is a straight line with a positive slope and a positive $y$-intercept. This corresponds to the graph shown in option $(B)$.

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