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(B) As we move along the conductor in the direction of the current,the potential $V$ decreases linearly due to the resistance of the conductor $(V = I

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(B) As we move along the conductor in the direction of the current,the potential $V$ decreases linearly due to the resistance of the conductor $(V = IR)$.When we cross the cell from the negative terminal to the positive terminal,the potential increases by an amount equal to the terminal voltage of the cell,which is given by $V = E - Ir$,where $r$ is the internal resistance.Since $Ir > 0$,the increase in potential is strictly less than the $e.m.f. \ E$.Therefore,the graph must show a linear decrease,followed by a sharp increase of magnitude less than $E$,and then another linear decrease.Comparing this with the given options,Graph $B$ correctly depicts this behavior.

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