Switch S of the circuit shown in fig is closed at t = 0. If e denotes the induced e.m.f. in L and i, the

current flowing through the circuit at time t, then which of the following graphs fig shows the variation of e with i

When switch S it closed, at that moment the current in the circuit is zero. So, the induced e.m.f. across the inductance is equal to e.m.f. E of the battery. As current increases, the induced e.m.f. in the inductance decreases. If at any instant t, the current in the circuit is i, then applying Kirchhoff's law, we have
$\mathrm{iR}+\mathrm{e}=\mathrm{E}\text{ or }\mathrm{e}=(\mathrm{E}-\mathrm{iR})$
So, the graph between e and i will be a straight line having negative slope.