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(B) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance. The terminal voltage $V$ across the resistance $R$ is g

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$2$

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(B) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance. The terminal voltage $V$ across the resistance $R$ is given by the equation: $V = E - Ir$,where $I$ is the current flowing through the circuit.For the first case: $I_1 = 0.3 \, A$ and $V_1 = 0.9 \, V$.Substituting these values into the equation: $0.9 = E - 0.3r$  --- (Equation $1$)For the second case: $I_2 = 0.25 \, A$ and $V_2 = 1.0 \, V$.Substituting these values into the equation: $1.0 = E - 0.25r$  --- (Equation $2$)Subtracting Equation $1$ from Equation $2$:$(1.0 - 0.9) = (E - 0.25r) - (E - 0.3r)$$0.1 = -0.25r + 0.3r$$0.1 = 0.05r$$r = \frac{0.1}{0.05} = 2 \, \Omega$.Thus,the internal resistance of the cell is $2 \, \Omega$.

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