Given is the graph of potential difference (V) across combination of three identical cells in series versus current (I). The emf of each cell would be

2 V

The correct answer is Option (1) → 2 V

The given graph is a straight line of terminal voltage $V$ versus current $I$.

At $I=0$, $V=6 \,$V ⟹ net emf of three identical cells in series is $6 \,$V.

At $I=1 \,$A, $V=0$ ⟹ total internal resistance $r_{total} = \frac{E}{I} = \frac{6}{1} = 6 \,\Omega$.

Since three identical cells are in series, each has emf $E_{cell}$ and internal resistance $r_{cell}$.

Total emf $= 3E_{cell} = 6 \,$V ⟹ $E_{cell} = 2 \,$V.

Total internal resistance $= 3r_{cell} = 6 \,\Omega$ ⟹ $r_{cell} = 2 \,\Omega$.

Therefore, the emf of each cell is $2 \,$V.