Three cells with internal resistance are connected as shown in the figure. The equivalent emf of the combination of three cells is

2 V

The correct answer is Option (1) → 2 V

Set left terminal potential $V_L=0$. The left $5\,$V cell makes the junction potential

$V_J = -5\,$V (open-circuit, no current through $1.5\ \Omega$).

Assume loop current $I$ around the bridge (junction $\to$ upper branch $\to$ right node $\to$ lower branch $\to$ junction). KVL gives

$4 + 3I + 3I + 10 = 0$

$\Rightarrow 6I + 14 = 0 \Rightarrow I = -\frac{7}{3}\,$A (negative sign means direction opposite to assumed).

Potential at right node via upper branch:

$V_R = V_J - 4 - 3I = V_J - 4 - 3\left(-\frac{7}{3}\right) = V_J - 4 + 7 = V_J + 3$

$V_R = -5 + 3 = -2\,$V

Equivalent emf (open-circuit) between left and right terminals, $V_{eq} = V_L - V_R$:

$V_{eq} = 0 - (-2) = 2\,$V

Therefore the equivalent emf of the combination is $2\,$V.