Two soap bubbles of unequal radii are connected by a narrow hollow...

The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \dfrac{4T}{R}$, where $T$ is the surface tension.

Let the radius of the smaller bubble be $r$ and the larger bubble be $R$, with $r < R$.

The pressure inside the smaller bubble is $P_1 = P_0 + \dfrac{4T}{r}$ and inside the larger bubble is $P_2 = P_0 + \dfrac{4T}{R}$.

Since $r < R$, we have $\dfrac{4T}{r} > \dfrac{4T}{R}$, which implies $P_1 > P_2$.

Air always flows from a region of higher pressure to a region of lower pressure. Therefore, air will flow from the smaller bubble to the larger bubble.

Answer: Air flows from the smaller bubble to the larger bubble.