Soap bubbles

This is how I solved it - Excess pressure ( Pressure inside - Pressure outside the bubble) = $\frac{4S}{R}$ , S is surface tension, R is the radius

Let The radius of the smaller bubble be R, Then excess pressure P1 in the bubble = $\frac{4S}{R}$

The radius of the larger bubble is 2R, Then P2 = $\frac{4S}{2R}$ = $\frac{2S}{R}$

Since Surface tensions S are equal P1 > P2.

After the obstruction is removed, the fluid ( here air ) will flow from region of high pressure to low pressure, so the air moves from the smaller bubble to the larger bubble.

Hence the size of the smaller bubble decreases, the size of the larger bubble increases

Excess pressure inside a soap bubble is $\frac{4T}{R}$ where $T$ is the surface tension and $R$ is the radius of the soap bubble. So, excess pressure is inversely proportional to $R$ $\implies$ The smaller bubble has greater pressure than the larger bubble.

As pressure is transferred from higher one to the lower one, it will be transferred from the smaller bubble to the larger bubble.

Thus, smaller bubble gets smaller and the larger one gets larger. As the smaller bubble gets smaller, pressure inside it increases and it is obvious that its rate of getting smaller increases as the process goes on.

try it practically