In air, a soap bubble of radius R breaks into 64 small soap bubbles of equal radius r . Assuming the surface tension remains unchanged, the ratio of excess pressure inside the big soap bubble to that

Question

In air, a soap bubble of radius R breaks into 64 small soap bubbles of equal radius r . Assuming the surface tension remains unchanged, the ratio of excess pressure inside the big soap bubble to that inside a small soap bubble is

TestHub · Accepted Answer

The volume of air is conserved when the bubble breaks up.

$\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3} \Rightarrow R^{3} = 64 r^{3} \Rightarrow R = 4 r$

A soap bubble has two free surfaces, so the excess pressure inside a bubble of radius $R$ is $Δ P = \frac{4 T}{R}$ .

$\frac{Δ P _{big}}{Δ P _{small}} = \frac{4 T / R}{4 T / r} = \frac{r}{R} = \frac{r}{4 r} = \frac{1}{4}$

Therefore, the required ratio is $1 : 4$ .

Physics - Fluid Question with Solution | TestHub

Options:

Doubts & Discussion