Question

A large soap bubble has surface tension S and radius R. On the surface of the bubble we place a small ring of radius r. (r < < R). Now we prick the film inside the ring. Air begins to flow out from the hole thus formed. Find the time in which the bubble will collapse completely. Assume the flow to be steady, non-viscous and incompressible. Take density of air as $\rho$.

Answer 1

The time in which the bubble will collapse completely is: $T = \frac{4 R^{7/2} \sqrt{\rho}}{7 r^2 \sqrt{2S}}$

Answer 2

The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$. When a hole of radius $r$ is pricked, air flows out due to this pressure difference. Assuming steady, non-viscous, and incompressible flow, the velocity of air exiting the hole can be found using Bernoulli's principle: $\frac{1}{2}\rho v^2 = \Delta P$, which gives $v = \sqrt{\frac{2\Delta P}{\rho}} = \sqrt{\frac{8S}{\rho R}}$.

The volume flow rate through the hole is $Q = A_{hole} \cdot v = (\pi r^2) \sqrt{\frac{8S}{\rho R}}$. The volume of the bubble is $V = \frac{4}{3}\pi R^3$. The rate of change of volume is $\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$. Since air is flowing out, $\frac{dV}{dt} = -Q$.

Thus, $4\pi R^2 \frac{dR}{dt} = - \pi r^2 \sqrt{\frac{8S}{\rho R}}$. Rearranging and integrating from $R$ to $0$ for radius and $0$ to $T$ for time: $4 R^2 dR = - r^2 \sqrt{\frac{8S}{\rho}} R^{-1/2} dt$ $\int_{R}^{0} R^{5/2} dR = - \frac{r^2}{4} \sqrt{\frac{8S}{\rho}} \int_{0}^{T} dt$ $[\frac{2}{7}R^{7/2}]_{R}^{0} = - \frac{r^2}{4} \sqrt{\frac{8S}{\rho}} T$ $-\frac{2}{7}R^{7/2} = - \frac{r^2}{4} \sqrt{\frac{8S}{\rho}} T$ $T = \frac{2}{7}R^{7/2} \cdot \frac{4}{r^2 \sqrt{8S/\rho}} = \frac{8 R^{7/2}}{7 r^2} \sqrt{\frac{\rho}{8S}}$ $T = \frac{8 R^{7/2} \sqrt{\rho}}{7 r^2 (2\sqrt{2}\sqrt{S})} = \frac{4 R^{7/2} \sqrt{\rho}}{7 r^2 \sqrt{2S}}$.