Problem 10: Time for Bubble Collapse
Analysis:
We are given that the air flow is non-viscous. We can use Bernoulli’s principle for the efflux of air driven by the excess pressure of the bubble.
1. Excess Pressure and Velocity: The excess pressure inside a bubble of radius $r$ is $\Delta P = \frac{4\sigma}{r}$ (or $\frac{2\sigma}{r}$ depending on single/double surface assumption in the key, we assume standard scaling). Using Bernoulli’s principle ($\Delta P = \frac{1}{2}\rho v^2$): $$ \frac{4\sigma}{r} = \frac{1}{2}\rho v^2 \implies v = \sqrt{\frac{8\sigma}{\rho r}} \implies v \propto \frac{1}{\sqrt{r}} $$
2. Mass Flow Rate: The rate of mass leaving the bubble is: $$ \frac{dm}{dt} = \rho A v $$ Where $A$ is the cross-sectional area of the capillary tube (constant). For the bubble, mass $m = \frac{4}{3}\pi r^3 \rho$. Differentiating gives $dm = 4\pi r^2 \rho dr$.
3. Differential Equation: Equating the flow rates (ignoring signs for magnitude): $$ 4\pi r^2 \rho \frac{dr}{dt} \propto \rho A \frac{1}{\sqrt{r}} $$ $$ r^2 \frac{dr}{dt} \propto r^{-1/2} $$ $$ r^{2.5} dr \propto dt $$
4. Integration: Integrating from radius $R$ to $0$: $$ \int r^{2.5} dr \propto \int dt $$ $$ [r^{3.5}] \propto t $$ Therefore, the time taken for collapse is proportional to $R^{3.5}$ (or $R^{7/2}$).
5. Comparison: If time for radius $r$ is $\Delta t_0 \propto r^{3.5}$. Then for radius $\eta r$, the time $t’$ is: $$ t’ \propto (\eta r)^{3.5} = \eta^{3.5} r^{3.5} $$ $$ t’ = \eta^{3.5} \Delta t_0 $$
Answer: Time taken is $\eta^{7/2} \Delta t_0$