Question

A soap bubble has radius $R$ and thickness $d(<<R)$ as shown. It collapses into a spherical drop. Find the ratio of excess pressure in the drop to the excess pressure inside the bubble is ${\left(\frac{R}{xd}\right)}^{\frac{1}{3}}$. Find $x$.

Answer 1

$\text{Let}r\text{be the radius of the water drop formed.}$

$\text{Since the volume of the water forming bubble and drop is same,}$

$\frac{4}{3}\pi (R^3-(R-d{)}^3)=\frac{4}{3}\pi r^3$

$⟹r^3\approx 3R^{2}d\left(\text{neglecting}{d}^2\text{and}{d}^3\right)$

$\text{Ratio of excess pressure in the drop to the excess pressure inside the bubble is given by,}$

$\text{Ratio}=\frac{2\sigma /r}{4\sigma /R}$

$\text{Ratio}=\frac{1}{2}\left(\frac{R}{r}\right)$

$\text{Substituting the value of r gives}{\left(\frac{R}{24d}\right)}^{1/3}$