Question

A soap bubble has radius $R$ and thickness $d(<<R)$ as shown. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

• A. ${\left(\frac{R}{3d}\right)}^{\frac{1}{3}}$
• B. ${\left(\frac{R}{6d}\right)}^{\frac{1}{3}}$
• C. ${\left(\frac{R}{24d}\right)}^{\frac{1}{3}}$
• D. None

Answer 1

The correct option is C

${\left(\frac{R}{24d}\right)}^{\frac{1}{3}}$

volume of soap solution in a bubble $=4\pi R^{2}d$ now we make a drop
$4\pi R^{2}d=\frac{4}{3}\pi r^2\Rightarrow r=(3R^{2}d{)}^{\frac{1}{3}}$
Excess pressure initially $=\frac{4S}{R}$
Excess pressure finally $=\frac{2S}{r}=\frac{2S}{(3R^{2}D{)}^{\frac{1}{3}}}$
Required ratio $={\left(\frac{R}{24d}\right)}^{\frac{1}{3}}$