Physics - Fluid Question with Solution | TestHub

Let
$r$
be the radius of the water drop. The volume of the shell equals the volume of the drop.
$\frac{4}{3}\pi [R^3-(R-d{)}^3]=\frac{4}{3}\pi r^3$
$r^3\approx 3R^{2}d(d\ll R)$
The ratio of excess pressures is
$\frac{2\sigma /r}{4\sigma /R}=\frac{R}{2r}$
$\frac{R}{2(3R^{2}d{)}^{1/3}}={\left(\frac{R}{24d}\right)}^{1/3}$