Question

In figure below, pressure inside a spherical drop is more than pressure outside.
If a liquid drop is in equilibrium, then the pressure difference between the inside and outside of the drop is

• A. $\frac{2 S_{l a}}{r}$
• B. $\frac{S_{l a}}{r}$
• C. $\frac{4 S_{l a}}{r}$
• D. $\frac{2 r}{S_{l a}}$

Answer 1

Answer: (A)

If a liquid drop is in equilibrium, then energy lost is balanced by the energy gain due to expansion under the pressure difference $\left(p_{i}-p_{o}\right)$ between the inside of the drop and the outside.
Initial surface area of liquid drop $=4 \pi r^{2}$
Final surface area of the liquid drop
$=4 \pi(r+\Delta r)^{2}$ $=4 \pi r^{2}+8 \pi r \Delta r$
( $\Delta r^{2}$ is very small and hence neglected)
Increase in the surface area of liquid drop
$=4 \pi r^{2}+8 \pi r \Delta r-4 \pi r^{2}=8 \pi r \Delta r$
External work done is increasing the surface area of the drop
$w=8 \pi r \Delta r S_{l a}\,\,\,...$(i)
where, $S_{l a}$ is the surface tension of liquid air interface.
However, work done is
$W=\left(p_{i}-p_{o}\right) 4 \pi r^{2} \Delta r \,\,\,\ldots$(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$p_{i}-p_{o}=\frac{2 S_{l a}}{r}$