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. $\frac{2S_{la}}{r}$
• B. $\frac{S_{la}}{r}$
• C. $\frac{4S_{la}}{r}$
• D. $\frac{2r}{S_{la}}$

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
$w=8\pi r\Delta r{S}_{la}...$(i)
where, $S_{la}$ is the surface tension of liquid air interface.
However, work done is
$W=\left(p_i-p_o\right)4\pi r^2\Delta r\dots$(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$p_i-p_o=\frac{2S_{la}}{r}$