A water drop is divided into 8 equal droples. The pressure difference between innere outer sides of the log drop:

Let the initial volume of water drope be V and

radius R

Let the final volume of water drop be V and

radius 9 .

$\begin{array}{rl}V& =8v\\ \frac{4}{3}\pi R^3& =8\times \frac{4}{3}\pi r^3\\ R^3& =8\times r^3\\ R& =2r\\ P_1-P_2& =\frac{4T}{R}[\text{ here, }T\text{ is surface ]}\\ P_1^{\prime }-P_2^{\prime }& =\frac{4T}{r}=\frac{4T\times 2}{R}\end{array}$

Clearly, pressure difference between outer and inner surface double it's value.