Question

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

• A. will be the same as for smaller droplet
• B. will be half of that for smaller droplet
• C. will be one-forth that for smaller droplet
• D. will be twice of that for smaller droplet

Answer 1

The correct option is D will be twice of that for smaller droplet

Let the initial volume of water drope be V and

radius R

Let the final volume of water drop be V and

radius 9 .

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

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