Question

A water drop is divided into eight equal droplets. The pressure difference between inner and outer sides of the big drop:

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

Answer 1

The correct option is B Will be half of that for smaller droplet

Suppose, $R=$ radius of water drop
and $r=$ radius of droplets
$\therefore \frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3$
(Since volume remains constant)
$\therefore r=\frac{R}{2}$
Since excess pressure inside drop $=\frac{2T}{R}$
$(T-surfacetension,R-radius)$
Therefore, pressure difference between inner and outer surface of big drop will be half of that for smaller droplet.