First slide

Surface tension

Question

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

Easy

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

Solution

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

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