First slide

Surface tension and surface energy

Question

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

Moderate

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)
$∴ r = \frac{R}{2}$
Excess pressure inside big drop $= \frac{2 T}{R}$
Excess pressure inside small drop $= \frac{2 T}{r} = \frac{2 T}{\frac{R}{2}} = \frac{4 T}{R}$
( 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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