First slide

Surface tension

Question

A water drop is divided into 8 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 $\frac{4}{3} {πR}^{3} = 8 \times \frac{4}{3} {πr}^{3}$
[Since, volume remains constant]
$r = \frac{R}{2}$
Since, excess pressure inside drop = $\frac{2 T}{R}$
[T-surface tension, R-radius]
Pressure difference between inner and outer surface of big drop will be half of that for smaller droplet.

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