Due to surface tension, the excess pressure inside a smaller drop is 9 units. If 27 smaller drops combine then excess pressure inside the bigger drop is

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Excess pressure inside a liquid drop of radius $r$ is given by $P = \frac{2 T}{r}$ .

For the smaller drop, $P_{1} = \frac{2 T}{r} = 9$ units.

When $27$ smaller drops combine to form a bigger drop of radius $R$ , the total volume remains conserved.

$\frac{4}{3} π R^{3} = 27 \times \frac{4}{3} π r^{3}$

$R^{3} = 27 r^{3} \Rightarrow R = 3 r$

The excess pressure inside the bigger drop is $P_{2} = \frac{2 T}{R}$ .

Substituting $R = 3 r$ , we get:

$P_{2} = \frac{2 T}{3 r} = \frac{1}{3} (\frac{2 T}{r})$

$P_{2} = \frac{1}{3} \times 9 = 3$ units.

Answer: $3$ units

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