Question

. Derive formula for pressure difference between inside & outside the bubble in air Pi -4T/r
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Answer 1

Answer:

Suppose that ABC is a half drop of a liquid, whose radius is r and surface tension is T. On the molecules of liquid towards the spart, the cohesive force F¹

would be downwards.

Suppose, excess pressure is P which acts upwards.

$therefore \: upwards \: force \: f {}^{2} \: = p \: a = p\pi \: r \: {}^{2}$

∴ For the equilibrium of the liquid drop

$f {}^{1} = f {}^{2} \\ \\ 2\pi \: r t \: = p\pi \: r {}^{2} \\ \\ p = \frac{2t}{r}$

Thus, the pressure inside the drop $\frac{2t}{r}$ more than the pressure outside the drop.