Question

derive an expression for excess pressure inside a drop and Bubble
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Answer 1

Answer:

This excess pressure acts outwards. Let the radius of the drop increase from r to r+dr r + d r under the excess pressure p . Neglect dr2 d r 2 because it is very small. Thus, the expression of excess pressure inside a liquid drop is p=2σr p = 2 σ r .

Explanation:

Hope it helps you

Answer 2

Answer:

p=2ò r

Explanation:

This excess pressure acts outwards. Let the radius of the drop increase from r to r+dr r + d r under the excess pressure p . Neglect dr2 d r 2 because it is very small. Thus, the expression of excess pressure inside a liquid drop is p=2σr p = 2 σ r .