Question

the excess pressure inside an air bubble of radius r just below the surface of water is p1 . the excess pressure inside a drop of same radius just out side the surface p2 .If T is the surface tension,then relation between p1 and p2 ....with reason plzzz ans

Answer 1

The relation between P1 and P2  is P2 = 2T / r

Explanation:

Since, an air bubble has 2 air-liquid interface.

Therefore the excess pressure inside a bubble is a result of 2 surfaces rather than just 1 in case of drop.

Excess pressure inside a bubble just below the surface of water "P1" = 2T/r

Excess pressure inside a drop P2 = 2T / r

P1 = P2

Thus the relation between P1 and P2  is P2 = 2T / r

Also learn more

What is air pressure?

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Answer 2

It has no component perpendicular to the horizontal surface. As a result, there is no pressure difference between the liquid side and the vapour side. ... Thus, there is always an excess of pressure on the concave side of a curved liquid surface over the pressure on its convex side due to surface tension.