Solveeit Logo
Login

Question

Question: The excess pressure inside an air bubble of radius r just below the surface of water is P<sub>1</sub...

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 the same radius just outside the surface is P2. If T is surface tension then

A

P1=2P2P _ { 1 } = 2 P _ { 2 }

B

P1=P2P _ { 1 } = P _ { 2 }

C

P2=2P1P _ { 2 } = 2 P _ { 1 }

D

P2=0,P1≠0P _ { 2 } = 0 , P _ { 1 } \neq 0

Answer

P1=P2P _ { 1 } = P _ { 2 }

Explanation

Solution

Excess pressure inside a bubble just below the surface of water P1=2TrP _ { 1 } = \frac { 2 T } { r } and excess pressure inside a drop ∴P1=P2\therefore P _ { 1 } = P _ { 2 }