Question

A small air bubble of radius 'r' is at a depth 'h' below the water surface (density of water= rho). Surface tension of water is T, atmospheric pressure is p_(0). Find pressure inside the air bubble for the condition r lt lt h

Answer 1

Explanation:

Given data

Radius of the bubble = r

Depth below water = h

Surface tension of water = T

Let pressure at h depth =  ph

Atmospheric pressure = P0

From surface tension formula we get  

Pressure in the bubble =2T/r

$apply\,pressure\,equation \\ P-{{P}_{h}}=\frac{2T}{r} \\ P=\frac{2T}{r}+{{P}_{h}} \\ P={{P}_{0}}+\rho gh+\frac{2T}{r}$

So pressure inside the bubble is ${{P}_{0}}+\rho gh+\frac{2T}{r}$.

Answer 2

The pressure inside the air bubble for the condition r<<h is P+ρgh+(2T/r)

• Let us go through the given data

       Radius=r

      Depth=h

      Density=ρ

     Surface tension=T

     Atmospheric pressure=Po

• The pressure outside the surface of air bubble would be P+ρgh(Where P is atmospheric pressure)

• The pressure due to the water on the air bubble would be (2T/r) as  water is in contact with the air bubble it exerts pressure on the bubble

• Total pressure inside the air bubble = P+ρgh+(2T/r)

• Therefore,The pressure inside the bubble is P+ρgh+(2T/r)