Question

There is an air bubble of radius r inside a drop of water of radius 3r. Find the ratio of gauge pressure at point b to that at point a.

Answer 1

Thank you for asking this question. Here is your answer:

We will let the pressure at A be P

and the pressure at B will be P'

And the pressure of atmosphere will be P₀

Now we will calculate the pressure difference relation:

P' - P₀ = 2s/3R

Now the gauge pressure  B = P' - P₀ = 2s/3R

P - P' = 2s/R

P - P₀ = 2S/R + 2S/3R

Now gauge pressure A = P - P₀

= 2S/R + 2S/3R = 8S/3R

The ratio of gauge pressure  = 8S/3R / 2S/3R

= 4:1

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