An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension $0.095\ \mathrm{J/m^2}$ and density $10^3\ \mathrm{kg/m^3}$. The difference between pressure inside the bubble and atmospheric pressure is ______ $\mathrm{N/m^2}$. (Take $g=10\ \mathrm{m/s^2}$)

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $P_{in}-P_{atm}=ho gh+\frac{2T}r=10^3	imes10	imes0.2+\frac{2	imes0.095}{10^{-3}}=2000+190=2190\ \mathrm{N/m^2}$. Answer: 2190.

Explanation

Reviewed by Dr. Abubacker

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