An air bubble of radius $$0.1\,cm$$  is in a liquid having surface tension $$0.06\,N/m$$   and density $${10^3}\,kg/{m^3}.$$   The pressure inside the bubble is $$1100\,N{m^{ - 2}}$$   greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid?
$$\left( {g = 9.8\,m{s^{ - 2}}} \right)$$

Solution :
Let depth of bubble below the liquid surface, $$h = ?$$
As we know, $${\rho _{{\text{Excess}}}} = h\rho g + \frac{{2s}}{r}$$
$$\eqalign{ & \Rightarrow 1100 = h \times {10^3} \times 9.8 + \frac{{2 \times 6 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}} \cr & \Rightarrow 1100 = 9800\,h + 120 \Rightarrow 9800\,h = 1100 - 120 \cr & \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m \cr} $$