There is an air bubble of radius 1.0 mm in a liquid of surface tension `0.075Nm^-1` and density `1000 kg m^-3`. The bubble is at a depth of 10 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? Take `g=9.8ms^-2`.

Explanation

Let the atmospheric pressure be `P_0`. The pressure of the liquid just outside the bubble is ure.
`P=P_0+hrhog`
The pressure inside the bubble is
`=P'+P+(2S)/r=P_0+hrhog+(2S)/r`
`P'-P_0`
`=(10cm)(1000 kgm^-3)(9.8ms^-2)+(2xx0.075Nm^-1)/(1.0xx10^-3m)`
`=980Nm^-2+150Nm^-2` ltbr. `=1130Pa.