Determine excess of pressure in side liquid drop

Explanation

Consider a liquid drop of radius r, having surface tension S. Letp be the excess pressure inside the liquid drop then the outward force `(F_(1))` acting.on the drop is given by
`F_(1)` Excess pressure `xx` surface area of drop
`F_(1)=pxx(pir^(2))" ".......(i)`
Now force `(F_(2))` due to the surface tension
`F_(2)`= Surface tension `xx` length = `Sxx` circumference of liquid.
`F_(2)=Sxx2pir" ".......(ii)`
This force acts towards the centre of the drop:
At equilibrium, `F_(1)=F_(2)`
From equations (i) and (ii), we have `pxxpir^(2)=Sxx2pir`
`p=(2S)/(r)`