Excess Pressure inside a Liquid Drop - QS Study
QS Study

Consider a liquid drop of radius r. The molecules on the surface of the drop experience a resultant force acting inwards due to surface tension. Therefore, the pressure inside the drop must be greater than the pressure outside it. The excess of pressure P inside the drop provides a force acting outwards perpendicular to the surface, to balance the resultant force due to surface tension. Imagine the drop to be divided into two equal halves.

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Considering the equilibrium of the upper hemisphere of the drop, the upward force on the plane face ABCD due to excess pressure P is P π r2 (Figure).

If T is the surface tension of the liquid, the force due to surface tension acting downward along the circumference of the circle ABCD is T 2πr.

At equilibrium, P πr2 = T2πr

P = 2T/r