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.

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 r^{2} (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 πr ^{2} = T2πr**

**P = 2T/r**