Force due to surface tension in calculation of excess pressure inside a liquid drop In the case of equilibrium of hemispherical drop, what i know is that the surface tension is responsible for holding the hemispherical drop with the other hemisphere thus forming a spherical drop and the surface tension also holds the liquid on the curved surface thus giving it the curved shape.
Why is force due to surface tension measured only along periphery(T2πr) whereas the surface tension exists also on the surface(2πr²) ?
Equilibrium condition: Atmospheric pressure pushing the curved surface along with the surface tension from the outside and pressure inside the liquid acting in the opposite direction to both these forces thus maintaining an equilibrium
 A: I think I understand your question. You wish to ask why the surface tension force acts only on the periphery (circumference) of the hemispherical water droplet. Well, it doesn't. It acts all over the surface, at every point. To understand the answer, draw any given line of a very small length (say $dx$) anywhere on the surface. The force $Tdl$ acts on it due to tension by its neighbouring molecules, say in a given direction (right, say). But it too exerts a net equal and opposite surface tension force $Tdl$ (towards the left). So the net force on the small line is zero. This argument can be extended to the whole surface. The only region where this line experiences a net non-zero force is the periphery. So the net force due to surface tension is exerted on the periphery as shown and is given by (assuming surface tension to be constant) $$F= \int{Tdl} = T \cdot 2\pi R$$  
A: Excess pressure inside a liquid drop
In the calculation you consider the balance of forces on a hemisphere of liquid. The only forces which are relevant are external forces which other objects exert on this hemisphere. These include the pull of surface tension $\sigma$ from the other hemisphere  at the rim where they join, and the pressure $p$ from the liquid inside the other hemisphere acting across the flat face of this hemisphere (the yellow area in the diagram). (Also the atmospheric pressure across the blue surface, which I have not shown. So in effect in my diagram $p$ is the excess pressure.)
It is not necessary to consider the surface tension forces at other points in the surface of this hemisphere (the blue area in the diagram) because these are internal forces. They always occur in equal and opposite pairs between adjacent elements of water which are both in this hemisphere. It is only along the circular rim of the hemisphere that the opposite paired surface tension forces are missing because they act on water which is in the other hemisphere not this one.

A: The surface acts like a stretched membrane. The attachment point of that membrane is the triple junction between the solid surface, the air, and the liquid surface. The summation of  the forces involved at that triple point is what results in the contact angle of the liquid on the surface. 
