# Surface Tension in a liquid placed in a container

Recently I was analysing a Question given by my teacher.

Q. Let a Cylindrical container of radius R be filled with water filled up to height $h_0$. Hypothetically consider the water alone divided by a imaginary rectangular surface into two equal halves of semicircular cross section. Find the net force on one half of water due to the other half.
Given the liquid has Surface Tension T, density $\rho$. Take Atmospheric pressure as $P_{0}$ and acceleration due to gravity as g.

My attempt: (Neglecting the water rise in container due to surface tension)
I found the Net Force due to pressure on the rectangular cross section by integrating forces on isobaric infinitesimal rectangular cross sections as $F_{pressure}=\int\limits_0^{h_0}(P_0+\rho gh) 2Rdh= R(2P_0h_0+\rho gh_0^2)$
For Force due to surface tension I considered the force to be acting along the outward perpendicular on all the four edges of rectangle and found it to be $F_{ST}=2(2R+h_0)T.$
So the Net force along the normal (to rectangle) into the water is $F_{net}=2P_0Rh_0+\rho gRh_0^2-2T(2R+h_0)$

But my teacher said one should not consider the Force surface tension to be acting on the side and the bottom edge of the rectangle as the Force of surface tension acts only in "Free Surfaces" and the answer should be $F_{net}=2P_0Rh_0+\rho gRh_0^2-2TR$

But my argument is that the force of surface tension arises due to intermolecular forces in the boundary surfaces of a liquid so it doesn't really matter what is after the boundary air or the container. Hence one must consider the force of surface tension to be acting on all the four edges of the rectangle and end up in my answer.