A glass capillary tube is of the shape of a truncated cone with an apex angle alpha so that it's two ends have cross sections of different radius. When dipped in water vertically, water rises to a height h , where the radius of cross section is b. If the surface tension of water is S, its density $\rho$ , and it's contact angle with glass is $\theta$, the value of $h$ will be..?
In the very first line of the solution to the problem, it is taken that $P_0 -P_1 = \rho gh$
Where $P_0$ is the pressure just outside and $P_1$ is the pressure inside
But I don't get it, $\rho g h$ is the pressure difference between the top most point of the beaker open and the bottom most point near which the edges converge, how does it also give the pressure difference between right outside and inside the water surface open to atmosphere?