Why is there a pressure difference of $h\rho g$ at the top of tube between inner and outer surface in capillary rise?

Question

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..?

Source

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?

Answer 1

As well as being the pressure in the air just above the meniscus, $P_\text 0$ is also the pressure at the water surface in the bowl (neglecting pressure changes in the air over height $h$). So the equation that you quote simply gives the hydrostatic pressure difference in the liquid column – and that is the same as the pressure difference across the meniscus!

Answer 2

There is a formula here

https://thefactfactor.com/facts/pure_science/physics/numerical-problems-on-capillary-action/5329/

That gives the height of the capillary column

$$h = \frac{2T\cos\theta}{r \rho g}\tag1$$

The surface tension will have to support the weight of the capillary column, that's where the pressure difference comes in, it's $h \rho g \times \pi a^2$ where $a$ is the radius at the bottom.

This is just an idea, but perhaps formula 1) came from this derivation, putting the up and downward forces equal $$\pi r^2 h \rho g = Tcos \theta 2 \pi r \tag 2$$

where the left $r$ will later be replaced with $a$ and the right $r$ replaced with $b$, although $a$ might need to be found in terms of $h$ and $\theta$...best of luck.

P.S. Think your question was on the internet, here ( https://www.concepts-of-physics.com/mechanics/capillary-rise.php ), still understood best by putting upward forces equal to downward forces, (their pressure method does seem confusing).

You'll also need to change $\theta$, (as it should be the angle to the vertical), to $\theta + \frac{\alpha}{2}$, then the answer will match, if the first $r$ is left as $b$ (although $a$ seems better, as the sides of the glass support the weight of the extra area), but at least that way you'll get a match to the given answer.