The capillary action formula is given as:
$$h=2T/ρgR$$
where h is the capillary rise, R is the radius of curvature of miniscus, ρ is density, g is force of gravity and T is the surface tension.
Now consider a capillary with a large base which gets narrower as we move along the top, such that the radius of miniscus at the top has a radius R,
now consider a tube of uniform radius made of same material such that it too allows a miniscus of radius R to form, looking at the formula both capillaries will have the same height of liquid in them but clearly the first one would have a greater weight of liquid suspended. What factors account for that? or is it that my utilization of the formula wrong, if so what is the proper interpretation?
[Physics] Doubt in capillary rise
capillary-actionsurface-tension
Related Solutions
The formula for capillary rise that most people know is easily derived through a pressure balance between the capillary pressure and the hydrostatic pressure. The hydrostatic pressure equals $$\Delta P_h=\rho g h$$ whereas the capillary pressure is $$\Delta P_c=\frac{2\gamma}{R}=\frac{2\gamma \cos \theta}{r}$$ So balancing these we get our 'famous' equation: $$h=\frac{2\gamma\cos\theta}{\rho g r} $$ Now we have a situation in which the height of our tube above the liquid, $h_{max}$ is smaller than $h$. For an equilibrium situation we still need the hydrostatic pressure and the capillary pressure to balance so we plug in the maximum height that we can get, $h_{max}$, and get: $$h_{max}=\frac{2\gamma \cos\theta_p}{\rho g r}$$ Note that I have changed $\theta$ into $\theta_p$ (see figure below), because that is in fact the only thing that can change, all the other parameters are fixed properties of the system.
It is not entirely clear how you define $A$, but if we define $A$ as the situation for which $\theta_p=\theta$ then changing $\theta_p$ would result in a gradual shift from $A$ to $B$ depending on the value of $h_{max}$. In fact, $B$ is the limit for $h_{max}=0$, because $\cos \pi/2=0$, in which case, as pointed out by Olin and can be seen from the capillary pressure equation, no hydrostatic pressure can be sustained. The fact that $\theta$ can become bigger (i.e. change into $\theta_p$) is caused by contact angle hysteresis at the rim of the capillary tube.
Your conjecture is correct: The rise of a liquid in a capillary is not just a function of the liquid-air surface tension but also the liquid-solid surface energy, AND this liquid-solid surface energy is present in the equation and its effect is represented by the contact angle parameter in the capillary rise equation.
Derivation of the capillary rise equation appears to be a bit involved, but there seems to be a good description here: Capillary Rise Equation Derivation
Note Figure 8.1 in the linked document: Even though it is the same liquid and same air for the two capillaries shown, in one capillary the rise is positive while in the other the rise is negative. The difference between the two capillaries? In one the contact angle is positive while in the other the contact angle is negative, presumably because the two capillaries are made of different materials so they have different liquid-solid surface energies.
Best Answer
Surface tension produces a pressure difference across the curved interface between the liquid and the air and it is this difference in pressure which results in the capillary rise.
The pressure is atmospheric where the dots are black and less than atmospheric by an amount $\frac {2T}{R}$ where the dots are red.
Here $T$ is the surface tension and $R$ is the radius of the tube at the air-liquid interface.
To ensure that the pressure where the bottom black dots are is atmospheric the liquid rises so that the increase in pressure due to the liquid level being higher is equal to the decrease in pressure at the top due to the effect of surface tension.
It is atmospheric pressure which keeps the columns of liquid in position with surface tension and the radius of the tube at the air-liquid interface which determines how high the column of liquid will rise.
Note that I have assumed that at the top the air-liquid interface is in contact with a vertical tube.