How is the water meniscus at the edge of a capillary tube

Suppose we have a capillary tube in which water can rise to a height of x cm. If we dip the tube such that the height above the surface is less than x, then how will the water meniscus be at the edge of the tube? Why?

Excuse the flat water at the surface near the base of the capillaries.

EDIT: @NewAlexandria Here's my reasoning.

A is the most likely case as once the water molecules at the inner circumference reach the edge, they cannot go any further up as there is no glass to give the needed normal reaction (cosine component). That's all I can think of.

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Sounds a bit like homework –  New Alexandria Dec 6 '13 at 14:48
It isn't, should I add my rationalizing –  user80551 Dec 6 '13 at 17:18
I'd probably make a more-rich question then, yes –  New Alexandria Dec 6 '13 at 17:35
Edited the question, is it ok now? –  user80551 Dec 6 '13 at 18:38

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.

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You should point out that theta can never be 90deg at any height above the regular water surface. At 90deg, the upward component of the pulling force is 0, so nothing would be holding the water up. The answer to the question as posed has to always be A. –  Olin Lathrop Dec 7 '13 at 13:57
I did point out that B can only occur for hmax equals 0. And I think you interpret A somewhat different than I do. You seem to interpret A as any downward curved interface, while I interpret it as the curve you get from the equilibrium contact angle. So e.g. A has a 45 degree angle, then there is depending on hmax, any value between 45 and 90 (B) possible. –  Michiel Dec 8 '13 at 9:37
Moreover, the question is posed as " If we dip the tube such that the height above the surface is less than x". Hmax is 0 still fits that description, so B is still an option –  Michiel Dec 8 '13 at 9:40
@OlinLathrop - although I agree that $h_{max}=0$ is a bit of a limiting case –  Michiel Dec 8 '13 at 14:42
@OlinLathrop As pointed, there is a capillary pressure $\Delta P_c$ Couldn't that hold the water up even if there is no force component in the upward direction making case B possible? –  user80551 Dec 8 '13 at 15:59

The answer is A. Think of what capillary action really is. It does not pull on the bulk of the water. It is a edge effect that pulls on the meniscus.

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