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enter image description here

For the above question, they've equated the excess pressure across the meniscus to the pressure due to height of liquid column in the capillary, as follows:

enter image description here

This clearly gives RH = constant, hence the graph shown in option (B) matches - a very familiar rectangular hyperbola.

However, there a certain physical aspects of this problem that I don't understand -

  1. Is the process quasi static?

  2. Why should the forces due to pressure be balanced, won't the column accelerate if pulled up?

  3. Will the contact angle change, practically, while the tube is being pulled up? Is assuming it to be constant simply an assumption to avoid complexity?

  4. How does the process actually happen, that is, what can be said about the flow of column in the tube if the tube is being pulled by a constant velocity, (say, v) or a constant acceleration?

Could someone please give a detailed explanation of the process, so that all the physical aspects are addressed as well? Simply a mathematical equation didn't really help me understand the process in question.

The equation RH = constant, is it correct or is it just an approximation? Also, if it is indeed right, or valid under certain assumptions and approximations, how can one intuitively work towards deducing it? The equation is that of a rectangular hyperbola, which is pretty familiar - so I guessed that there must be some intuitive or non mathematical approach to the problem.

Thanks in advance.

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  • $\begingroup$ I have the same problems with this that you have. What they are saying doesn't make sense to me, particularly the equation for the pressure difference as a function of the radius of curvature, which is valid strictly for a case in which the contact angle is zero; yet they are saying that the radius of curvature is not equal to r. $\endgroup$ – Chet Miller Feb 20 '18 at 13:31
  • $\begingroup$ Exactly. How can we possibly find the answer to this question? Many people saw it, none gave an answer. $\endgroup$ – arya_stark Feb 20 '18 at 13:58
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Maybe they (a) relax the requirement that the contact angle be zero except in the final equilibrium position and (b) assume that the interface is spherical, with the radius of curvature equal to R. Under these circumstances, the contact angle would be given by: $$\cos{\phi}=\frac{r}{R}$$And the quasi static equilibrium of the column would be: $$(2\pi r \cos{\phi})T=\rho g H$$So, under these assumptions, combining these two equations gives $$RH=\frac{2\pi r^2T}{\rho g}$$And, when R decreases to r, the contact angle would be zero (and the column would stop rising).

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  • $\begingroup$ This does not assume that the contact angle is constant. It assumes that the radius of curvature R is decreasing as the column height is increasing, so that the contact angle starts out large and decreases to zero as time progresses. This does not give the time progression of height and radius of curvature. It just assumes that a quasi static condition exists as time progresses. If you want to include the time history in the model, you need to include the force balance on the fluid, and the viscous drag on the flow, and you need to include possible non-sphericity of the free surface. $\endgroup$ – Chet Miller Feb 20 '18 at 16:03

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