Capillary action column height in a tube fitted inside another tube?

I was thinking about how would capillary action change in a tube (classic example) and in a tube fitted inside another tube (considering water as the liquid involved).

Height of liquid column: where:

$\gamma$ = liquid-air surface tension

$\theta$ = contact angle

$\rho$ = density of liquid

$g$ = gravity acceleration

$r$ = radius

I tried my best to draw the examples I'm interested in order to help my explanation.

I didn't consider the capillarity inside the smaller tube in both example #2 and #3 because I'd like to assume that "a/2" in example #1 is close to "c" in example #2 and #3 (drawings not to scale). Since from what I understand the column height is given, among other things (most of which can't be changed, like liquid-air surface tension, contact angle, density of liquid and gravity acceleration), by the tube radius, I'd like to know if "c" in example #2 can be considered as "a/2" in example #1 to calculate column height using above formula.

Also I'd like to know how having beads of slightly smaller diameter than "c" between the two tubes (example #3) would affect the column height.

If said beads were less dense than water, could they still improve column height or would they just form a floating mat on top of 1 unit thickness?

What'd be the column height of example #2 and #3 assuming "c" as 1mm ?

I'm quite sure that given the same reached height "h" in example #2 and #3, "c" of #2 has to be smaller than "c" in #3, though it's apparently the opposite (beads lower the water column) and pointed at sciencedirect article by another user.

Edit: I've been told that If $c\ll R$ the radius of your outer tube, the total curvature is approximately $\cos \theta/c$, so you will get $$h = \frac{\gamma}{\rho g} \frac{\cos \theta}{c}.$$

Beads will usually lower the apparent surface tension, so you'll get a lower column, although the amount of that depends on their wetting properties and of their arrangement (packing)

However by $c \ll R$ which orders of magnitude/fractions are we talking about?

I guess $c$ still has to be smaller than water's capillary length (about 2.7mm), right?

If $c$ is slightly smaller than the outer tube radius (like 1/6), how'd total curvature and hence Jurin's law (above formula) be affected?

Thank you very much

• Assuming $\theta$ the same for each cylinder wall in example #2, I derive a height two times the last you give, i.e., Jurin's law with $c$ instead of $r$. Vertical force is $2 \pi a \gamma \cos \theta + 2 \pi (a-c) \gamma \cos \theta$, and that keeps the weight of a volume of water that is $\rho g {V} = \rho g \pi [a^2-(a-c)^2]h$. – Andrestand Jan 25 '15 at 21:23