What Happens to Liquid Height in Short Capillary Tubes?

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The discussion focuses on the behavior of liquid height in short capillary tubes, specifically when the tube length is less than or equal to the height of the liquid (l ≤ h). The height of the liquid in the tube is determined by the formula h = (2T cos θ) / (ρgR), where T is surface tension, θ is the contact angle, ρ is liquid density, and R is the tube radius. When the tube is insufficiently long, the liquid does not spill over, indicating that adhesive forces are insufficient to pull the liquid further up. The angle of contact remains unchanged despite the reduced volume of liquid in the tube.

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I don't know if there is a special place for fluids. so i'll post it here...

When a tube of small radius is placed in a liquid, the liquid rises or falls in the tube, due to adhesive and cohesive forces. also the height of the liquid in the tube is given by

h = \frac { 2 T cos\theta}{\rho g R}.

where T is surface tension, \theta is the angle of contact, \rho is the density of the liquid, R is the radius of the capillary tube..

However, if a tube a insufficient length is taken, i.e l <= h then something happens. i am not sure what...but i think, the liquid does NOT spill over, i.e the full height of the liquid is not reached... is it because there is no more adhesive forces to pull ther liquid further up the tube?? in such a case, does the angle of contact remain the same??

is it true?
please explain why, too.
thank you.
 
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geniusprahar_21 said:
I don't know if there is a special place for fluids.

Yeah, it's a pity there is no specific place here for fluids but there are places for strings, membranes and so on , even you can check there are several posts about fluids dispersed in PF.

When a tube of small radius is placed in a liquid, the liquid rises or falls in the tube, due to adhesive and cohesive forces. also the height of the liquid in the tube is given by

h = \frac { 2 T cos\theta}{\rho g R}.

where T is surface tension, \theta is the angle of contact, \rho is the density of the liquid, R is the radius of the capillary tube..

.

That is a steady solution. If let's say, once reached the dynamical equilibrium you cut the top of the tube such as you cut some portion of the meniscus, some amount of liquid will flow out the tub in order to satisfy the constraint of mechanical equilibrium and the angle of contact \theta. The height of the meniscus would be the same but with some less fluid inside the tube.
 
thnx, thts just what i needed...
 

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