Capillary Tube and Surface Tension

[zorro]

Homework Statement

A glass capillary sealed at the upper end is of length 0.11m and internal diameter 2 x 10^-5 m. The tube is immersed vertically into a liquid of surface tension 5.06 x 10^-2 N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same? What will happen to the water level inside the capillary if the seal is now broken?

Homework Equations

The Attempt at a Solution

Let A be the area of cross section of the tube. Initially, the volume of air inside the tube is AL. Here pressure is Patm.
Where L is the length of the tube.
When is it dipped, capillary rise takes place till some height.
Let it be dipped through a distance 'x' when the level of water equalises outside level. Volume of air present= A(L-x)
I don't know what will be the pressure at this stage. Is it equal to Patm?

When the seal is broken, the water will rise upwards and adjust its meniscus ( I calculated the capillary rise as 1.03m). What is the formula for the radius of meniscus?

 

[rl.bhat]

When the tube is dipped, a bubble is formed at the tip. Excess pressure inside the bubble is 2T/R. The water will start rising when ρ*g*h1 = 2T/R, where h is the depth to which the tube is dipped. At this stage pressure P inside the tube is (Po + ρ*g*h1) and volume V = A*L.
When the levels of the water inside and outside are equal, you can apply P1V1 = P2V2
(Po + ρ*g*h1)* A*L. = [Po + ρ*g*(h1 + h2)]*A*(L.- h2)

 

[zorro]

When the tube is dipped, a bubble is formed at the tip. Excess pressure inside the bubble is 2T/R.

What is h2 in your expression?
Is the formation of bubble alone responsible for capillary action? I think water rises even when there is no bubble formed.

When the tube is dipped, a bubble is formed at the tip. Excess pressure inside the bubble is 2T/R. The water will start rising when ρ*g*h1 = 2T/R, where h is the depth to which the tube is dipped

In the formula for capillary rise,
H=2Tcos(theta)/rρg, it doesnot depend on the height to which the tube is dipped. Then how can you say that water will start rising when ρ*g*h1 = 2T/R?


Thanks for your reply

 

[rl.bhat]

When you try to dip the closed tube in the water, air pressure will not allow the water to rise. So only the middle portion of the water in the tube will rise forming a hemispherical liquid drop, with excess pressure 2T/R. When you dip the tube further, due to increased air pressure radius of the meniscus decreases. At a certain depth the liquid surface becomes flat. At that stage ρgh = 2T/R. If you dip further the liquid will start rising.

Till h1 the liquid is not rising in the tube. When you dip further till h2, the liquid level inside and outside will be the same.

 

[zorro]

Till h1 the liquid is not rising in the tube. When you dip further till h2, the liquid level inside and outside will be the same.

I still don't understand the difference between h1 and h2. Till h1 the liquid is not rising in the tube, which means that the level of water in the tube and outside is same. What I understood is that h1=h2!

 

[rl.bhat]

I still don't understand the difference between h1 and h2. Till h1 the liquid is not rising in the tube, which means that the level of water in the tube and outside is same. What I understood is that h1=h2!

Take a test tube and dip in the water up side down. And push it inside gently. You can see that level of water inside and out side are different.

Similarly in the case of closed capillary tube, because of the surface tension, water will not rise in tube until pressure due the depth water overcome the pressure due to surface tension, which is 2T/R. In the problem I have equated it to ρ*g*h1, where h1 is the length of the tube in side the water. When you push the tube further inside the water, the water will start rising up. At the depth of (h1 + h2) the level of water inside the tube will be in level with outside water.

 

[zorro]

On solving your equation, I got
h2 = L - h1 - Po/ρg
h1=2T/rρg h1=1.03 m
As L=0.11, It means that the tube is dipped far below the surface of water.
on Solving I got h2 as -10.92m, which seems absurd to me.

 

[rl.bhat]

My observations about this problem.
1) If the given capillary tube is open and dipped in the tube, the water will rise up to the top and stop there. There will be no over flow.
2) If you dip the closed tube, the pressure inside the tube be the atmospheric pressure. But the pressure at the bottom due to the meniscus will be less then the atmospheric pressure. So the water will not rise in the tube.
3) by dipping the tube further, within 11cm it may not be possible to achieve the equal level inside and outside.