# Surface tension - liquid rising in capillary tube

1. May 29, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Honestly, I don't have any idea on this. As far as I remember, the usual practice is to equate the excess pressure equal to $2T/R$ (where T is the surface tension and R is the radius of curvature, this may be wrong though) but I don't know how should I calculate the excess pressure here.

At height h below the surface of liquid, is the pressure $P_o+\rho_1gh$ or $P_o+\rho_2gh$? ($P_o$ is the atmospheric pressure)

Any help is appreciated. Thanks!

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2. May 29, 2013

### haruspex

What do you think the pressure is:
- at the liquid boundary?
- inside the tube at the bottom?
- inside the tube just below the surface?
The 2T/R formula is correct.

3. May 29, 2013

### Saitama

I am still confused.

Atmospheric pressure, $P_o$?

I don't know about this as I said before, is it $P_o+\rho_1gh$ or $P_o+\rho_2gh$?

4. May 29, 2013

### haruspex

No, I meant at the boundary between the two liquids.
If it were different from the pressure at the boundary between the two liquids (which is at the same horizontal height as the bottom of the tube) what would happen?

5. May 29, 2013

### Saitama

Is it any different from $P_o$? Why?

I am still clueless. :(

6. May 30, 2013

### haruspex

Because that boundary is a distance h down from the top of the lighter liquid. So the pressure there is P0+what?
This applies everywhere at that boundary, which includes right next to the lower end of the tube. If the pressure were any different just a fraction away, directly in the lower end of the tube, wouldn't you expect something to flow?

7. May 30, 2013

### Saitama

$P_o+\rho_2gh$?

In case I am misinterpreting your replies, I will show you by an image what I am thinking.
You ask me the pressure at point A. It is $P_o+\rho_2gh$. What about the pressure at point B? Is pressure here also equal to $P_o+\rho_2gh$? Why it can't be $P_o+\rho_1gh$ as the point is height $h$ below the liquid of density $\rho_1$?

Sorry if I am missing something obvious.

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8. May 30, 2013

### haruspex

That's where you are going wrong. You are forgetting that it will be affected by the surface tension at the liquid/air boundary. This is why I was trying to get you to work from the other end, answering for the pressure at B first. The answer for that cannot be affected by the presence of the heavier liquid.

9. May 30, 2013

### Saitama

Thank you haruspex for the explanation.

Is pressure at A equal to $P_o+\rho_2gh+\frac{2T}{R}$? And the last step is to equate pressures at A and B?

10. May 30, 2013

### Tanya Sharma

Hi Pranav...

The pressure at A will be $P_o+\rho_2gh-\frac{2T}{R}$.

The pressure just below the top of capillary at the air liquid interface will be $P_o-\frac{2T}{R}$.
The pressure at height h below , i.e point A will be $P_o+\rho_2gh-\frac{2T}{R}$.

Now equate this pressure with that at point B and you get the answer .

Last edited: May 30, 2013
11. May 30, 2013

### Saitama

How do you get the minus sign there?

12. May 30, 2013

### Tanya Sharma

The shape of the liquid meniscus will be concave .The excess pressure 2T/R is always on the concave side of the meniscus.The pressure just above the air liquid interface is P0 .Hence,pressure just below the air liquid interface will be P0 - 2T/R .

13. May 30, 2013

### Saitama

Sorry, this is a dumb question but what does this statement mean?

14. May 30, 2013

### Tanya Sharma

Whenever you place a capillary in a liquid ,the liquid surface will either have a concave or a convex surface depending on the intermolecular forces between the liquid and the surrounding solid .If it has concave surface then the liquid rises.In case it is convex,liquid depresses within the tube .In this case we can see since the denser liquid has risen in the tube,we may infer that the meniscus is concave and the excess pressure is on the concave side i.e above the air liquid interface .

Last edited: May 30, 2013
15. May 30, 2013

### haruspex

To put it in less technical terms, if the surface goes up at the sides it's because surface tension is pulling it up the sides of the tube. This pull is opposing atmospheric pressure from above, so the pressure in the liquid just below it is less than atmospheric.

16. May 31, 2013

### Saitama

Thank you both for the help.