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Surface tension - liquid rising in capillary tube

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  • #1
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1. Homework Statement
attachment.php?attachmentid=59131&stc=1&d=1369856718.jpg



2. Homework 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##? :confused: (##P_o## is the atmospheric pressure)

Any help is appreciated. Thanks!
 

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  • #2
haruspex
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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
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I am still confused.

- at the liquid boundary?
Atmospheric pressure, ##P_o##?

haruspex said:
- inside the tube at the bottom?
I don't know about this as I said before, is it ##P_o+\rho_1gh## or ##P_o+\rho_2gh##? :confused:
 
  • #4
haruspex
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Atmospheric pressure, ##P_o##?
No, I meant at the boundary between the two liquids.
I don't know about this as I said before, is it ##P_o+\rho_1gh## or ##P_o+\rho_2gh##? :confused:
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
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No, I meant at the boundary between the two liquids.
Is it any different from ##P_o##? Why? :confused:

haruspex said:
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?
I am still clueless. :(
 
  • #6
haruspex
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Is it any different from ##P_o##? Why? :confused:
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
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Because that boundary is a distance h down from the top of the lighter liquid. So the pressure there is P0+what?
##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##?
attachment.php?attachmentid=59145&stc=1&d=1369891679.png

Sorry if I am missing something obvious.
 

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  • #8
haruspex
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You ask me the pressure at point A. It is ##P_o+\rho_2gh##.
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
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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.
Thank you haruspex for the explanation. :smile:

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
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Hi Pranav...

The pressure at A will be [itex]P_o+\rho_2gh-\frac{2T}{R}[/itex].

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

Now equate this pressure with that at point B and you get the answer .
 
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  • #11
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Hi Pranav...

The pressure at A will be [itex]P_o+\rho_2gh-\frac{2T}{R}[/itex].

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

Now equate this pressure with that at point B and you get the answer .
How do you get the minus sign there? :confused:
 
  • #12
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The shape of the liquid meniscus will be concave :smile: .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
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The excess pressure 2T/R is always on the concave side of the meniscus.
Sorry, this is a dumb question but what does this statement mean? :confused:
 
  • #14
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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 .
 
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  • #15
haruspex
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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
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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.
Thank you both for the help. :smile:
 

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