
#1
Nov1112, 03:23 PM

P: 42

As the question states I have hard time understanding that how does a liquid reach equilibrium in case of capillary depression (mercury for example ). I know how it goes down but not how it 'stops'.
like in case of liquids which have more force of adhesion ,as the liquid goes up because of the forces, it's balanced by it's own weight. what balances the force pulling mercury down? details please? 



#2
Nov1112, 03:41 PM

P: 3,554





#3
Nov1112, 03:46 PM

P: 42

can you please elaborate a bit on it?




#4
Nov1112, 07:34 PM

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Mechanics behind capillary depression.
Notice the angle that the water makes with the wall of the capillary compared to the mercury. Each fluid has its own characteristic contact angle. In the case of a wetting fluid like water, the contact angle is positive, and, in the case of a nonwetting fluid like mercury, the contact angle is negative. (Actually, the contact angle depends on the surface involved, and the presence of contaminants in the liquid). If the contact angle is positive, the fluid will rise, and if the contact angle is negative, the fluid level will fall. Do the force balance and see how it works.




#5
Nov1212, 01:42 AM

P: 42

actually I tried the force balance but where I got stuck was that which force balanced the declining mercury in the capillary?




#6
Nov1212, 03:33 AM

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#7
Nov1212, 07:18 AM

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#8
Nov1212, 02:41 PM

P: 42





#9
Nov1212, 07:11 PM

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#10
Nov1312, 03:43 PM

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#11
Nov1312, 04:43 PM

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#12
Nov1512, 05:32 AM

P: 42

okay folks, after racking my brains I think I know what's happening but I just wanted to see if my assumptions are right I'm looking at it this way.
a beaker contains mercury and a capillary is introduced. now due to surface tension and the convex meniscus, there's excess pressure at a point b just below the level of meniscus in the capillary. at the same level outside the pressure is say P. taking this as point a pressure at point a = P pressure at point b = P + 2*sigma/R where R is the radius of the meniscus. so in order to balance this pressure, the mercury is pushed down till it reaches a point where the outside pressure too is P+2*sigma*R. 



#13
Nov1512, 06:39 AM

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Yes. That sounds right.




#14
Nov1512, 08:59 AM

P: 42

yes. but now I think I found another doubt, now in the way that I'm imagining it, there can be two paradoxes.
1)what happens to the pressure of the air column. does it change? or is it just negligible? 2)when the liquid in capillary goes up/down doesn't the height of the liquid outside changes? or is it again too small for considerations. 



#15
Nov1512, 03:12 PM

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Answer to #2: What really matters is the difference in height between the mercury outside the capillary and the mercury outside the capillary, and not the absolute heights. 



#16
Nov1512, 05:06 PM

P: 42




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