Calculating Pressure at Point A Using Excess Pressure Formula

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Homework Help Overview

The discussion revolves around calculating the pressure at a specific point in a capillary tube using the excess pressure formula. Participants are exploring the implications of surface tension and the geometry of the liquid surface in relation to atmospheric pressure.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the use of the excess pressure formulas and the relationship between the pressure at point A and atmospheric pressure. There are discussions about whether the pressure should be calculated as P₀ - hdg or if it should simply be P₀ due to the exposure to the atmosphere. The role of surface tension and curvature of the liquid surface is also examined.

Discussion Status

The conversation is active, with participants seeking clarification on the problem setup and the physical principles involved. Some have offered insights into the effects of surface tension on pressure differences, while others are still grappling with the implications of the given information.

Contextual Notes

The problem involves a capillary tube of radius r, and there are references to specific equations related to excess pressure in liquid drops and bubbles. The exact diagram mentioned is not provided, which may be affecting the clarity of the discussion.

Suraj M
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Homework Statement


Find the pressure at A.

WIN_20150212_161705.JPG

Homework Equations


Excess pressure(drop) = 2S/r
Excess pressure (bubble) = 4S/r

The Attempt at a Solution


Shouldn't it be = P₀ - hdg??
the answer is Option C.. but how?? why have they used the pressure in a drop of liquid?
 
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Well, it's pretty hard to tell what is going on. Have you got some more explanation of what that diagram represents?
 
Actually the question also says that it is a capillary tube of radius r. (it must be open)
so what i said : P≠P₀-hdg
Thats all the question says, DEvens.
since its just a point exposed to the atmosphere it should be P₀. But why subtract 2S/r ?
 
Suraj M said:
Actually the question also says that it is a capillary tube of radius r. (it must be open)
so what i said : P≠P₀-hdg
Thats all the question says, DEvens.
since its just a point exposed to the atmosphere it should be P₀. But why subtract 2S/r ?
Because the surface is curved, and there is surface tension acting within the surface. To get the surface to curve that way, the pressure on the concave side of the surface must be higher than on the convex side of the surface. The difference is 2S/r. This value follows from an equilibrium force balance, either on a part of the surface or on the entire surface.

Chet
 
Oh okay, understood. Thank you!
 

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