Calculating Pressure at Point A Using Excess Pressure Formula

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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!