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Pressure at a point

  1. Feb 12, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    Find the pressure at A.

    WIN_20150212_161705.JPG
    2. Relevant equations
    Excess pressure(drop) = 2S/r
    Excess pressure (bubble) = 4S/r

    3. 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?
     
  2. jcsd
  3. Feb 12, 2015 #2

    DEvens

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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?
     
  4. Feb 12, 2015 #3

    Suraj M

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    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 ?
     
  5. Feb 12, 2015 #4
    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
     
  6. Feb 13, 2015 #5

    Suraj M

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    Oh okay, understood. Thank you!
     
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