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What is the mass of the water in the stirrer supported by surface tension?

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A hollow stirrer that has a diameter of 2.0 mm is inserted into a cup of water at 20∘C. The surface tension of water at that temperature is γ(surface tension) = 7.28×10−2 N/m.

    What is the mass of the water supported by the surface tension in the stirrer?

    Given:
    d=2 mm
    p(rho): 1000 kg/m^3
    y(surface tension): 7.28E-2 N/m

    Find: Mass

    Conversions:
    d=.002 m r=.001

    Problem: I tried this equation, but I keep on getting the wrong answer, I am not sure if I am using the correct formula.

    2. Relevant equations
    Surface tension: deltaP=2y/r
    Pressure: deltaP=F/A

    3. The attempt at a solution
    detlaP=2y/r
    deltaP=F/A
    F=mg
    A=(pi)r^2
    Plug them in:
    mg/(pi)r^2=2y/r
    Change the equation to equal mass (m):
    m=2y(pi)r/g
     
    Last edited by a moderator: Jan 16, 2017
  2. jcsd
  3. Jan 16, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    Your work looks good. What numerical answer did you get for m?
     
  4. Jan 16, 2017 #3
    Nevermind, I got the right answer just a minor mistake. Sorry about that. Thanks for the help!
     
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