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Depth of lake (Pressure/fluids)

  1. Apr 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1.6 m long glass tube closed at one end
    is weighted and lowered to the bottom of a
    freshwater lake. When the tube is recovered,
    an indicator mark shows that water rose to
    within 0.32 m of the closed end.
    Assume the temperature is constant, the
    acceleration of gravity is 9.8 m/s2, the density
    of the glass is 3400 kg/m3 , the density of the
    lake water is 1000 kg/m3, and air pressure is
    1.013 × 105 Pa.

    Determine the depth of the lake. Answer in
    units of m.


    2. Relevant equations
    P = PA + PLake
    Plake = rho * g * h (h = depth of lake)
    rho = m / V
    V = pi * r2 * L1 (L1 = 1.6 m)

    3. The attempt at a solution
    So I tried setting up a proportion:

    L1 / L2 (L2 = 0.32 m ) = PA / PL

    (L2PA) / L2 = PL = rho * g * h
    From there, h seems to be easy to figure out, but I don't think my process is right because I ignored the density of the glass given.

    Can someone help me?
     
  2. jcsd
  3. Apr 23, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Looks like your method is almost OK.

    Isn't the key relationship

    P1*V1 = P2*V2

    1 atm * 1.6 = Pbottom *.32

    ρ *g * h = 1.6 / .32 atm

    They gave you the density of glass but not the weight of the weight used to submerge it? To what purpose? Maybe not all givens are always necessary?
     
  4. Apr 23, 2009 #3
    Ah, ok.

    I forgot P(bottom) was 1 atm + pgh.

    Thank you!
     
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