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Determining the Depth of a Lake

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A 2.1 m long glass tube that is 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 0.43 m of the closed end. Determine the depth of the lake. Assume constant temperature.

    2. Relevant equations
    Honestly...I have no idea how to approach this problem.
    The only relevant equations I know for this sort of problem are:
    (height)=2(surface tension)/(density)(g)(radius)
    But neither of these equations seem to get me anywhere...

    3. The attempt at a solution
    Like I said, I really don't have any idea how to approach this problem.
    The furthest I got was to calculate the total height of the water in the tube: 1.67m
    Using those equations I posted, I could calculate the radius of the tube or the initial pressure. But that information is irrelevant to the question. Any sort of information to help point me in the right direction would be greatly appreciated. Thanks in advance.
  2. jcsd
  3. Nov 23, 2008 #2
    how about calculating the water pressure at the bottom of the lake? that will probably get us somewhere.

    P_1V_1 = P_2V_2
    The pressure that the air inside comes to at the bottom is the same as the water pressure since at equal pressure that's where they would hit equilibrium

    ah then use your first equation, look up the density of water, and make sure all your units are correct
  4. Nov 23, 2008 #3
    Oooh I see where you're going with this...So calculate the pressure of the air after the tube is lowered, which would be the same as the water pressure? I'm not really sure what I'd use for P1 though. Would I use the pressure of air at sea level?
  5. Nov 23, 2008 #4
  6. Nov 23, 2008 #5
    Nope...didn't work.
    I tried P1=101,325 Pa, V1=(pi)(r)^2(h)=pi(0.008795)^2(2.1)=0.000528
    Then V2=(pi)(0.008795)^2(0.43)=0.000108
    So P2=53.55/0.000108 = 494843
    Plugging that into my first equation, 494843=(10^3)(9.8)(h)

    It wasn't correct :(
  7. Nov 23, 2008 #6


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    The water pressure is not rho*g*h at the bottom of the lake. Do you see what is missing?
  8. Nov 23, 2008 #7
    Should it be rho*g*h + the atmospheric pressure?
  9. Nov 23, 2008 #8


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  10. Nov 23, 2008 #9
    So (494843 + 101325)=(10^3)(9.8)(h)

    Still not right...
  11. Nov 23, 2008 #10
    If we equate the pressure due just to the water with the formula with density of water and gravity and height

    P_{water}=(494843 - 101325)= 393,518 \frac{N}{m^2}
    393,518 \frac{N}{m^2}=(10^3 \frac{kg}{m^3})(9.8 \frac{m}{s^2})(h)
    Last edited: Nov 23, 2008
  12. Nov 24, 2008 #11


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    I think you're close.

    Is the air in the tube (that we got the pressure for) located exactly at the bottom of the lake?
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