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I have a couple of brain blocks.

  1. Sep 5, 2004 #1
    I've got a U-tube filled with Mercury and water. If I add 11.2 cm of water to the left side, how much does the mercury rise on the right side?

    For simplicity's sake, D will stand for density, p0 for initial pressure, w = water, m = mercury, h = 11.2 cm for the water and d = the distance the mercury rises, and g is gravity.

    The equation I think I'm supposed to use is

    p0 + Dw(g)(1/2 * h) = p0 + Dm(g)(d)

    --> Dw(1/2 * h) = Dm(d)

    d = [ Dw(1/2 * h) ] / Dm = [ 1,000 kg/m^3 ( 1/2 * 11.2 cm ) ] / 13.6*10^3 kg/m^3

    d = .412 cm which is what the book says, but my question is why did I use (1/2 * h) instead of h? Is there another way to do this problem?
  2. jcsd
  3. Sep 5, 2004 #2
    I believe the correct answer depends on whether it's mercury or water that occupies the lower, bent portion of the U-tube. Let's assume however that mercury fills the bottom. Then the water-mercury interface is on the left side (water side) at a point where the tube remains vertical. The importance of the interface position is that for a height change [tex]\inline{dx}[/tex] of the mercury surface level, the distance change of the mercury surface to water-mercury interface is [tex]\inline{2dx}[/tex]. Thus, if we add 11.2 cm of water to the left-hand side, then [tex]\inline{ \rho_w g [11.2 \text{ cm}]= \rho_m g [2dx] }[/tex].
    Last edited: Sep 5, 2004
  4. Sep 6, 2004 #3
    If the mercury level falls by x cm on the left side. then the mercury on the right side rises by x cm too relative to the initial level.

    See the attached file. Do you know why now?

    The density of mercury is greater than water, so it is always beneath water.
  5. Sep 6, 2004 #4
    Thanks for the help; you cleared that up pretty well. :biggrin:
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