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Equilibrium Levels of Mercury and Water

  1. Nov 5, 2007 #1
    You have a "U-shaped" container with a valve at the bottom of a negligible volume tube. The left side is filled with water and the right side is filled with mercury. The valve is opened and since the two are immiscible there is a fluid level differentiation. Determine the fluid level (distance from the bottom of the container) if the initial heights of both fluids are 1.00 meters.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how much "research I hsould be doing online to find out densities and such or if I even need them for this equation. But I found the density of water is 1.000kg/meters^3 and mercury's density is 13.6x10^3kg/meters^3

    I looked up this formula in my book hoping it might help me. Pressure 2 = Pressure 1 + (Density)(Gravity)(Height). So am I correct in assuming that the pressure of water equals mercury's pressure + mercury's density x gravity x 1.00 meters?

    If so, then I calculated the pressure of the water to be 133280? If that's right how do I make the jump from knowing the water's equilibrium pressure to finding out the height of the water?

    I know that it's going to be higher than the mercury but how much higher?

    Thank you for any help you can give me!
  2. jcsd
  3. Nov 5, 2007 #2


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    Staff: Mentor

    The weights of the columns of both sides much equilibrate. The mecury height decreases by h and the water side increases by h assuming constant cross-sectional area and incompressible fluid.

    1 cm of Hg and the mass of 13.6 cm of water for the same cross-sectional area. Conversely 1 m of water has the same mass/weight at 0.07353 m of Hg.

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