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Posted by on Tuesday, April 10, 2012 at 4:14pm.

A U-shaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.46 m and h2 = 0.25 m, what is the density of the liquid?

on the left side of the tube is the liquid h2= 0.25 m, and h1= 0.46, and on the right side is the water and its 0.45 m
How can you calculate the density of the liquid?

Thank you.

  • physics - , Tuesday, April 10, 2012 at 4:46pm

    Supposing that ρ1 is the density of the unknown liquid, ρ is the density of the water, and g is the gravity acceleration; let's choose a random intersection face at the bottom of the U-tube's bottom. Since the pressures from the both sides balance with each other, we can introduce the equation:
    ρ1•h2•g+ ρ •(h1-h2) •g= ρ•0.45•g.
    Replace the symbols h1 and h2 with 0.46 m and 0.25 m, respectively,
    we can conclude that
    ρ1=0.96 ρ =0.96• 1000 = 960 kg/m^3.

  • physics - , Tuesday, April 10, 2012 at 5:10pm

    Thank you.I don't know why, but it says the answer is wrong even though i have to answer it as g/cm^3 which would be 960000 g/cm^3 i assume, but even that's wrong! :/ I am really confused I don't know why its wrong.

  • physics - , Tuesday, April 10, 2012 at 5:25pm

    I'm sure that my solution is right.
    Check your given data.
    And more, 960 kg/m^3 = 960 000 g/m^3 =
    =0.96 g/cm^3 (1 cm^3 =10^-6 m^3)

  • physics - , Tuesday, April 10, 2012 at 5:35pm

    Thank you. It was because of the units my units should have been as g/cm^3, so I got it now. Thank you, I really appreciate it :)

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