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Hydrostatics and work done by a gas in a situation similar to a siphon

  1. Dec 2, 2011 #1
    It isn't actually homework, it was just a problem proposed and then the answer was also given, but I don't understand it and I would like some help in trying to understand.

    1. The problem statement, all variables and given/known data
    In a cylindrical container the water level is at 30 cm. If you float a glass bowl in it, the water level will rise by 5,4 cm. What will be the water level if the glass bowl is drowned in the container? Density for glass = 2700 kg/m3, for water = 1000 kg/m3

    2. Relevant equations
    Here is the given answer which I don't understand:

    [tex]
    \begin{split}
    &h_0=0,3m\\
    &h_1=0,054m\\

    &h ?\\
    \\
    &h_2ρ_{glass}=h_1ρ_{glass}-h_1ρ_{water}\\
    &h_2=h_1(ρ_{glass}-ρ_{water})/ρ_{glass}\\
    &h_2=3,4 cm\\
    &h=h_0+h_2\\
    &h=33.4cm\\
    \end{split}[/tex]

    I don't have the slightest idea why are the densities and heights multiplied together and how does it give the answer. If anyone can shed some light on this I would be grateful.


    3. The attempt at a solution

    What I tried to do is that first I assumed the bottom of the container to have an area of [itex]S (cm^{2})[/itex]. Then in order for the water level to rise 5,4 cm, the volume of water displaced must be [itex]S*5,4 (cm^{3})[/itex]. For the bowl to displace that much water, it must weight as much as the water displaced, ie [itex]m=S*5,4*1(g/cm^{3})=5,4S (g)[/itex]. For a glass object weighing that much, it's volume must be [itex]5,4S/2,7=2S (cm^{3})[/itex]. Since when something is submerged underwater, it displaces water equal to the volume of the object, so the water level rises by 2 cm. What did I do wrong?


    Edit: sorry for the wrong title, I originally wanted to post two problems but I then decided to only post the hydrostatic buoyancy/water displacement problem.
     
  2. jcsd
  3. Dec 2, 2011 #2
    I don't understand their solution. I agree with your solution. You do a free body of a floating bowl: FB=h1*area*ρwater*g=mglass*g solve for mglass

    With submerged bowl:mglassglass*volbowl where volbowl=h2*area. Area is the area of bottom of cylinder

    Set the mglass equal to each other and you get h2=2 cm

    If you find the answer let me know
     
  4. Dec 14, 2011 #3
    It turned out that the answer calculated by you and me was correct and that they had revised their answer.
     
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