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

AI Thread Summary
The discussion revolves around a hydrostatics problem involving a glass bowl floating in water and the subsequent water level changes when the bowl is submerged. The original solution provided was confusing, particularly regarding the calculations involving densities and heights. Participants clarified that the correct approach involves equating the mass of the displaced water to the mass of the submerged bowl to find the new water level. After recalculating, it was determined that the correct rise in water level when the bowl is submerged is 2 cm, rather than the initially stated 3.4 cm. Ultimately, the participants confirmed that their revised calculations were accurate, resolving the confusion around the problem.
chingel
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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.

Homework Statement


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

Homework Equations


Here is the given answer which I don't understand:

<br /> \begin{split}<br /> &amp;h_0=0,3m\\<br /> &amp;h_1=0,054m\\<br /> <br /> &amp;h ?\\<br /> \\<br /> &amp;h_2ρ_{glass}=h_1ρ_{glass}-h_1ρ_{water}\\<br /> &amp;h_2=h_1(ρ_{glass}-ρ_{water})/ρ_{glass}\\<br /> &amp;h_2=3,4 cm\\<br /> &amp;h=h_0+h_2\\<br /> &amp;h=33.4cm\\<br /> \end{split}

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.

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 S (cm^{2}). Then in order for the water level to rise 5,4 cm, the volume of water displaced must be S*5,4 (cm^{3}). For the bowl to displace that much water, it must weight as much as the water displaced, ie m=S*5,4*1(g/cm^{3})=5,4S (g). For a glass object weighing that much, it's volume must be 5,4S/2,7=2S (cm^{3}). 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.
 
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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
 
It turned out that the answer calculated by you and me was correct and that they had revised their answer.
 
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