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

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SUMMARY

The discussion centers on a hydrostatics problem involving a glass bowl floating in water and the subsequent calculations for water displacement when the bowl is submerged. The initial water level is 30 cm, and the bowl causes a rise of 5.4 cm. The correct final water level, after considering the densities of glass (2700 kg/m³) and water (1000 kg/m³), is determined to be 33.4 cm. The confusion arises from the application of buoyancy principles and the relationship between the densities and heights of the displaced water and submerged bowl.

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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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