- #1
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.
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
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.
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.
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:
[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.
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.