(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[This probably seems really easy to the physics whizzes out there but...*blushes*]

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.30x10^{3}kg/m^{3}. The area of each base is 4.00cm^{2}, but in one vessel the liquid height is 0.854 m and in the other it is 1.560 m. Find the work done by the gravitational force in equalising the levels when the two vessels are connected.

2. Relevant equations

p_{2}=p_{0}+ [tex]\rho[/tex]gh

p=F/A

[tex]\rho[/tex]=m/V

p=[tex]\rho[/tex]gh

3. The attempt at a solution

Well, it seems that this question has gone beyond the knowledge imparted earlier in the chapter this is in - either that or I have a lack of grey matter =]. However, I think I'm nearly there. The solution is as follows.

Find the pressures in the first and second cylinders.

p_{1}= 1.3x10^{3}x 9.8m/s^{2}x 0.854 m = 10,880 Pa

p_{2}= 1.3x10^{3}x 9.8m/s^{2}x 1.56 m = 19,874 Pa

[tex]\Delta[/tex]p = 8,994 Pa

F= p x A = 8,994 Pa x 4x10^{-4}m^{2}= 3.60 N

Well, I've found the force due to the pressure difference. I'm not quite sure whether that is the correct approach though.

Now the work done by the gravitational force must be found. This is where I'm stuck. Is it as simple as multiplying the force - calculated above - by the change in height of the water? Or should I use U=-GMm/r?

I calculated the height of the equalised water to be 0.7 m. I did this by finding the height at which it would have to be for the water to cause a pressure of 8,994Pa.

EDIT: I suspect the height I have calculated is wrong.

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# Fluids at rest: two connected cylinders

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