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two identical cylinderical vessels with bases at the same level each contain a liquid of density d. the height of liquid in one vessel is h1 and that of other vessel is h2. the area of either base is A. the work done by gravity in equastising the levels when the two vessels are connected is ?

two identical cylinderical vessels with bases at the same level each contain a liquid of density d. the height of liquid in one vessel is h1 and that of other vessel is h2. the area of either base is A. the work done by gravity in equastising the levels when the two vessels are connected is ?

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1 Answers

Abhishek
9 Points
9 years ago
PE = mghcm=dAhgh/2=dAgh2/2 (taking base as reference)
so PE initial = dAg(h12+h22)/2
    
in final position h= (h1+h2)/2         height of centre of mass=(h1+h2)/4
mass=d(2A)((h1+h2)/2) =dA(h1+h2)
so PE final = mghcm=dAg(h1+h2)2 /4
 
change in pe = -work done by conservative force
work = -change in pe
         = PE initial -PE final
         = dAg(h12+h22-2h1h2)/4
         =dAg(h1-h2)2/4

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