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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 div 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 ?

harsh , 10 Years ago
Grade
anser 1 Answers
Abhishek

Last Activity: 10 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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