Two identical cylindrical vessels with their bases at same level each contains a liquid of density ρ. The height of the liquid in one vessel is h1 and that in the other vessel is h2. The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected, is

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(h1−h2)gρ

(h1−h2)gAρ

12(h1−h2)2gAρ

14(h1−h2)2gAρ

answer is D.

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Detailed Solution

If h is the common height when they are connected, by conservation of massρA1h1+ρA2h2=ρh(A1+A2)h=(h1+h2)/2 [as A1=A2=A given]As (h1/2) and (h2/2) are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the system Ui=(h1Aρ)gh12+(h2Aρ)h22=ρgA(h12+h22)2 ...(i)When vessels are connected the height of centre of gravity of liquid in each vessel will be h/2, i.e. ( (h1+h2)4 [as h=(h1+h2)/2]Final potential energy of the system UF=(h1+h2)2Aρgh1+h24=Aρg(h1+h2)24 …(ii)Work done by gravityW=Ui−Uf=14ρgA[2(h12+h22)−(h1+h2)2]=14ρgA(h1~h2)2

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