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Question: Two identical cylindrical vessels with their bases at same level each contains a liquid of density d...

Two identical cylindrical vessels with their bases at same level each contains a liquid of density d. The height of the liquid in one vessel is h1h _ { 1 } and that in the other vessel is h2h _ { 2 } . The area of either vases is A. The work done by gravity in equalizing the levels when the two vessels are connected, is

A

(h1−h2)gd\left( h _ { 1 } - h _ { 2 } \right) g d

B

(h1−h2)gAd\left( h _ { 1 } - h _ { 2 } \right) g A d

C

12(h1−h2)2gAd\frac { 1 } { 2 } \left( h _ { 1 } - h _ { 2 } \right) ^ { 2 } g A d

D

14(h1−h2)2gAd\frac { 1 } { 4 } \left( h _ { 1 } - h _ { 2 } \right) ^ { 2 } g A d

Answer

14(h1−h2)2gAd\frac { 1 } { 4 } \left( h _ { 1 } - h _ { 2 } \right) ^ { 2 } g A d

Explanation

Solution

Potential energy of liquid column is given by

mgh2=Vdgh2=Ahdgh2m g \frac { h } { 2 } = V d g \frac { h } { 2 } = A h d g \frac { h } { 2 } =12Adgh2= \frac { 1 } { 2 } A d g h ^ { 2 }

Initial potential energy =12Adgh12+12Adgh22= \frac { 1 } { 2 } A d g h _ { 1 } ^ { 2 } + \frac { 1 } { 2 } A d g h _ { 2 } ^ { 2 }

Final potential energy = 12Adgh2+12Adh2g=Adgh2\frac { 1 } { 2 } A d g h ^ { 2 } + \frac { 1 } { 2 } A d h ^ { 2 } g = A d g h ^ { 2 }

Work done by gravity = change in potential energy

W =[12Adgh12+12Adgh22]−Adgh2= \left[ \frac { 1 } { 2 } A d g h _ { 1 } ^ { 2 } + \frac { 1 } { 2 } A d g h _ { 2 } ^ { 2 } \right] - A d g h ^ { 2 }

=Adg[h122+h222]−Adg(h1+h22)2= A d g \left[ \frac { h _ { 1 } ^ { 2 } } { 2 } + \frac { h _ { 2 } ^ { 2 } } { 2 } \right] - A d g \left( \frac { h _ { 1 } + h _ { 2 } } { 2 } \right) ^ { 2 } [As h=h1+h22h = \frac { h _ { 1 } + h _ { 2 } } { 2 }]

=Adg4(h1−h2)2= \frac { A d g } { 4 } \left( h _ { 1 } - h _ { 2 } \right) ^ { 2 }