MCQ
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 $ h_1$ and that in the other vessel is $h_2$ . The area of either base is $A$ . The work done by gravity in equalizing the levels when the two vessels are connected is
  • A
    $(h_1 -h_2)gd$
  • B
    $(h_1 -h_2)gAd$
  • C
    $\frac {1}{2}(h_1 -h_2)^2gAd$
  • $\frac {1}{4}(h_1 -h_2)^2gAd$

Answer

Correct option: D.
$\frac {1}{4}(h_1 -h_2)^2gAd$
d
Work done = initial potential energy -final potential energy 

or $W{\text{ }} = {\text{ }}Adg\,\left[ {\frac{{h_1^2 + h_2^2}}{2} - {{\left( {\frac{{{h_1} + {h_2}}}{2}} \right)}^2}} \right]\, = \,\frac{{Adg}}{4}{({h_1} - {h_2})^2}$

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