Question

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 bases 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. $\frac{1}{2}(h_1-h_2{)}^{2}gAd$
• D. $\frac{1}{4}(h_1-h_2{)}^{2}gAd$

Answer 1

The correct option is D

$\frac{1}{4}(h_1-h_2{)}^{2}gAd$

Potential energy of liquid column is given by $mg\frac{h}{2}=Vdg\frac{h}{2}=Ahdg\frac{h}{2}=\frac{1}{2}Adg{h}^2$

Initial potential energy $=\frac{1}{2}Adg{h}_1^2+\frac{1}{2}Adg{h}_2^2$

Final potential energy $=\frac{1}{2}Adg{h}^2+\frac{1}{2}Ad{h}^{2}g=Adg{h}^2$

Work done by gravity = change in potential energy

$W=[\frac{1}{2}Adg{h}_1^2+\frac{1}{2}Adg{h}_2^2]-Adg{h}^2$

$=Adg[\frac{h_1^2}{2}+\frac{h_2^2}{2}]-Adg{\left(\frac{h_1+h_2}{2}\right)}^2[Ash=\frac{h_1+h_2}{2}]$

$=Adg[\frac{h_1^2}{2}+\frac{h_2^2}{2}-\left(\frac{h_1^2+h_2^2+2h_1{h}_2}{4}\right)]=\frac{Adg}{4}(h_1-h_2{)}^2$