# Two identical cylindrical vessels with their bases at the same level each contain a liquid of density ρ (rho). The height of the liquid in one vessel is h1 and in other is h­2. The area of either base is A. What is the work done by gravity in equalizing the levels when the tow vessels are connected?

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Hello Student,
P.E. of liquid in cylinder 1
U1 = (m) g h1 / 2 = ( ρ x A x h1 ) g h1 / 2 = ρAgh21 / 2
Note :
[The total mass can be supposed to be concentrated at the center of the filled part which will be at height h­1 / 2]
Similarly P.E. of liquid in cylinder 2 U2 = ρAgh22 / 2
∴ Total P.E. initially U = U1­ + U2 (h21 + h2­2)
After the equalising of levels.
P.E. of liquid in cylinder 1 U1 = mg h / 2 = ρAg / 2 h2
P.E. of liquid in cylinder 2 U2’ = mg h / 2 = ρAg / 2 h2
∴ Total P.E. finally U’ = U1’ + U2’ = ρAgh2
The change in P.E.
∆U = U – U’ = ρAg [h21 / 2 + h22 / 2 – h2]
Total volume remains the same.
Ah1 + Ah2 = Ah + Ah
⇒ h = h1 + h2 / 2
Therefore,
∆U = ρAg [h21 / 2 + h22 / 2 – (h1 + h2 / 2)2]
= ρAg / 4 (h1 – h2)2
This change in P.E. is the work done by gravity
Thanks
Kevin Nash