Question

two identical cylindrical vessels with their bases at same level each contains a liquid of density p. 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:

Answer 1

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

Answer 2

Explanation:

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