Question

# Two identical cylindrical vessels with their bases at same level each contains a liquid of density ρ. 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

A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is C If h is the common height when they are connected, by conservation of mass ρA1h1+ρA2h2 = ρh(A1+A2) h = (h1+h2)2 [as A1 = A2 = A given] As (h12) and (h22) are heights of initial centre of gravity of liquid in two vessels, the initial potential energy of the system Ui = (h1Aρ)g h12+(h2Aρ)h22 = ρgA(h21+h22)2 ....(i) When vessels are connected the height of centre of gravity of liquid in each vessel will be h2, i.e. (h1+h2)4[as h = (h1+h2)2] Final potential energy of the system Uf = [(h1+h2)Aρ2]g(h1+h24) = Aρg[(h1+h2)24] ....(ii) Work done by gravity W = Ui−Uf = 14ρgA[2(h21+h22)−(h1+h2)2] = 14ρgA(h1∼h2)2

Suggest Corrections
11
Related Videos
The Law of Conservation of Mechanical Energy
PHYSICS
Watch in App