JEE MAINS PHYSICS QUESTION SEPTEMBER 2020
Answers
Answer:
option 3 is the answer for your question as per my thinking
Q - > Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x₁ and in the other, x₂. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in
energy of the system in the process is ....
solution : it has given that, initial height of containers of same base area are x₁ and x₂ respectively. now when valve is opened, at certain time water level becomes same for both containers.
initial potential energy = U₁ + U₂
= mgh₁ + mgh₂
= ρAx₁ × (x₁/2)g + ρAx₂ × (x₂/2)g [ h₁ and h₂ are distance of centre of mass of water from the bottom for each of them]
= ρAg [(x₁² + x₂²)/2 ]
after opened the valve,
l = (x₁ + x₂)/2
and h = (x₁ + x₂)/4
so, final potential energy = U₁' + U₂'
= ρA(x₁ + x₂)/2 g × (x₁ + x₂)/4 + ρA (x₁ + x₂)/2 g × (x₁ + x₂)/4
= ρAg(x₁ + x₂)²/4
we have to find the loss in potential energy when water level becomes same.
now loss in potential energy = -∆U
= U_i - U_f
= ρAg[(x₁² + x₂²)/2 - (x₁ + x₂)²/4]
= ρAg((x₁ - x₂)²/4
Therefore the correct option is (1) ρAg((x₁ - x₂)²/4