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 :
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 :
Option 1 -
1/4 gdS(x₂ − x₁)²
Option 2 -
gdS(x₂ + x₁)²
Option 3 -
1/2 gdS(x₂ − x₁)²
Option 4 -
gdS(x₂² + x₁²)/2
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1 Answer
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Correct Option - 3
Detailed Solution:sx? dg (x? /2) + sx? dg (x? /2)
Uf = (S (x? +x? )/2)gd (x? +x? )/4) x 2
= S (x? + x? )²gd/4
U? -Uf = (Sgd/4) {2x? ² + 2x? ² − (x? + x? )²}
= (Sgd/4) (x? -x? )²
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