Two identical cylindrical vessels are kept on the ground and each… | JEE Main 2020 (4 Sep, Shift 2)

Q18JEE Main 2020 · 4 SEP · SHIFT 2Medium

Pressure in a Fluid

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:

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Answer: (c)

PE of a liquid column of height $x$ : $U = \tfrac{1}{2} dgSx^2$ .
Initial: $U_i = \tfrac{1}{2}dgS(x_1^2 + x_2^2)$ . Final height $h = \tfrac{x_1+x_2}{2}$ , so $U_f = dgS h^2 = \tfrac{1}{4}dgS(x_1+x_2)^2$ .
$U_i - U_f = \tfrac{1}{4}dgS\left[2x_1^2 + 2x_2^2 - (x_1+x_2)^2\right] = \tfrac{1}{4}dgS(x_2 - x_1)^2$
So the answer is (c).

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Pressure in a Fluid question from JEE Main 2020 (4 Sep, Shift 2)

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