JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Fluid Mechanics

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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

x_{1}

and in the other,

x_{2}

. 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:

(A) $\frac{3}{4} g d S (x_{2} - x_{1})^{2}$
(B) $g d S (x_{2} + x_{1})^{2}$
(C) $\frac{1}{4} g d S (x_{2} - x_{1})^{2}$ ✓ Correct
(D) $g d S (x_{2}^{2} + x_{1}^{2})$

Answer: (C)

Solution

Taking each column of liquid to act at its own centre of mass, the initial potential energy is

U_{i} = \frac{S g d}{2} (x_{1}^{2} + x_{2}^{2})

. After the vessels are connected the liquid settles to a common height

\frac{x _{1} + x _{2}}{2}

in each, so

U_{f} = 2 \cdot S \frac{x _{1} + x _{2}}{2} d g \frac{x _{1} + x _{2}}{4} = \frac{S g d}{4} (x_{1} + x_{2})^{2}

. The energy therefore falls by

U_{i} - U_{f} = \frac{S g d}{4} [2 (x_{1}^{2} + x_{2}^{2}) - (x_{1} + x_{2})^{2}] = \frac{1}{4} g d S (x_{2} - x_{1})^{2}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Fluid Mechanics

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