Final answer:
Option A: ρA(h1 - h2/2)² g.
In a system of two interconnected hydraulic cylinders with equal cross-sectional areas, the work done by gravity during the equalization of liquid levels is computed by the change in potential energy, which is the product of the density of the liquid, gravitational acceleration, area, and square of half the difference in initial heights.
Step-by-step explanation:
The question involves the concept of conservation of energy in a fluid system, specifically relating to hydraulics and the work done by the force of gravity when two cylinders with liquid are interconnected and reach an equilibrium. When the cylinders are interconnected, and the liquids' levels equalize, the work done by gravity is related to the change in potential energy of the liquid.
To calculate this, consider that the liquid in the first cylinder falls a height of (h1 - h)/2, where h is the new height, and the liquid in the second cylinder rises by (h - h2)/2. Since the cross-sectional area of the two cylinders is the same (A), the volume of the liquid that moves is equal to A times the difference in height before and after the process.
Thus, the work done by gravity is equal to the change in potential energy, which is:
Work = potential energy change = mass * g * height change
The mass of the fluid can be obtained by multiplying the density (ρ) with the volume of the fluid that has moved, and the height change can be taken as half the difference between the initial heights because the cross-sectional areas are equal.
The correct expression for the work done is:
Work = ρA(g)((h1 - h2) / 2)²
Thus, the right answer to the question is Option A: ρA(h1 - h2/2)² g.