A

Step 1: Consider the potential energy of the ball when it is raised through a height $h$. The increase in potential energy of the ball is given by:
$$ PE_{ball} = mgh $$.
Step 2: When the ball is immersed in the liquid, it displaces a volume of liquid equal to its own volume $v$. The potential energy of this displaced liquid decreases as it moves downwards by height $h$ and is given by:
$$ PE_{liquid} = v \rho gh $$
where $\rho$ is the density of the liquid.
Step 3: The total potential energy change of the system (ball + liquid) is:
$$ \Delta PE_{total} = PE_{ball} - PE_{liquid} = mgh - v \rho gh $$
Step 4: For equilibrium to hold (the ball to float), the upward buoyant force exerted by the liquid on the ball must equal the weight of the ball. Thus, Archimedes's principle indicates that:
$$ F_{buoyant} = v \rho g = mg $$
which implies that the weight of the liquid displaced (the downward potential energy loss) equals the weight of the ball (the upward potential energy gain). This balance leads to Archimedes's law: The weight of the liquid displaced is equal to the weight of the ball.
Therefore, A.