In a large vessel with water floats in a vertical position a thin-walled beaker in which a certain amount of water is poured. The difference between the levels of water in the vessel and the beaker is $x$. How will this difference change if a cork is put into the beaker?
Decision:
Let $m$ be the mass of the empty beaker, $S$ be its cross-sectional area, $V$ be the volume of water poured into the beaker, $V_{1}$ be the volume of the submerged part of the beaker, $\rho$ be the density of water, and $g$ be the acceleration of free fall. The beaker is in equilibrium, i.e., the force of gravity acting on it is balanced by the expulsive force:
$mg+\rho gV= \rho gV_{1}$.
Obviously, $V_{1} = V + Sx$. Then
$x=\frac{m}{\rho S}$
is independent of the amount of water poured into the glass.
Suppose a cork is thrown into the glass, $M$ is its mass and $V_{2}$ is the volume of the part of the cork immersed in water. By Archimedes' law we have:
$Mg = \rho gV_{2}$.
If now the cork is replaced by the volume of water $V_{2}$ (it has mass $M$), the water level in the beaker will not change. The depth of immersion of the beaker will also not change. Hence, the difference in the levels of water in the beaker and the vessel will remain unchanged. Since the difference of levels does not depend on the amount of water in the beaker, it is clear that the value of $x$ will not change when the stopper is lowered into the beaker.
Decision:
Let $m$ be the mass of the empty beaker, $S$ be its cross-sectional area, $V$ be the volume of water poured into the beaker, $V_{1}$ be the volume of the submerged part of the beaker, $\rho$ be the density of water, and $g$ be the acceleration of free fall. The beaker is in equilibrium, i.e., the force of gravity acting on it is balanced by the expulsive force:
$mg+\rho gV= \rho gV_{1}$.
Obviously, $V_{1} = V + Sx$. Then
$x=\frac{m}{\rho S}$
is independent of the amount of water poured into the glass.
Suppose a cork is thrown into the glass, $M$ is its mass and $V_{2}$ is the volume of the part of the cork immersed in water. By Archimedes' law we have:
$Mg = \rho gV_{2}$.
If now the cork is replaced by the volume of water $V_{2}$ (it has mass $M$), the water level in the beaker will not change. The depth of immersion of the beaker will also not change. Hence, the difference in the levels of water in the beaker and the vessel will remain unchanged. Since the difference of levels does not depend on the amount of water in the beaker, it is clear that the value of $x$ will not change when the stopper is lowered into the beaker.