A washer of mass $M$, having the form of a cylinder with the area of the base $S$ and height $h$, floats on the interface of two immiscible liquids with densities $\rho_{1}$ and $\rho_{2}$ ($\rho_{1} < \rho_{2}$).The base of the washer is parallel to the liquid interface.Find the depth of immersion of the washer in the lower fluid.
Decision:
Obviously, the lighter liquid is on top (otherwise the equilibrium of the system is unstable and any arbitrarily small fluctuations of the interface will bring the system to a stable equilibrium state). Let us denote the density of the substance of the puck by $\rho$.According to the problem condition $\rho = \frac{M}{Sh}$.
If $\rho_{2} < \rho_{1}$, the puck floats. If $\rho > \rho_{2}$, the puck sinks. Therefore, it follows from the problem condition that $\rho_{1} < \rho < \rho_{2}$. Let's denote the height of the part of the washer in the lower fluid,by $h_{2}$.Let $p_{1}$ be the pressure in the liquid at the level of the upper base of the puck, $p_{2}$ - at the level of the lower base, $p_{2}$ exceeds $p_{1}$ by an amount numerically equal to the weight of the corresponding column of liquid with unit cross-sectional area: $p_{2}=p_{1}+g(\rho_{1}(h-h_{2})+\rho_{2}h_{2})$, where $g$ is the acceleration of free fall.
The hydrostatic pressure force $F_{A}$ acting on the puck and pushing it upward can be written as
$F_{A}=(p_{2}-p_{1})S=gS(\rho_{1}(h-h_{2})+\rho_{2}h_{2})$.
Since the puck is in equilibrium, $F_{A} = Mg$. Consequently, the quantity we are looking for
$h_{2}=h \frac{\rho-\rho_{1}}{\rho_{2}-\rho_{1}}$
From the obtained expression, we can see that when $\rho$ changes from $\rho_{2}$ to $\rho_{1}$, the depth of immersion in the lower fluid changes from $h$ to 0.
Decision:
Obviously, the lighter liquid is on top (otherwise the equilibrium of the system is unstable and any arbitrarily small fluctuations of the interface will bring the system to a stable equilibrium state). Let us denote the density of the substance of the puck by $\rho$.According to the problem condition $\rho = \frac{M}{Sh}$.
If $\rho_{2} < \rho_{1}$, the puck floats. If $\rho > \rho_{2}$, the puck sinks. Therefore, it follows from the problem condition that $\rho_{1} < \rho < \rho_{2}$. Let's denote the height of the part of the washer in the lower fluid,by $h_{2}$.Let $p_{1}$ be the pressure in the liquid at the level of the upper base of the puck, $p_{2}$ - at the level of the lower base, $p_{2}$ exceeds $p_{1}$ by an amount numerically equal to the weight of the corresponding column of liquid with unit cross-sectional area: $p_{2}=p_{1}+g(\rho_{1}(h-h_{2})+\rho_{2}h_{2})$, where $g$ is the acceleration of free fall.
The hydrostatic pressure force $F_{A}$ acting on the puck and pushing it upward can be written as
$F_{A}=(p_{2}-p_{1})S=gS(\rho_{1}(h-h_{2})+\rho_{2}h_{2})$.
Since the puck is in equilibrium, $F_{A} = Mg$. Consequently, the quantity we are looking for
$h_{2}=h \frac{\rho-\rho_{1}}{\rho_{2}-\rho_{1}}$
From the obtained expression, we can see that when $\rho$ changes from $\rho_{2}$ to $\rho_{1}$, the depth of immersion in the lower fluid changes from $h$ to 0.