# Why does the pressure change on uniformly mixing two liquids? The two cylinders are connected the upper cylinder has a cross-section of A and the lower one has a cross-section of 2A (I've taken the cross-sections to be A and 2A as they are easier to work with than the radii r and 2r).

Initially the liquid present in the upper cylinder has a density of $$\rho$$ and the lower one has a density of $$2\rho$$. The pressure at the bottom in this scenario can be calculated as:

$$P_1=\rho\cdot g\cdot 2h+2\rho\cdot g \cdot h=4\rho\cdot g\cdot h$$

Now, if we were to mix the two liquids present in the cylinders to create a new solution with a uniform density of $$\frac32\rho$$ (The volume of both the cylinders is the same so we can simply take the mean of the two densities).

The pressure in this case can be calculated as:
$$P_2=\frac32\rho\cdot g \cdot 3h= \frac92\cdot g \cdot h$$

This was one approach to calculate the pressures another method is by simply calculating $$\frac{F}{A}$$

Since the entire liquid system is in equilibrium and the only external force balancing the gravitational force is the normal applied at the bottom surface the pressure will be:

$$P=\frac{Mg}{A}$$
$$P=\frac{\rho\cdot 2h \cdot A\cdot g + 2\rho\cdot 2A \cdot h \cdot g}{2\cdot A }= 3\rho \cdot g \cdot h$$

Which is matching with neither of the cases but i can't see why this is wrong.

First consider the case of a cylinder of uniform cross section $$A$$, in which liquids of density $$\rho_1,\rho_2,$$ with corresponding volumes $$V_1,V_2,$$ are present. Using the hydrostatic equation, the pressure at the bottom is $$P_1=\rho_1gh_1+\rho_2gh_2=\rho_1g(V_1/A)+\rho_2g(V_2/A)=(\rho_1V_1+\rho_2V_2)g/A$$. If after mixing the total volume doesn't change then mass conservation says that the final density should be $$\rho_f=(\rho_1V_1+\rho_2V_2)/(V_1+V_2)$$. The corresponding pressure at the bottom will be: $$\rho_fgh_f=\rho_fg(h_1+h_2)=\rho_fg(V_1+V_2)/A=(\rho_1V_1+\rho_2V_2)g/A,$$ the same as before. Therefore, when the cylinder has uniform cross-section, mixing doesn't change the pressure at the bottom, which is logical since the same weight of fluid is supported by the bottom in both cases.
In your problem the cross-section varies. Let $$\rho_1,V_1,A_1$$ and $$\rho_2,V_2,A_2$$ be the density, volume and cross-section of the two fluids. The interface between the two fluids lies exactly where the cross-section changes. Using the hydrostatic equation, the pressure is: $$P_1=\rho_1gh_1+\rho_2gh_2=\rho_1g(V_1/A_1)+\rho_2g(V_2/A_2)$$, which can't be simplified further. After mixing the density becomes $$\rho_f=(\rho_1V_1+\rho_2V_2)/(V_1+V_2)$$ as before. But now the pressure at the bottom changes to $$\rho_fgh_f\neq P_1,$$ as you can verify by substitution. Since the same weight of fluid is being supported before and after mixing, is the change in pressure a contradiction? No, it isn't.
Above result derived using hydrostatic equation is correct. If you want to derive the pressure on the bottom using force balance then you must recognise that in the case where cross-section of the container varies, weight of the fluid is not the only force acting on the bottom. There is also a downward reaction force due to the horizontal wall where the cross-section changes (denoted by black arrows in the figure below). This reaction force is equal to the product of hydrostatic pressure at the horizontal wall (equal to $$\rho_1gh_1$$ before mixing and $$\rho_fgh_1$$ after mixing) and its area. When this is accounted for you get the same answer as above. 