Mechanics of Liquids and Gases I am solving a problem about the mechanics of liquids and gases. 
Here we see water (blue) and unknown (purple) liquid. I want to calculate the density of the unknown (purple) liquid. We know that $\rho_w=1000$ $kg$/$m^3$ and $h_w=39$ $cm=0,39$ $m$. Let $q_p$ be the unknown density. We also know $h_p=30$$cm$ = $0,3$$m$. I guess that the hydrostatic pressure at the bottom of the test-tube is equal for both liquids. I can't convince myself that this is true. Can you help me?

 A: For the diagram shown, the static pressure of the two liquids have to be the same at the bottom, or
$$ρ_{w}gh_{w}=ρ_{l}gh_{l}$$
where $l$ and $w$ are the unknown liquid and water, respectively.
From that can calculate the density $ρ_l$ of the unknown liquid.
Hope this  helps.
A: If the pressure from each leg of the manometer was not equal at the bottom of the tube, there would be a pressure differential that would cause flow in the tube.  The fact that the liquids on each side of the tube have come to rest verifies that no such pressure differential exists.
A: Though it is not easy to convey this literally but.. 
 
What you probably need is an understanding of density, more dense liquid implies that it holds the mass in less volume when compared to the other so essentially what is being implied is that two masses are providing equal normal reaction at the bottom we know that the bottom part is at equilibrium as both liquids apply equal force which is only possible if they have equal masses but their densities are different thus different sized columns must exist for equilibrium 
As for the question you can equate the forces as told accurately by Bob D which should provide result as 1300 kg/m3 which may indicate that the substance might be liquid argon.
Hope that answered you queries
