Solved University-Physics Problems

Physics 2.-  Fluid Statics


Javier Montenegro Joo


   Return to:  Index Page of Physics 2  

Return to   VirtualDynamics University     home page

  Fluid Statics.-  

(1) An open U-shaped tube has branches of section  A and, it contains a liquid of a certain density.  Determine the volume of another liquid having a density which is half the density of the first liquid, which must be poured in a branch of the tube so that the first liquid rises ho  in the other branch.  Both liquids are immiscible.


The figure shows the two acts of this problem:  In the first act appears only the first liquid. In the second act, the second liquid has been poured in the left branch of the tube, being h  the height of this second fluid. For the first liquid to raise a height ho in the right hand branch, the interface level must descend  ho  in the left branch. Under these conditions, at the level of the interface, the pressure is the same in both branches:




 (3) A long hose containing two immiscible liquids is vertically bent as shown in the sketch. A piston is making a pressure Po at the left extreme of the hose, while the extreme at the right side is closed with a stopper.  Determine the pressure at the stopper.

Solution .-

Note that when varying the pressure Po exerted by the piston at the left extreme of the hose, the pressure changes along the entire system, hence in order to find the pressure at any point, the connected pressures starting at the left extreme must be considered.  Bear in mind that at the level of two contiguous interfaces the pressure is the same. If the piston were retired the pressure at its position would be that of the atmosphere.


(4) In the figure, a container contains liquid of density D. (a) Determine the pressures at points   a, b, c, and d.  (b) Demonstrate that the pressure at  point b is greater than that at point a. (c) Demonstrate that if the pressures at points  a and b  were equal, then the liquid would be at the same level at both points.


         The sketch below shows the container and the interfaces I1 and I2.

         At the place where the container is open, the pressure is the atmospheric pressure Po.

         Pressures are always positive or zero. Negative pressures are impossible.

         At the apparently empty regions a and b, there is air exerting pressures Pa  and  Pb., respectively.

         At points a and b, if the corresponding pressures Pa and Pb were higher, the levels of the liquid would be lower at those points and, vice versa.

         If points a and b were at the same level, then Pa = Pb.

         At the bottom of the tank, the pressures are the same: Pc = Pd.

         At the level of each interface, the pressures are the same.


Computational & Simulational Physics Experiments with the Chaos Virtual Lab:

Experimental Findings in the Chaotic Nonlinear Damped and Forced Oscillator




(6) In the figure, a balloon full of gas at a pressure PG is connected to a U-shaped tube which contains two liquids, A and B, whose densities are shown. Determine:  (a) the height H (b) the pressure of the gas in the balloon for liquid B to be leveled in both branches of the U-tube.

Solution .-

At the level of the interface between the gas and the liquid B, the pressure is the same. On the other hand, if liquid B is leveled in both branches of the U-tube:   H = 0.

  [10] The figure shows a vessel containing gas, water and oil. Determine the pressures at points 1, 2, 3 and 4.


 At the level of the interface the pressures are equal.


Return to   VirtualDynamics University     home page