Posted by Sonya on .
The crosssectional area of the Utube shown below is uniform and is equal to one square centimeter (1.00cm^2). One end is open to the atmosphere. Atmosphere pressure may be taken as P1=1.01 * 10^5 Pa. The other end of the Utube is connected to a pipe in which a constant gas pressure of P2=1.25 atmospheres is maintained. The Utube contains a liquid which has a density of 1.0*10^4kg/m^3.
(I) Calculate the difference in height, h, in centimeters, of the eater levels in the two arms of the tube.
(II) 10 cm^3 of a liquid which has a density of 0.85 times that of the first liquid are now poured into the arm which is open to the atmosphere (assume that the liquids do NOT mix) What is the new difference between the liquid levels in the two tubes, assuming the pressures p1 and p2 remain constant?
(III) What the second liquid still in the arm that is open to the atmosphere, the pressure p2 is now reduced until it is equal to atmospheric pressure. What now is the difference (if any) between the liquid levels in the two arms of the Utube?

Physics _ Please please help 
Damon,
Bernoulli:
p + (1/2) rho v^2 + rho g h = constant
in this problem v = 0 so
p + rho g h = constant
say height = 0 at high pressure end and H at one atmosphere end
1.01*1^5 + rho g H = 1.12*1.01*10^5 + 0
.12 * 1.01* 10^5 = 1*10^4 (9.8) H
solve for H
Just keep that up 
Physics _ Please please help 
Sonya,
I got 25.5 for the (I) one
Could you help me with the (II) and (III)