A11. Calculate the difference in height, h, in centimeters, of the water levels

Question

A11. Calculate the difference in height, h, in centimeters, of the water levels in the two arms of the tube.

A12. 10 cm^3 of a second 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?

A13. With 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 U-tube?

Explanation / Answer

A11. From the Bernoulli theorem:
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*10^5 + rho g H = 1.25*1.01*10^5 + 0

0.25 * 1.01* 10^5 = 1*10^4 (9.8) H

H = 0.257 m...........Ans.

A12. if density is changed then

0.25 * 1.01* 10^5 = 0.85*1*10^4 (9.8) H

H = 0.303 m...........Ans.

A13. if P2 is changed then

1.01*10^5 + rho g H = 1.01*10^5 + 0

0 * 1.01* 10^5 = 0.85*1*10^4 (9.8) H

H = 0 m...........Ans.

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