# In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m^3, ρwater = 1000 kg/m^3).

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In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m^3, ρwater = 1000 kg/m^3).

(a) 1260%

(b) 92.64 %

(c) Remains unchanged (0%)

(d) 13.6%

This question was addressed to me in an interview for job.

I need to ask this question from Manometer in chapter Buoyancy and Floatation of Fluid Mechanics

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Right choice is (a) 1260%

To explain I would say: Since the gauge pressure remains the same ρ*(h2 – h1) = constant. The height difference in mercury manometer is 0.30 m and that in a water manometer is 4.08 m. Percent change is thus, 1260%. Be careful about the denominator used for computing percent change.

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