Question

In the manometer given above, 2 immiscible fluids mercury (ρ = 13600 kg/m^3) and water (ρ = 1000 kg/m^3) are used as manometric fluids. The water end is exposed to atmosphere (100 kPa) and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. Ignore the width of the manometer tube and the radius of curvature. The value of h is found to be 9.45 m. The height of the mercury column is given to be 75 cm. Find the gauge pressure of the gas. (g = 9.8 m/s^2)

(a) 100 kPa

(b) 50 kPa

(c) 200 kPa

(d) 0 kPa

This question was addressed to me in class test.

My question is from Manometer in section Buoyancy and Floatation of Fluid Mechanics

Answer 1

Right option is (d) 0 kPa

To explain: Height of water column = 0.75 + 9.45 = 10.2 m. We equate the pressures at the bottom most point. Pa + ρw.g.(10.2) = Pg + ρm.g.(0.75). We find, Pg = 100 kPa = Absolute pressure. Hence, gauge pressure will be 0.