One limb of a mercury manometer is connected to a gas chamber and the other limb is open to the atmosphere. The mercury level on the gas side is 15 cm lower than the mercury level on the atmospheric side. Taking ρ(Hg) = 13600 kg/m³ and g = 10 m/s², what is the gauge pressure of the gas?
Options

Question

One limb of a mercury manometer is connected to a gas chamber and the other limb is open to the atmosphere. The mercury level on the gas side is 15 cm lower than the mercury level on the atmospheric side. Taking ρ(Hg) = 13600 kg/m³ and g = 10 m/s², what is the gauge pressure of the gas?

Options

G1 Physics · Accepted Answer

Because the mercury level is lower on the gas side, the gas pressure exceeds atmospheric pressure. The gauge pressure is P(gauge) = ρ(Hg)gh. Substitute ρ(Hg) = 13600 kg/m³, g = 10 m/s², h = 0.15 m. Therefore, P(gauge) = 13600×10×0.15 = 20400 Pa = 20.4 kPa. Correct option: B.