[tracy123]

An unknown liquid is used to fill a closed-tube manometer. The atmosphere is found to produce a height difference of 4.19 m in this manometer at the same time that a mercury manometer gives a displacement of 729.2 mm. What is the density of the unknown liquid, in units of g/cm^3?

I think the formula's d=P/(hG), where G=9.8m/s^2... but my units don't work out .

[Donaldson Tan]

P = h.p.g where
1. h is the difference in height
2. p is the density
3. g is gravitational constant

let component 1 be the unknown liquid.
let component 2 be mercury

density of mercury = p2 = 13600kg/m3

since pressure is the same in both cases, then
h1.p1.g = h2.p2.g => h1.p1 = h2.p2
density of unknown liquid = p1 = h2.p2/h1 = (729.2E-3)(13600)/(4.19) = 2370kg/m3