objects A and b are submerged at a depth of 1m in a liquid with a specific gravity of 0.877. Given that the density of object B is one third that of Object A and that the gauge pressure of object A is 3atm, what is the gauge pressure of object B?
assume atmospheric pressure is 1atm and g = 9.1m/s/s
P_gauge = P_0 + (rho)gz - P_atm
The Attempt at a Solution
Because this is 2 objects submerged in one liquid I am assuming that absolute pressure, P_0 = ambient pressure, P_atm.
So equation is just P_gauge = (rho)gh
And gauge pressure depends on density of fluid not the object submerged, so they will have same gauge pressure.
Also that S.G. = rho of fluid over rho of water, so density of fluid = .877(1000) = 877kg/m^3
But I don't understand why the calculation doesn't work:
P_gauge = (rho)gh
(877kg/m^3) (9.8m/s/s)(1m) should = 3atm ?
any help is really appreciated t..thanks