What is the issue with calculating the gauge pressure of object B?

[ppppparker]

Homework Statement

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

Homework Equations

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

 

[SteamKing]

Homework Statement

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

Homework Equations

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

Couple things here:
1. g was given as 9.1 m/s² in the problem statement. Why did you use 9.8 m/s²? Is one a typo?

2. The pressure differential due to the fluid depth is ΔP = ρ ⋅ g ⋅ Δh.

3. A gauge pressure reading is set it so that 1 atm. abs. = 0 atm. gauge. A depth of 1 meter of fluid which is less dense than water is clearly an insufficient depth to provide 2 atm. of pressure difference. What can you conclude about the gauge reading for object A?