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## 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