Calculating Gauge Pressure at the Bottom of a Test Tube

[DrMcDreamy]

Homework Statement

A test tube standing vertically in a test tube rack contains 3.8 cm of oil, whose density is 0.81 g/cm3 and 6.4 cm of water. What is the gauge pressure on the bottom of the tube? The acceleration of gravity is 9.8 m/s². Answer in units of Pa

Homework Equations

P₁-P_o=$$\rho$$_oilgh₁+$$\rho$$_H2Ogh₂

The Attempt at a Solution

P₁-P_o= (810 kg/m³)(9.80 kg m /s²)(0.038 m) + (1000 kg/m³)(9.80 kg m/s²)(0.064 m)

P₁-P_o= 929 Pa

Is the work and answer right? TIA

 

[Rick88]

They are indeed correct.

 

[DrMcDreamy]

Thank you!

 

[Rick88]

Actually, I wonder whether you should also consider atmospheric pressure.

Also, why is it P1-P0 and not +?

 

[DrMcDreamy]

^It came out right with 929 Pa.
The way the prof had shown it in class is P1-P0:

P₂=P_o+$$\rho$$_oilgh₁

P₁=P₂+$$\rho$$_H2Ogh₂

P₁=P_o+$$\rho$$gh₁+$$\rho$$gh₂

P₁-P_o=$$\rho$$gh₁+$$\rho$$gh₂

 

[Rick88]

Oh ok, it was down to notation, then.
When I thought you might need the atmospheric pressure is because I was calculating P₁, but you needed P₁-P₀. :)