Pressure in fluids question

Oct 2017
8
0
A glass tube of cross-sectional area 1.0x10^-4 m^2 is partially filled with water. An oil with a density of 800kg/m^3 is slowly poured into the tube and floats on top of the water. The height of the oil above the water surface is 8cm.
A) what is the change in pressure at a depth of 10 cm below the water surface?
b) would the pressure at this depth increase or decrease if you added more oil to the tube? Why?

My work:
a) Poil = P0+pgh
Poil = 1.013e5 +800x9.8x0.08
Poil = 101927.2 Pa

Pwater = 1.013e5 + 1000x9.8x0.18
Pwater = 103064
0.18 is when I added 0.08 +0.10 because that is the depth of the water? I think.

Then I subtracted the Pwater from the Poil to get 1136.8 Pa. I don't think that is right at all b/c someone told me you only need to calculate pressure of oil??? What? Also, I don't know how would I use the cross sectional area for this question, is it just extra useless info? Sorry for the formatting.

b) Increase because more oil = greater depth = greater pressure.
 
Last edited:
Apr 2017
518
125
A) wants the Change in pressure caused by adding the oil ...800 x 9.81 x 8 = 62,784 Pa ... ( that's the sort of figure we should expect since 1atm is 10meters of water = 100,000Pa)

No need for the Area of tube or the depth of water ....

B) ...Sure the pressure will increase ... correct answer
 
Oct 2017
8
0
A) wants the Change in pressure caused by adding the oil ...800 x 9.81 x 8 = 62,784 Pa ... ( that's the sort of figure we should expect since 1atm is 10meters of water = 100,000Pa)

No need for the Area of tube or the depth of water ....

B) ...Sure the pressure will increase ... correct answer
For a), isnt the formula for pressure: P0 + pgh. Why do you not place in your equation the atmospheric pressure (P0)
 
Apr 2017
518
125
For a), isnt the formula for pressure: P0 + pgh. Why do you not place in your equation the atmospheric pressure (P0)
Because the question asks for the CHANGE in pressure when the oil is added.
 
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