TPR Workbook Physics Passge 14, Q 5 HELP

[Radioactive1112]

---

Members don't see this ad.

Hello

I need help understanding the answer to this question. First, fluids annoy me. So if anybody has any good websites to practice fluid problems please list below. Anyway, the passage states...

A vat is filled with equal amounts of three diff liquids which separate into three distinct layers.

Layers
P1 density 736
P2 density 1000
P3 density 1490

Each has a thickness of .3m

A solid ball density Po is dropped into the vat and settles between layers 2/3 (p2 and p3) at the interface.

Question 5: What is the approximate difference in pressure between Points X (interface of p1) and Y (interface of 2/3) ?

A. 3.0 kpa
B. 4.5 kpa
C. 5.2 kpa
D. 10.0 kpa

The answer is 5.2 kpa with the explanation that gauge pressure is only the gauge pressure at Y. Ok that makes sense to me because the height of X is zero. However, to calculate gauge pressure of Y they used the height of .3m. I thought you would use .6m since its the second layer down? Can somebody explain why this is? I can try to upload the picture if this doesnt make sense.

 

[Cawolf]

I don't have the problem, but likely it is due to both of the most dense layers being under the top layer. Therefore, the pressure exerted by the layer of rho = 736 is exerted across both bottom layers and does factor into the pressure differential.

All you need to do is calculate the pressure exerted by the middle layer on the bottom layer for relative difference.

 

[Labrat07]

Here's how I look at this. We know the ball is between 1000 and 1490. So use (1490-1000)/2 = 250 ish.

Then you calculate the pressure of 1000- 736 = 264 ish
264 ish + 250 ish = 520 ish. In term of conceptual, the top layer 736 pressure = 1000 -736. Then you have to add the pressure of water between that layer and the ball. We got all the pressure of water on the pall.