Help with this Gauge Pressure problem - Kaplan's answer seems incorrect

[1bls2g]

Kaplan's answer is B, and their explanation is basically that the Pg at the bottom equals 8atm-3atm-1atm. However, I thought the definition of Pg is P (hydrostatic/absolute pressure) - Patm, rather than P - Po. Po for the bottom liquid with higher density is Patm + p1gh, which is 4 atm; however, Pg should be 7 atm at the bottom of the tank, right? So the ratio should be 3atm:7atm, which isn't even an answer choice...

Isn't the point of a "gauge" to be calibrated to read 0atm at atmospheric pressure so that only the additional pressure is read? Or am I wrong in thinking that? Kaplan says two opposing things in their review notes. First they define Pg = P - Patm, and they state that when P0 = Patm, Pg = pgh. That makes sense. However, in this answer they're implying that Pg = P - P0 = pgh; they're implying that no matter what P0 is (even if it's Patm + some other pressure), Pg = pgh. Isn't that false?

See below for my diagram of it:

Why would the answer be 3:4 instead of 3:7?

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[SuperSneaky24]

Because one fluid lies ontop of the other, you have to remember that hydrostatic pressure at the bottom of the second fluid is P = Patm + P1gh + P2gh. So it's actually 8 = 1 + 3 + p2gh, where p2gh = gauge pressure of the second fluid. Therefore you'd get p2gh = 4. That makes the answer 3/4.

You may have overthinked this question?

Edit: Didn't see your middle explanation til I just saw it. All I remember about hydrostatic pressure is that it's the total pressure felt from gravity + the fluid. Gauge pressure simply just measures the pressure exerted by only the fluid portion.

 

[1bls2g]

From what I remember being taught (as well as the review notes' formula in Kaplan's book, which is then contradicted by this question's answer being B) the gauge pressure is always defined as Hydrostatic pressure - Patm. It's only ever defined as the pressure exerted just by the fluid portion if the P0 = Patm.

 

[SuperSneaky24]

OOoooooooo I see where you are confused now. I believe that formula is only for single fluid problems, as it doesn't make sense to use it in this problem.

It should be Ptotal = Pgauge1 + Pgauge2 + Patm. That's what I used to solve the problem.

 

[1bls2g]

That kind of makes sense and kind of doesn't, though, right? I understand that if we were to keep the two liquids in separate containers rather than on top of one another, the first would have 3atm gauge pressure and the second would have 4 atm gauge pressure. But the question is asking about the ratio of the actual gauge pressure readings, right? So if one fluid generates 3 atm of pressure, then you submerge further into the second one that (on its own) would have created 4 atm of gauge pressure, shouldn't the total gauge pressure reading be 7atm at the bottom?

I think maybe the question isn't asking what it's meant to be asking to arrive at 3:4. Maybe it should be asking what is the ratio of individual gauge pressures created by each layer of fluid instead of the actual (cumulative) gauge pressures. The gauge pressure at the bottom of the second fluid should be Pgauge1 + Pgauge2 (3+4)

 

[1bls2g]

Plus, to further add to my point, since the volume of each fluid of different densities is the same and they're both in the same container, it wouldn't make any sense for the gauge pressure reading at the bottom of the heavier fluid to read only 4 atm, because that would mean that the pressure generated by the heavier fluid was only 1 atm even though it's got the same height (because of same volume) as the lighter fluid. That would actually imply that the bottom fluid is 1/3 the density of the top one, so it should actually be on top.

 

[SuperSneaky24]

That kind of makes sense and kind of doesn't, though, right? I understand that if we were to keep the two liquids in separate containers rather than on top of one another, the first would have 3atm gauge pressure and the second would have 4 atm gauge pressure. But the question is asking about the ratio of the actual gauge pressure readings, right? So if one fluid generates 3 atm of pressure, then you submerge further into the second one that (on its own) would have created 4 atm of gauge pressure, shouldn't the total gauge pressure reading be 7atm at the bottom?

I think maybe the question isn't asking what it's meant to be asking to arrive at 3:4. Maybe it should be asking what is the ratio of individual gauge pressures created by each layer of fluid instead of the actual (cumulative) gauge pressures. The gauge pressure at the bottom of the second fluid should be Pgauge1 + Pgauge2 (3+4)

Yeah you're right. They probably what they should have said for clarity. I guess they assumed some ppl would get 3:7 so they didn't include that as an answer choice so it wouldn't confuse people.

Plus, to further add to my point, since the volume of each fluid of different densities is the same and they're both in the same container, it wouldn't make any sense for the gauge pressure reading at the bottom of the heavier fluid to read only 4 atm, because that would mean that the pressure generated by the heavier fluid was only 1 atm even though it's got the same height (because of same volume) as the lighter fluid. That would actually imply that the bottom fluid is 1/3 the density of the top one, so it should actually be on top.

I don't think they thought about the gauge reading device, but rather about the individual gauge pressure of the fluids. But if the individual gauge reading for the bottom fluid was 4atm, that would make sense density-wise.

This was just a poorly worded question and you caught on perfectly, so nice! Hopefully come test day these ambiguous questions won't show up.

 

[1bls2g]

Thanks! 🙂 I feel better about it now hahah. Kaplan writers can be quite inconsistent some times 🙁