Gauge Pressure

[qtsjoe]

The question is sparked from a question in my TPR science workbook.

A beaker with liquids of different densities A,B,C

Given:

density/rho -> A<B<C

each layer is .3m of fluid (In total .9m depth)

What is the difference in gauge pressure of a point (1) at the surface of A (in between air and the surface of A) compared to a point (2) in between C and B?

Relevant formula :

Pgauge= rho g d

----
This is my confusion

We know that point A has no depth (0) therefore the difference is equal to the gauge pressure at point 2.

P= (rho A+ rho B) (10) ( .3m)

my intuition was to take into account both the distance of liquid A as well as B (=.6m) but the solution manual says as stated above.

---

Members don't see this ad.

 

[SaintJude]

I'm gonna go ahead and say.... it was a typo. It should be 0.6.

 

[qtsjoe]

I'm gonna go ahead and say.... it was a typo. It should be 0.6.

i thought so too... hopefully we are right.

 

[MedPR]

Why doesn't A have a depth?

Edit: Oh, I see. The book is saying A has no depth and that's why you are thinking there is a problem. I agree.

 

[empo]

The solution manual is right.


---------------------- 0
A
---------------------- (rho A)*g*0.3m
B
---------------------- ((rho A)*g*0.3m) + ((rho B)*g*0.3m)
C

- - - - - - - - - - - -

If you want you can use the total depth of A + B, but then you need to use the average density of the two liquids. (rho A + rho B)/2 which will give you the same answer.