Gauge pressure and mercury question

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Homework Help Overview

The discussion revolves around a gauge pressure problem involving a column of mercury and water. The original poster presents a multiple-choice question regarding the gauge pressure at a specific depth in a glass filled with these two liquids, highlighting the densities of water and mercury as key factors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the concept of gauge pressure, noting the need to consider the depth of the liquids and their respective densities. There is a focus on the formula for static pressure in fluids and how to apply it to the given scenario. Questions arise about the interpretation of the depth in relation to the different liquid layers.

Discussion Status

Some participants have provided insights into the calculations involved, while others seek further clarification on the steps necessary to arrive at the gauge pressure. The discussion includes attempts to break down the problem into manageable parts, with varying levels of understanding among participants.

Contextual Notes

The problem involves specific measurements and densities, which are critical to the calculations but may not be fully understood by all participants. There is an emphasis on the need for clarity in applying the relevant formulas and concepts.

stormnebula
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Gauge pressure question - please help

I came across this multiple choice question while revising for my exams but I can't get any of the chioces. I've spent over an hour on this but can't figure out. Can someone please help me because it's starting to annoy me.

What is the gauge pressure at a depth of 6cm in a glass filled with 4cm of mercury and 4cm of water? Water has a density of 1000kg/m^3, and mercury has a density 13.6 times as great.
(a)3.1kPa
(b)5.6kPa
(c)5.8kPa
(d)310kPa
(e)560kPa

Any help will be great.
 
Last edited:
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Gauge pressure is the absolute pressure minus the ambient pressure, so the problem asks for the pressure in the column simply due to the depth of the liquids.

One has 4 cm of Hg and 4 cm of water, but the problem asks for pressure at a depth of 6 cm (from the upper surface). Hg is denser to it has to be on the bottom.

So what is the pressure underneath 4 cm water and 2 cm Hg.

Static pressure in a fluid is given by:

[tex]P = \rho\,g\,h[/tex] where [tex]\rho[/tex] is fluid density, g = acceleration of gravity (~9.81 m/s2), and h is height of fluid.

Consider 1 atm = 14.7 lbf/in2 = 760 mm Hg = 101.325 kPa = (760x13.6) mm water.

Final the partial pressures due to depths of each fluid and that will be the gauge pressure.


See also - http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp
 
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Thanks, I figured out the answer now. Your were a big help :smile:
 
I had the same question. Can you please go step by step on how you acquired the answer, because I don't understand how to "final" the different liquids and getting a gauge pressure.
 
Welcome to PF!

Hi Blackplague! Welcome to PF! :smile:

If you don't understand the formula, look at it this way …

https://www.physicsforums.com/library.php?do=view_item&itemid=80" = force/area

the gauge pressure at the bottom of the glass is the weight of all the liquid in a vertical cylinder, divided by the area of that cylinder …

so if you have a vertical cylinder of cross-section area A, what is the weight of water and mercury in that cylinder? :wink:
 
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