Gauge pressure and mercury question

AI Thread Summary
The discussion revolves around calculating gauge pressure at a depth of 6 cm in a glass containing 4 cm of mercury and 4 cm of water. The key formula for static pressure in a fluid, P = ρgh, is emphasized, where ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid. Participants clarify that the gauge pressure must consider the contributions of both mercury and water, with mercury being denser and thus affecting the pressure more significantly. After some back-and-forth, the original poster finds the answer with assistance from others. The conversation highlights the importance of understanding fluid dynamics principles to solve such problems effectively.
stormnebula
Messages
3
Reaction score
0
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:
Physics news on Phys.org
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:

P = \rho\,g\,h where \rho 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
 
Last edited:
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:
 
Last edited by a moderator:

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Solving an Air Pressure Question: Who Is Right?
Replies
5
Views
2K
Gauge pressure and atmospheric pressure
Replies
2
Views
2K
What is the issue with calculating the gauge pressure of object B?
Replies
1
Views
3K
Calculating Liquid Pressure: Water and Mercury
Replies
1
Views
2K
Calculating Work Done by Gas in a Mercury-Filled U-Tube with Right Angle Bends"
Replies
10
Views
2K
What is the solution to the straw paradox?
Replies
9
Views
3K
How do I find the gauge pressure?
Replies
1
Views
9K
Rising of a liquid during rotataion in a cylinder
Replies
1
Views
1K
Change in gauge pressure using viscous flow through a pipe.
Replies
6
Views
8K
When is gauge pressure absolute pressure?
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top