# Finding change in pressure when mercury level drops

1. Aug 24, 2016

### confusedperson

I don't know what I did wrong for this question, but I assume something is wrong as my solution is extremely simple and this is a 20 marks question...
1. The problem statement, all variables and given/known data

2. Relevant equations
Pressure = h ( density in kg/m^3 ) ( gravitational acceleration)

3. The attempt at a solution
Pressure change at A = -0.04 ( 13600 ) ( 9.81 )

Final gauge pressure = 150.3kPa - 0.04 (13600) (9.81) - 1.025 * 10^5 Pa

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2. Aug 24, 2016

### BvU

Hi conf,

You want to make a real balance equation and take the density of water into account as well...

3. Aug 24, 2016

### confusedperson

Sorry I do not understand what is the real balance equation...

4. Aug 24, 2016

### BvU

4 cm of Hg took the place of 4 cm of water. That 4 cm of water exercised a pressure as well

5. Aug 24, 2016

### confusedperson

Thank you. I'm not entirely sure how that works but I assume I have to add 4cm of water back into the final gauge pressure?

6. Aug 24, 2016

### Staff: Mentor

And don't forget the 4 cm of mercury that were there. The mercury level changes on both sides, the water level changes on one side.

7. Aug 24, 2016

### confusedperson

Huh. That was not expected...I simply have not come across a problem of this type before. There goes at least 10 marks for my final exam. However, lesson learnt. Thank you guys for enlightening me on this new knowledge.

8. Aug 24, 2016

### BvU

That's the idea.
I thought he/she did take the Hg, but only forgot the water ? So with (13600 - 1000) instead of just 13600 kg/m3 it would have been alright ?

9. Aug 24, 2016

### haruspex

@mfb is pointing out that if the level of Hg on the right drops by x then the level on the left rises by x, so the difference in heights drops by 2x. The pressure exerted comes from the height difference.

10. Aug 24, 2016

### BvU

Oh boy, have I been sleeping !! Fixed on the water Thanks for the wake-up call ! But nevertheless I'm off to bed.