Finding change in pressure when mercury level drops

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confusedperson
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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. Homework Statement

Refer to uploaded image.

Homework Equations


Pressure = h ( density in kg/m^3 ) ( gravitational acceleration)

The Attempt at a Solution


Pressure change at A = -0.04 ( 13600 ) ( 9.81 )[/B]
Final gauge pressure = 150.3kPa - 0.04 (13600) (9.81) - 1.025 * 10^5 Pa
 

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Sorry I do not understand what is the real balance equation...
 
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?
 
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.
 
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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.
 
That's the idea.
mfb said:
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.
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 ?
 
BvU said:
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 ?
@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.
 
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