
#1
Sep2106, 10:59 PM

P: 8

A Ushaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 20.0 cm deep.
How far upward from its initial position does the mercury in the right arm rise? So far, this is my work: Using pressure = density(water) x g x height(water column) and neglecting atmospheric pressure as both sides would experience it, I equated that equation to p = density(Hg) x g x height as they would equal on the horizontal axis. I found the height to be 14.7 mm but that answer is not correct. I am confused and have gone at this in a number of ways. I know that the value 14.7 is not the difference between the new position and the initial position and so I've kind of hit a road block. Any input or thought or help would be appreciated! 



#2
Sep2206, 03:30 AM

P: 3,172

i think it should be:
[tex]\rho_{w}h'g+\rho_{Hg}h'g=\rho_{Hg}hg[/tex] where h'=20 cm and h is the height of the right arm. in order to find the difference i think you need this equation: xh'=hx but im not sure im right here, ive done this stuff in highschool two years ago. 


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