## pressure in u-tube of mercury with added water

1. The problem statement, all variables and given/known data
A simple open U-tube contains mercury. When 11.8 cm of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

Hint: At the level of the interface between the water and the mercury, the pressure on the left must balance the pressure on the right. In each the pressure is the gauge pressure of a column of liquid standing above that level.

2. Relevant equations
P=pgh

3. The attempt at a solution

the pressure at the surface of interaction in the right arm should equal the pressure, at the same height, of the mercury in the left arm. therefore:

p(water)gh(water)=p(mercury)gh(mercury)
1*10*11.8=13.534*10*h
11.8/13.534=h

.872=h

But the online thing says that isn't the right answer. where am i going wrong?
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 Recognitions: Homework Help The amount of the mercury iremains the same. If its surface is depressed by d in the right arm, it should rise by d in the left arm, so you have h=2d length of mercury balancing the water column, but the change of height is d, half of your value. ehild

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