posted by nathan .
A u-shaped tube, open to the air on both ends,contains mercury. Wather is poured into the left arm until the water column is 10.0 cm deep. How far upward from its initial position does the mercury in the right arm rise?
(1/2)*10.0 cm/13.6 = 0.37 cm
13.6 is the specific gravity of mercury.
The added weight of water on one side will then be balanced by the higher column of Hg on the other side. The 1/2 factor is there because it rises on one side and falls on the other
Give us written solution plz anyone