Mercury is poured into an uneven u-shaped tube. The cross-sectional area of the left column is , while the cross-sectional area of the right column is . Then 300 grams of water is poured into the right column. Given that the density of water is and the density of mercury is , calculate the displaced height of the mercury in the left column.
Let be the pressure at the base of the water column. Since both sides of the tube are open to the atmosphere,
Subtract the atmospheric pressure on both sides. Then divide the gravitational acceleration on both sides. Thus,
It is known that
but both heights are not known. We cannot solve for , however there is enough information to solve for . Using the mass of the added water, we can exploit the fact
to solve for the height of the water column.
So . Since
Let be the displaced height of the mercury in the left column, and be the height from the dotted line to the base of the water column. It can be deduced that
So far it appears we have reached a dead end. There is only one equation, but there are two unknowns. However, we should note the volume of water below the dotted line is the same volume of the displaced mercury above the dotted line. Hence, if represents the volume of mercury displaced above the dotted line, then
Now we can solve for .
Thus, the displaced height of the mercury in the left column is .