A column of mercury of 10cm length is contained in the middle of a narrow horizontal 1m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will the column of mercury be displaced if the tube is held vertically?

Explanation

If initially the length of air column on each sides is `L`, according to the given problem,
`2L + 10 = 100`, i.e., `L = 45`…..(1)
Now if the tube is held veritcal, the `Hg` column will be displaced downward byy such that `P_(B) + 10 = P_(A)` ......(2)
applying Boyle's law to air enclosed inside `A`,
`P_(0)LA = P_(A)(L-Y)A,` i.e., `P_(A) = (LP_(0))/((L - y)`......(3)
While for air enclosed inside `B`,
`P_(0)LA = P_(B)(L + y)A`, i.e., `P_(B) = (LP_(0))/((L + y))` .....(4)
Substituting the values of `P_(A)` and `P_(B)` from equation `(3)` and (4) in `(2)`, with `L = 45` and `P_(0) = 76 cm` of `Hg`, we get
`(45 xx 76)/((45 - y)) - (45 xx 76)/((45 + y)) = 10 , y^(2) + 684y - (45)^(2) = 0`
or `y = ([-684pmsqrt((684)^(2)+4(45^(2)))])/2`
or `y = -342 + 345 = 3cm`