A column of Hg of 10cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both ends. Both the halves of the tube contain air at a pressure 76 cm of Hg. By what distance will the column of Hg be displaced, if the tube is held vertical? (Assume temperature to be constant)

A column of Hg of 10cm length is contained in the middle of a narrow h

1.	A column of Hg of 10cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both ends. Both the halves of the tube contain air at a pressure 76 cm of Hg. By what distance will the column of Hg be displaced, if the tube is held vertical? (Assume temperature to be constant)
Answer» SOLUTION :If initially the length of air column on each side is L, ACCORDING to the given problem, `2L+10=100,i.e.,L=45cm""....(1)` Now if the tube is held vertical, the Hg column will be displaced downward by y such that `P_(B)+10=P_(A)""....(2)` Now applying Boyle.s law to air enclosed in side A, `P_(0)LA=P_(A)(L-y)A,i.e.,P_(A)=(LP_(0))/((L-y))""...(3)` while for air enclosed in side B, `P_(0)LA=P_(B)(L+y)A` i.e., `P_(B)=(LP_(0))/((L+y))" "...(4)` Substituting the value of `P_(A)andP_(B)` from equation (3) and (4) in (2), with L = 45 and `P_(0)` = 76 cm of Hg, we GET `(45xx76)/((45-y))-(45xx76)/((45+y))=10,ory^(2)+684y-(45)^(2)=0` or `y=[-684pmsqrt((684)^(2)+4(45)^(2))]//2` or `y=-342+345=3cm`