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 ?

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 ?
Answer» Solution :If initially the length of air column on each side is L, according tothe given problem, 2L + 10 = 100, i.e., L = 45 cm `"…(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 gas equation to air enclosed in side A, `(P_(0)LA)/(RT)=(P_(A)(L-y)A)/(RT)," i.e., "P_(A)=(LP_(0))/((L-y))"...(3)"` while for air enclosed in side B, `(P_(0)LA)/(RT)=(P_(B)(L+y)A)/(RT)," 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 anhd `P_(0)=76CM` of Hg we get `(45 times 76)/((45-y))-(45 times 76)/((45+y))=10" (or) "y^(2)+684-(45)^(2)=0" (or) "y=-342+345=3cm`