A column of Hg of 10 cm 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 conl.ain 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 10 cm length is contained in the middle of a narrow

1.	A column of Hg of 10 cm 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 conl.ain 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 = 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 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) L A = P_(B) (L + y) A ,` i.e, `P_(B) = (LP_(0))/((L + y)) `....... (4) Substituting the values of `P_(A) and P_(B) ` from euqation (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 , " or " y^(2) + 684y- (45)^(2)` = 0 or= y `[ - 684 PM sqrt((684)^(2) + 4 (45)^(2) ) ] `/2 or y = - 342 + 345= 3 cm