A column of Hg of 10 cm in length is contained in the middle of the narrow 1 m long tube which is Closed at both ends .Both the halves of the tube Contained air at a pressure 76 cm of Hg .By What distance will be the column of Hg be displaced if the tube is held verticall?

Dear student 

M is the mid-point of tube AB.
At equilibrium
p1 x A + mg = p2 x A
p1 x A + 10 x A x dHg = p2 x A
⇒ p1 + 10dHg x g = p2
For air present in column AP
p x 45 x A = p1 x (45 + x ) x A
⇒ p1 = 45 / 45 + x x 76 dHg x g …..(ii)
For air present in column QB
p x 45 x A = p2 x (45 – x ) x A
⇒ p2 = 45 / 45 - x x 76 dHg x g (iii)
From (i), (ii) and (iii)
45 x 76 x dHgg / 45 + x + 10 dHg x g = 45 / 45 – x x 76 x dHg x g
⇒ 45 x 76 / 45 + x + 10 = 45 x 76 / 45 – x
x = 2.95 cm.