A horizontal tube of 1m is closed at both ends. A mercury column of 20 cm is contained in the middle of the tube and two equal lengths of this tube contain air at an atmospheric pressure of 76cm. The tube is now placed vertically. Calculate the distance through which the mercury column will be displaced at a constant temperature.

A horizontal tube of 1m is closed at both ends. A mercury column of 20

1.	A horizontal tube of 1m is closed at both ends. A mercury column of 20 cm is contained in the middle of the tube and two equal lengths of this tube contain air at an atmospheric pressure of 76cm. The tube is now placed vertically. Calculate the distance through which the mercury column will be displaced at a constant temperature.
Answer» Solution : Let M and N are two parts of tube at the LEFT and right sides of O. Case (i) : When tube is horizontal, Lengths of M and N =40cm Pressure of M and=76 cm of Hg Case (II): Let the tube is inverted and placed in vertical position. Let the mercury column is displaced by a distance y by placing tube vertically Lengths of M and N are (40-y) and (40+y) respecively. `P_(1)`= Preessure of par M `P_(2)`= Pressure of PART N As TEMPERATURE is constant Therefore, we can write PV=constant For M: `P_(0) (40 A)=P_(1) (40-y) A` `rArr P_(1)=(P_(0) (40))/((40-y)) .......(i)` For N: `P_(0) (40 A)=P_(2) (40+y) A` `rArr P_(2)=(P_(0)40)/((40+y)) ...........(ii)` Here A= area of cross section of tube Pressure at lower face `(P_(1))="Pressure at upper face "(P_(2))+` Pressure due to mercury column of height 20 cm. `P_(1)=P_(2)=20pg ........(iii)` Use (i) and (ii) in (iii) `P_(0)(40)/((40-y)) =(P_(0)40)/((40+y)) +20pg` `(40P_(0))/((40-y)) -(40P_(0))/((40+y)) =20pg` `40 XX 76 pg [(2y)/(1600-y^(2))]=20pg` `(52) (2y)=1600-y^(2)` `304 y=1600-y^(2)` `y^(2)+304 y-1600=0` Solving for y, we get y=5.18 cm