Let
A cm2 be the area of cross-section of the tube. When the tube is in the horizontal position, as shown in the figure (a), the length of the air column on the left hand side region (I) and right hand side region (II) of the mercury column is 45 cm each, the total length of the tube being 100 cm with the mercury column length 10 cm in the middle. The pressure of air in each half is P = 76 cm of mercury and the volume of air in each half is
V=(45A) cm3.
When the tube is turned to vertical position, as shown in figure (b), the mercury column is displaced downwards. Let
x cm be the length of air column I, when the mercury column attains equillibrium. Volumn of column is
V1=Ax cm3 and let
P1 cm of Hg be the pressure of air in column I. The length of column II is
(90−x) cm and its volume is
V2=(90−x)A cm3. Let the pressure of air in column II be
P2 cm of Hg.
∴P1−P2=10 cm of Hg
⋯(i) Initially (when the tube in the horizontal position) the pressure and volume of column I are P = 76 cm of Hg and V = 45A
cm2 respectively. Applying Boyle's law, we have, for column I.
P1V1=PV or
P1=PVV1=76 cm Hg×45A cm3Ax cm3 P1=76×45x cm of Hg
⋯(ii) Similarly, for column II, we have
P2V2=PV or
P2=PVV2=76 cm of Hg×45A cm3(90−x)A cm3 =76×45(90−x)cm of Hg
⋯(iii) Using equation (ii) and (iii) in (i), we get
76×45x−76×45(90−x)=10 which gives
x2−774x+30780=0 the two roots of this equation are x = 731.95 cm and 42.05 cm.
The first value 731.95 cm is obviously impossible.
Hence
x=42.05 cm≈42 cm ∴ Displacement of mercury column
=45 cm−42 cm=3 cm