An open glass tube is immersed in mercury in such a way that a lenth of 8cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm. What will be length of the air column above mercury in the tube now?
(Atmosphere pressure =76cm of Hg)

Explanation

Figure-2.16(a) shows the ttube in initial sttate then it is closed at top and raised up so that its length above mercury becomes `8+44=52` cm as shown in figure-2.16(b). As initially tube is open to atmosphere, pressure of air inside is 76 cm of Hg and its vol ume is 8A if A is the cross-sectional area of the tube.
When its upper end is closed and raised up, situtation is shown in figure-2.16 (b). Let the air column be of length x then its final volume will be xA. Pressure at the mercury level in the container is equal to atmospheric pressure 76 cm of Hg and the air column is separated by another mercury column of height 52-x above this level which will oppose the atmospheric pressure thus pressure inside the column will be `76-(52-x)=(24+x)` cm of mercury.
As during the process temperature of system remains constant, thus we can use Boyle's law for the intial and final states as
`P_(1)V_(1)=P_(2)V_(2)`
or `" "76xx8A=(24+x)xx xA`
or `" "608=24x+x^(2)`
or `" "x^(2)+24x-608=0`
On solving, we get x = 15.4 cm or x = -39.4 cm
Thus the acceptable value of final air column is 15.4 cm.