A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?

Explanation

When the tube is held horizontally it is as shown in figure.
Density of mercury `= 13.6 xx 10^(3) k g m^(-3)`
Area of cross-section of the tube = A
Pressure of air `P_(1) = 76 cm` of Hg
Volume `= A xx l = A xx 15 cm^(3)`
When the tube is made vertical, some mercury flows out. Let h be the height of mercury. The pressure exerted by air column `P_(2)` and the mercury column of height h balances the atmospheric pressure .
` P_(2) + h` = atmospheric pressure = 76 cm of Hg
`:. P_(2) = 76 - h`
and `V_(2) = A ( 100 - h )`
from Boyle.s law, if temperature is constant

`P_(1) V_(1) = P_(2) V_(2)`
`76 xx A xx 15 = ( 76 - h ) xx A xx ( 100 - h )`
`1140 = 7600 - 76 h - 100 h+ h^(2)`
`h^(2) - 176 h + 6460 =0`
`h = ( + 176 +- sqrt( 176^(2) - 4 xx 6460 xx 1))/( 2 xx 1 )`
`= ( 176 +- sqrt( 30976 - 25840))/( 2)`
= 123.8 cm or 52.2 cm
The length of mercury column is 76cm only so h = 52.2cm. The length of mercury column that has flown out.