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

If a is the area of cross- section of the tube (a),

Inilial volume of air, `V_(1) = 15 ` a
initial pressure of air,
`p_(1) ` 76 cm (of He column)
On holding this tube of narrow bore vertically (b), let h be the vertical distance through which the mercury in the tube gets depressed to balance the atmospheric pressure. Obviously, h denotes the final pressure of air, i.e., `P_(2) = h `.
It is to be noted that the length of the air column left
= length of the tube - length of Hg column
`100 - (76- h) = (24 + h)`
Final volume , `V_(2) = (24 + h)a `
Applying Boyle.s law,
`P_(1) V_(1) = P_(2) V_(2) " or " 76 xx 15a ` = h (24 + h)a
or `1140 = 24h + h^(2) " or " h^(2) + 24h - 1140 = 0`
or `h = (-24 + sqrt((24)^(2) + 4 xx 1 xx 1140))/(2)`
since h cannot be negative tube, h = 23.8 cm .
Thus, in the vertical position of lhe tube, 23.8 cm of rnercury flows out and the remaining 52.2 cm of n1ercury thread plus the 48 cm of air above it remain in equilibriun with the outside atmospheric pressure.