A metre long narrow bore held horizontally (and close 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, the mercury thread of length of air `= 15 cm`. A lenght of `9 cm` of the tube will be left at teh open end, fig., The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of teh tube be `1 sq. cm`,
`:. P_(1) 76 cm` and `V_(1)=15cm^(3)`
When the tube is held vertically, `15 cm` air gets another `9 cm` of air (filled in the right handside in the horizontal position) and let `h cm` of mercury flow out to balance the atmospheric pressure. Then the heights of air column and mercury column are `(24+h)cm` and `(76-h) cm` respectively.
the pressure of air = `76-(76-h)=h cm` of mercury.
`:. V_(2) =(24+h) cm^(3)` and `P_(2) = h cm`
If we assume that temperature remains constant, then
`P_(1)V_(1) =P_(2)V_(2) or 76 xx 15 = h xx (24+h)` or `h^(2) + 24h - 1140 = 0 or h= -24 +- sqrt((24)^((2)+4 xx 1140))/(2) = 23.8 cm or -47.8 cm`.