When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end, figure (a). The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.
∴P1=76cm and V1=15cm3
When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure (b) 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
∴V2=(24+h)cm3 and P2=hcm
If we assume that temperature remains constant, then P1V1=P2V2
or 76×15=h×(24+h) or h2+24h–1140=0
or h=

−24±√(24)2+4×1140

2

=23.8cm or −47.8cm
Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical position of the tube, 23.8 cm of mercury flows out.