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. $\therefore P_1 = 76 \text{ cm}$ and $V_1 = 15 \text{ cm}^3$ 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$\therefore V_{2} = (24 + h) \text{ cm}^3$ and $P_{2} = h \text{ cm}$If we assume that temperature remains constant, then $P_{1} V_{1} = P_{2} V_{2}$or $76 \times 15 = h \times (24 + h)$ or $h_2 + 24h - 1140 = 0$or $h = \frac{-24 \pm \sqrt{(24)^{2} + 4 \times 1140}}{2} = 23.8 \text{ cm}$ or $-47.8 \text{ cm}$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.