$A$ 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?

(D) Length of the narrow bore,$L = 100 \; cm$.
Length of the mercury thread,$l = 76 \; cm$.
Length of the air column between mercury and the closed end,$l_a = 15 \; cm$.
When the tube is held vertically with the open end at the bottom,the air column length increases because some mercury flows out. Let $h \; cm$ of mercury flow out.
The final length of the air column becomes $l_2 = 15 + (100 - 76 - 15) + h = 24 + h \; cm$.
The final length of the mercury column becomes $76 - h \; cm$.
Initial pressure $P_1 = 76 \; cm$ of Hg,Initial volume $V_1 = 15 \; cm$ (proportional to length).
Final pressure $P_2 = 76 - (76 - h) = h \; cm$ of Hg.
Using Boyle's Law,$P_1 V_1 = P_2 V_2$:
$76 \times 15 = h(24 + h)$
$h^2 + 24h - 1140 = 0$.
Solving for $h$ using the quadratic formula: $h = \frac{-24 + \sqrt{576 + 4560}}{2} = \frac{-24 + 73.05}{2} \approx 24.5 \; cm$.
Thus,$24.5 \; cm$ of mercury flows out,and the final air column length is $24 + 24.5 = 48.5 \; cm$.