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?

Length of the narrow bore, L = 1 m = 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 
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is,
 100 – (76 + 15) = 9 cm 
Hence, the total length of the air column = 15 + 9
                                                              = 24 cm 
Let h cm of mercury flow out as a result of atmospheric pressure. 
∴ Length of the air column in the bore = 24 + h cm 
And, length of the mercury column = 76 – h cm 
Initial pressure, P₁ = 76 cm of mercury 
Initial volume, V₁ = 15 cm³ 

Final pressure, P₂ = 76 – (76 – h) = h cm of mercury 
Final volume, V₂ = (24 + h) cm³ 

Temperature remains constant throughout the process. 
∴                  P₁V₁ = P₂V₂ 

              76 × 15 = h (24 + h) 
h² + 24h – 1140 = 0


Height cannot be negative.
Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it.
The length of the air column will be 24 + 23.8 = 47.8 cm.