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?

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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, la = 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)...
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Lengthof 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,la= 15 cm Since theboreis 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 Lethcm of mercury flow out as a result of atmospheric pressure. ∴Length of the air column in thebore= 24 +hcm And, length of the mercury column = 76 –hcm Initial pressure,P1= 76 cm of mercury Initial volume,V1= 15 cm3 Final pressure,P2= 76 – (76 –h) =hcm of mercury Final volume,V2= (24 +h) cm3 Temperature remainsconstant throughout the process. ∴P1V1=P2V2 76 × 15 =h(24 +h) h2+ 24h– 1140 = 0 = 23.8 cm or –47.8 cm Height cannot be negative. Hence, 23.8 cm of mercurywill flow out from theboreand 52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm. read less
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