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, la= 15 cm
Since the boreis 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 + 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) = 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
∴ h = - 24 +- underroot [(24)2 + 4 x 1 x 1140] / 2 x 1
= 23.8 cm or -47.8 cm
Height cannot be negative.
Hence, 23.8 cm of mercurywill flow out from the boreand 52.2 cm of mercury will remain in it.
The length of the air column will be 24 + 23.8 = 47.8 cm.
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