A metre long narrow bore held horizontally (and close 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?

Explanation

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.
`therefore`Length of the air column in the bore = 24 + h cm
And, length of the mercury column = 76 – h cm
Initial pressure, `P_(1) `= 76 cm of mercury
Initial volume, `V_(1) = 15 cm^3 `
Final pressure, `P_2 `= 76 – (76 – h) = h cm of mercury
Final volume,` V_(2) = (24 + h) cm^3 `
Temperature remains constant throughout the process.`therefore P_1V
_1=P_2V_2`
76x15=h(24+h)
`h^2+24h-1140=0`
`therefore h=-24pmsqrt((24)^2+4xx1xx1140)/(2xx1)`
23.8 cm or-47.8 cm
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.