Question -
Answer -
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 atthe bottom, the mercury length that occupies the air space is:
= 100 тАУ (76 + 15)
= 9 cm
Therefore,
The total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure
So,
Length of the air column in the bore = 24 + h cm
And,
Length of the mercury column = 76 тАУ h cm
Initial pressure, V1┬а= 15 cm3
Final pressure, P2┬а=76 тАУ (76 тАУ h)
= h cm of mercury
Final volume, V2┬а=(24 + h) cm3
Temperature remains constant throughout the process
Therefore,
P1V1┬а= P2V2
On substituting, we get,
76 x 15 = h (24 + h)
h2┬а+ 24h тАУ 11410= 0
On solving further, we get,
= 23.8 cm or -47.8 cm
Since height cannot be negative. Hence, 23.8 cm of mercury willflow out from the bore
Length of the air column = 24 + 23.8 = 47.8 cm