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?

Asked by Pragya Singh | 1 year ago |  153

##### Solution :-

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)

= 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 – h cm

Initial pressure, 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

Therefore,

P1V1 = P2V2

On substituting, we get,

76 x 15 = h (24 + h)

h2 + 24h – 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 will flow out from the bore

Length of the air column = 24 + 23.8 = 47.8 cm

Answered by Abhisek | 1 year ago

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