Question

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?

Answer 1

Here,
Total leni of the tube = 1 m or 100 cm

For horizontal position
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Length of mercury thread in the tube = 76 cm
Length of air column between closed end and mercury thread = 15 cm
So, Length of remaining part of the tube = 100 - ( 76 + 15)
= 9 cm will be air space in the tube .
Hence, total length of air column = 15 + 9 = 24 m { see figure }
When the tube vertical 9 cm of air + some mercury flow out .
Let h is the length of mercury column flowing out ,
Then, length of air in column = (15 + 9 + h) = (24 + h)

And Length of mercury = 100 - (24 + h)
= (76 - h)
Pressure (P1 )in horizontal position = 76 cm of Hg
Pressure ( P2) in vertical position = 76 - ( 76 - h) = h cm of Hg

Volume of air ( V1) in horizontal position = Length of air × area of tube
= 15A cm³
Volume of air ( V2) in vertical position
= (24 + h)A cm³

Use Boyle's Law ,
P1.V1 = P2.V2
76 × 15A = h × (24 + h)A
1140 = 24h + h²
Use quadratic formula ,
h = { -24 ± √(24² + 4 × 1140)}/2
= 23.8 cm or -47.8 cm
h ≠ -47.8 cm
So, h = 23.8
Hence, new length of mercury = ( 76- 23.8) = 52.2 cm