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Question

A 10.0 cm column of air is trapped by a column of 4 cm long Hg in a capillary tube of uniform bore when the tube is held horizontally at 1 atm. The length of the air column when the tube is held vertically with the open end up is:
Given: Patm=76 cm of Hg


A
3.53 cm
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B
10.52 cm
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C
9.50 cm
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D
4.61 cm
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Solution

The correct option is C 9.50 cm
When open end is up, the weight of mercury represents an increase in the pressure on the gas.
4 cm Hg=40 mm Hg

Thus, final pressure =760+40=800 mm Hg

Also, Volume=Area×length
P1V1=P2V2 (by Boyle's law)
P1×A×l1=P2×A×l2
(A= area of cross section = constant)
So, P1l1=P2l2
760×10=800×l2
l2=9.50 cm
Length of air column =9.50 cm

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