Answer :

Let the curved surface area of tube be A.


Volume =areaheight


Given


Initial length of trapped air=20cm =0.2m


Length of mercury column=10cm=0.1m


So, Mercury column pressure =0.1g Pa


Initial volume of air trapped V1=0.2A


Atmospheric pressure =75cm of Hg=0.75m of Hg


1mm of Hg = hg Pa


Where h= height of mercury column =1mm=0.001m


= density of mercury


g= acceleration dur to gravity


So, atmospheric pressure= 0.75m of Hg= 0.75g Pa


Let the pressure of the trapped air when closed end of the tube is upward be P1. Now, pressure of the mercury and trapped air will then be equal to atmospheric pressure.


P1+0.1g=0.75g


P1=0.65g


When the tube is inverted such that closed end is downward then pressure of trapped air will be P2.


P2=atmospheric pressure + mercury column pressure


P2=0.75g+0.1g=0.85g


Let length of air column at P2=x


Volume of air trapped will be V2=A x


Now temperature in both cases will remain same, as no heat is being either added or abstracted from tube.


Applying boyle’s law,


P1V1=P2V2


0.65g0.2A=0.85gAx



The length of the air column trapped when the tube is inverted so that the closed-end goes down is 0.15m=15cm.


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