A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.

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.