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Let curved surface area of tube =A

Given

Length of air column=43cm=0.43m

Length of mercury column=20cm=0.20m

Pressure due to mercury column==0.2m of Hg

Atmospheric pressure=P_a=0.76m of Hg

Let the pressure of air column before titling =P₁

So P₁=P_a +P_H

P₁=0.76+0.2=0.96m of Hg

Volume =areaheight

Volume of trapped air V₁=Alength of air column=0.43A

If the tube is titled through an angle 60^o only pressure of mercury column will get affected and not the atmospheric pressure.

So, change in P_H will be

P’_H=P_Hcos60^o = 0.20.5 =0.1m of Hg

So now the pressure of air column will become P₂

P₂=P_a+P’_H=0.76+0.1=0.86m of Hg

Then volume will change. Let it now be V₂=lA where l is new length of air column.

It is given in question that the temperature remains same. So, according to boyle’s law which states that PV=constant when temperature is constant, we can write,

P₁V₁=P₂V₂

Length of the air column will become 0.48m=48cm