Answer :


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=Pa=0.76m of Hg


Let the pressure of air column before titling =P1


So P1=Pa +PH


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 60o only pressure of mercury column will get affected and not the atmospheric pressure.


So, change in PH will be


P’H=PHcos60o = 0.20.5 =0.1m of Hg


So now the pressure of air column will become P2


P2=Pa+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


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