a 10cm column of air is trapped by an 8cm long column of Hg in a capillary tube horizintally fixed at 1atm pressure.What will happen to Hg column if the tube is held vertically with open end down ?

upward pressure on Hg column=760mm=P1,downward pressure=80mm+10cm of air=P2, which is obviously less than P1.

So the mercury column will be pushed up by the air below it and compresses the trapped air column till its pressure increases to the point where equilibrium is established.

P1= P (atm)P(atm)=P2+density of Hg ×g× hAs n1=n2P1V1=P2V2P(atm)×(area×l1)=P(atm)-density of Hg ×g ×h×(al2)Therefore l2=h of Hg×l1÷(h of Hg -h)l2= 76×10÷76-8l2 =760÷68L2 = 11.1764