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Answer 1 · 2018-01-17T15:57:54

Let the cross section of the barometric tube be #acm^2# and atmospheric pressure is #p# cm of Hg.

So initial volume of air column #V_1=10acm^3#

And initial pressure of the air column #P_1=(p-72)# cm of Hg.

Again final volume of air column #V_2=8acm^3#

And initial pressure of the air column #P_2=(p-71)# cm of Hg.

By Boyle's law we can write.

#(p-72)*10a=(p-71)*8a#

#=>10p-720=8p-568#

#=>10p-8p=720-568#

#=>2p=152#

#=>p=152/2=76# cm of Hg.