Assuming the gas to be an ideal mono-atomic gas ∴ Cv=23R and area of cross section to be A. Initially P04A=0.1R×300 (For both) Let piston shifts by × meter and final temperature is T. Then P2A(4−x)=0.1RT∴P2A=4−x0.1RT=10(4−x)RT And P1A=10(4+x)RT But finally P2A−P1A=83=10RT(4−x1−4+x1)⇒10RT(16−x22x)=830.2Cv×300+mg⋅x=0.2CvT⇒23R×102×300+83x=102×23RT⇒(10900R+83x)=3(10RT)(10300R+383x)=10RT(30R+383x)(16−x22x)=83⇒83(3+3x)(16−x22x)=83⇒39+x16−x22x=1⇒18x+2x2=48−3x2⇒5x2+18x−48=0∴x=1.78≈2 So, the distance from top is 6m.