Assuming the gas to be an ideal mono-atomic gas ∴ Cv=
3
2
R 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=
0.1RT
4−x
=
RT
10(4−x)
And P1A=
RT
10(4+x)
But finally P2A−P1A=83=
RT
10
(
1
4−x
−
1
4+x
) ⇒
RT
10
(
2x
16−x2
)=83 0.2Cv×300+mg.x=0.2CvT ⇒
3
2
R×
2
10
×300+83x=
2
10
×
3
2
RT ⇒(
900R
10
+83x)=3(
RT
10
) (
300R
10
+
83
3
x)=
RT
10
(30R+
83x
3
)(
2x
16−x2
)=83 ⇒83(3+
x
3
)(
2x
16−x2
)=83 ⇒
(9+x)
3
(2x)
(16−x2)
=1 ⇒18x+2x2=48−3x2 ⇒5x2+18x−48=0 ∴x=1.78≈2 So, the distance from top is 6m.