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JEE Advanced 2026 · Paper 1, Physics question 11

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition (P1)(P_1) and a freely movable but thermally insulated piston (P2)(P_2). The partition P1P_1 with thermal conductivity KK, cross sectional area AA and width xx divides the container into two sections, S1S_1 and S2S_2, each containing one mole of a monoatomic gas. The piston P2P_2 moves freely such that the gas in S2S_2 is always at the atmospheric pressure. Initially, the difference between the temperatures of S1S_1 and S2S_2 is ΔT0\Delta T_0. The time it takes for the temperature difference to become ΔT02\dfrac{\Delta T_0}2 is nxR/KAnxR/KA, where RR is the universal gas constant. The value of nn is:
[Given: ln⁡2≈0.7\ln2\approx0.7]

A horizontal insulated container: section S₁ on the left, a hatched fixed partition P₁ of width x, section S₂, then a movable piston P₂ on the right (open beyond).
Answer and worked solution

Official answer: 0.63

Heat capacities.

  • S1S_1 is closed by fixed walls: constant volume, CV=32RC_V=\frac32R.
  • S2S_2 stays at atmospheric pressure: CP=52RC_P=\frac52R.

Heat flow. H=KAxΔTH=\dfrac{KA}{x}\Delta T from the hotter section to the colder one, so
d(ΔT)dt=−KAxΔT(23R+25R)=−16KA15xRΔT.\frac{d(\Delta T)}{dt}=-\frac{KA}{x}\Delta T\left(\frac{2}{3R}+\frac{2}{5R}\right)=-\frac{16KA}{15xR}\Delta T.

Halving time.
t=1516ln⁡2⋅xRKA≈1516(0.7)xRKA=0.656xRKA.t=\frac{15}{16}\ln2\cdot\frac{xR}{KA}\approx\frac{15}{16}(0.7)\frac{xR}{KA}=0.656\frac{xR}{KA}.
So n≈0.66n\approx0.66.