Verified Solution • 2025–26 Session moderate QJEEAdv21Phba7e: JEE Advanced 2021 Physics - Class 11 JEEAdv21Phba7e Solution The value of QRT0{Q \over {R{T_0}}}RT0Q is Multiple Choice Options A 4(22+1)4(2\sqrt 2 + 1)4(22+1) B 4(22−1)4(2\sqrt 2 - 1)4(22−1) C (52+1)(5\sqrt 2 + 1)(52+1) D (52−1)(5\sqrt 2 - 1)(52−1) Step-by-Step Verified Solution Q=ΔU=ΔU1+ΔU2Q = \Delta U = \Delta {U_1} + \Delta {U_2}Q=ΔU=ΔU1+ΔU2 ΔU2=1×2R×(2T0−T0)+2RT0(2−1)\Delta {U_2} = 1 \times 2R \times (\sqrt 2 {T_0} - {T_0}) + 2R{T_0}(\sqrt 2 - 1)ΔU2=1×2R×(2T0−T0)+2RT0(2−1) For Δ\DeltaΔ U2 : Using mole conservation of left side. p0V0RT0=22p03V02RTL⇒TL=32T0{{{p_0}{V_0}} \over {R{T_0}}} = {{2\sqrt 2 {p_0}{{3{V_0}} \over 2}} \over {R{T_L}}} \Rightarrow {T_L} = 3\sqrt 2 {T_0}RT0p0V0=RTL22p023V0⇒TL=32T0 ΔU1=1×2R×(32T0−T0)=2RT0(32−1)\Delta {U_1} = 1 \times 2R \times (3\sqrt 2 {T_0} - {T_0}) = 2R{T_0}(3\sqrt 2 - 1)ΔU1=1×2R×(32T0−T0)=2RT0(32−1) Q=2RT0(2−1)+2RT0(32−1)=2RT0(42−2)Q = 2R{T_0}(\sqrt 2 - 1) + 2R{T_0}(3\sqrt 2 - 1) = 2R{T_0}(4\sqrt 2 - 2)Q=2RT0(2−1)+2RT0(32−1)=2RT0(42−2) ΔQRT0=2(42−1){{\Delta Q} \over {R{T_0}}} = 2(4\sqrt 2 - 1)RT0ΔQ=2(42−1) Related Learning Resources & Exam Preparation JEE Advanced PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring