In steady state ⇒(dtdQ)1=(dtdQ)2=(dtdQ)3L2KA(3 T−T1)=LKA( T1−T2)=L2KA( T2−T)6 T−2 T1=T1−T2=2 T2−2 T6 T−2 T1=T1−T26 T+T2=3 T1....(i) T1−T2=2 T2−2 T2 T−3 T2=−T1.....(ii) Multiply equation (2) By ' 3 ' 6 T−9 T2=−3 T1....(iii) Equation (1) - (3) 10 T2=6 T1 T2 T1=35
NEET UG 2025 — Physics Thermodynamics
2025mcqmedium
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T1 and that at the right junction is T2. The ratio T1/T2 is 
Official previous-year question
Held on 30 Apr 2025 · Verified 9 Jul 2026.
Options
- A
23
- B
34
- C
35
- D
45
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