T1−T2=120oC
We define thermal resistance
R=KAL
Given circuit can be reduced to

Equivalent resistance Req=58R
Thermal current i=58RT1−T2
TP−TQ=(58R)T1−T2(53R)
=83×120=45
JEE Main 2019 — Physics Thermodynamics
Temperature difference of 120oC is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 23L, is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to:

Held on 9 Jan 2019 · Verified 6 Jul 2026.
45oC
35oC
60oC
75oC
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Work through every JEE Main Thermodynamics PYQ, year by year.