QAC+QDC=QCB
2ℓKA(200−t)+ℓKA(125−t)=2ℓ(t−100)KA
400−2t+125−t=2t−200
725=5t
t=145ºC
QCD=ℓKA(145−125)=KAℓ20=2W
A rod CD of thermal resistance 10.0 KW^-1 is joined at the middle of an identical rod AB as shown in figure. The ends A,B and D are maintained at…
JEE Main 2021 — Physics Thermodynamics
2021integermedium
A rod CD of thermal resistance 10.0KW−1 is joined at the middle of an identical rod AB as shown in figure. The ends A,B and D are maintained at 200∘C,100∘C and 125∘C respectively. The heat current in CD is PW. The value of P is
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Held on 27 Aug 2021 · Verified 6 Jul 2026.
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