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JEE Main 2021 · Physics · Calorimetry and Heat Transfer · easy

Two rods of the same cross-section are joined end to end. The first has length L and conductivity K; the second has length 2L and conductivity 3K.…

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Two rods of the same cross-section are joined end to end. The first has length LL and conductivity KK; the second has length 2L2L and conductivity 3K3K. The free end of the first is held at 100∘C100^\circ\mathrm{C} and the free end of the second at 0∘C0^\circ\mathrm{C}. In steady state, with no loss from the sides, the temperature of the junction is
  1. A. 30∘C30^\circ\mathrm{C}
  2. B. 40∘C40^\circ\mathrm{C}
  3. C. 50∘C50^\circ\mathrm{C}
  4. D. 60∘C60^\circ\mathrm{C}
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Answer: (B) 40∘C40^\circ\mathrm{C}

Thermal resistances are R1=LKAR_1=\dfrac{L}{KA} and R2=2L3KAR_2=\dfrac{2L}{3KA}, so R1:R2=3:2R_1:R_2=3:2. The same heat current flows through both, so the temperature drops divide in the ratio 3:23:2: the drop across the first rod is 35×100=60∘C\dfrac35\times100=60^\circ\mathrm{C}. The junction is at 100−60=40∘C100-60=40^\circ\mathrm{C}.
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