JEE Main 2022 · Physics · Calorimetry and Heat Transfer · easy

Two uniform rods of the same cross-section are joined end to end. Rod 1 has length 20 cm and thermal conductivity 2K; rod 2 has length 30 cm and…

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Two uniform rods of the same cross-section are joined end to end. Rod 1 has length

20\ \mathrm{cm}

and thermal conductivity

2K

; rod 2 has length

30\ \mathrm{cm}

and conductivity

K

. The free end of rod 1 is held at

80^\circ\mathrm{C}

and the free end of rod 2 at

20^\circ\mathrm{C}

. In steady state with no loss from the sides, the temperature of the junction is

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Answer: (C) $65^\circ\mathrm{C}$

In steady state the heat current is the same through both rods:

\dfrac{(2K)A(80-T)}{20}=\dfrac{KA(T-20)}{30}

. Simplifying,

\dfrac{80-T}{10}=\dfrac{T-20}{30}

, so

3(80-T)=T-20

, i.e.

240-3T=T-20

and

T=65^\circ\mathrm{C}

. Equivalent method: thermal resistances are in the ratio

\frac{20}{2K}:\frac{30}{K}=1:3

, so the larger temperature drop,

45\ \mathrm{K}

of the

60\ \mathrm{K}

total, is across rod 2.

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