A) \[40{}^\circ C\]
B) \[60{}^\circ C\]
C) \[50{}^\circ C\]
D) \[35{}^\circ C\]
Correct Answer: C
Solution :
Let the temperature of junction be\[\theta \]. In equilibrium, rate of flow of heat through rod 1 is equal to sum of rate of flow of heat through rods 2 and 3,\[ie\], \[{{\left( \frac{dQ}{dt} \right)}_{1}}={{\left( \frac{dQ}{dt} \right)}_{2}}={{\left( \frac{dQ}{dt} \right)}_{3}}\] \[\therefore \,\,\,\frac{KA(\theta -20)}{l}\,=\frac{KA(60-\theta )}{l}+\frac{KA(70-\theta )}{l}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\theta -20=130-2\theta \] or \[3\theta =150\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\theta ={{50}^{\text{o}}}C\]
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