A) \[55{}^\circ C~\]
B) \[60{}^\circ C\]
C) \[75{}^\circ C\]
D) \[50{}^\circ C\]
Correct Answer: A
Solution :
Let the temperature of the junction be \[\theta .\] Heat flow through the rod is given by \[Q=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})t}{d}\] Here, \[{{K}_{1}}(100-\theta )={{K}_{2}}(\theta -25)\] \[\Rightarrow \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{\theta -25}{100-\theta }\] But \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{2}{3}\] (given) \[\therefore \] \[\frac{2}{3}=\frac{\theta -25}{100-\theta }\] \[\Rightarrow \] \[3\theta -27=200-2\theta \] \[\Rightarrow \] \[5\theta -275\] \[\theta -{{55}^{o}}C\]
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