A) 80°C
B) 20°C
C) 60°C
D) 40°C
Correct Answer: A
Solution :
Temperature of interface q = \[\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\] \[(\because \,\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{1}{4}\Rightarrow \]If K1 = K then K2 = 4K) Þ \[\theta =\frac{K\times 0+4K\times 100}{5\,K}\]= 80°C
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