A) \[\sqrt{2}K\]
B) \[3K\]
C) \[\frac{4}{3}K\]
D) \[\frac{2}{3}K\]
Correct Answer: C
Solution :
Equivalent thermal conductivity of the compound slab, \[{{K}_{eq}}=\frac{{{l}_{1}}+{{l}_{2}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2K}}\] \[=\frac{2l}{\frac{3l}{2K}}=\frac{4}{3}K\]
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