A) \[\frac{{{d}_{1}}+{{d}_{2}}}{{{K}_{1}}+{{K}_{2}}}\]
B) \[\frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}\]
C) \[\frac{{{K}_{1}}{{d}_{1}}+{{K}_{2}}{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}}\]
D) \[\frac{{{d}_{1}}+{{d}_{2}}}{\left( \frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}} \right)}\]
Correct Answer: D
Solution :
: Let\[\theta =\]temperature of common interface of the two slabs. \[\therefore \] \[Q=\frac{{{K}_{1}}A({{\theta }_{1}}-\theta )t}{{{d}_{1}}}\]for\[{{1}^{st}}\]slab Also \[Q=\frac{{{K}_{2}}A(\theta -{{\theta }_{2}})t}{{{d}_{2}}}\]second slab \[\therefore \] \[{{\theta }_{1}}-\theta =\frac{Q{{d}_{1}}}{{{K}_{1}}At}\] \[\theta -{{\theta }_{2}}=\frac{Q{{d}_{1}}}{{{K}_{2}}At}\] Or \[({{\theta }_{1}}-{{\theta }_{2}})=\frac{Q{{d}_{1}}}{{{K}_{1}}At}+\frac{Q{{d}_{2}}}{{{K}_{2}}At}\] Or \[({{\theta }_{1}}-{{\theta }_{2}})=\frac{Q}{At}\left( \frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}} \right)\] ?..(i) For composite slab of thickness \[({{d}_{1}}+{{d}_{2}}),\] \[({{\theta }_{1}}-{{\theta }_{2}})=\frac{Q}{Qt}\left( \frac{{{d}_{1}}+{{d}_{2}}}{K} \right)\] ?.(ii) \[\therefore \] \[\frac{({{d}_{1}}+{{d}_{2}})}{K}=\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}}\] Or \[K=\frac{({{d}_{1}}+{{d}_{2}})}{\left( \frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}} \right)}\]
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