A) \[\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{\theta }_{1}}+{{\theta }_{2}}}\]
B) \[\frac{{{K}_{1}}{{\theta }_{1}}{{d}_{1}}+{{K}_{2}}{{\theta }_{2}}{{d}_{2}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}\]
C) \[\frac{{{K}_{1}}{{\theta }_{1}}{{d}_{2}}+{{K}_{2}}{{\theta }_{2}}{{d}_{1}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}\]
D) \[\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]
Correct Answer: C
Solution :
For first slab e Heat current. \[{{H}_{1}}=\frac{{{K}_{1}}({{\theta }_{1}}-\theta )A}{{{d}_{1}}}\] For second slab, Heat current, \[{{H}_{2}}=\frac{{{K}_{2}}(\theta -{{\theta }_{2}})A}{{{d}_{2}}}\] As slabs are in series\[{{H}_{1}}={{H}_{2}}\] \[\therefore \] \[\frac{{{K}_{1}}({{\theta }_{1}}-\theta )A}{{{d}_{1}}}=\frac{{{K}_{2}}(\theta -{{\theta }_{2}})A}{{{d}_{2}}}\] \[\Rightarrow \] \[\theta =\frac{{{K}_{1}}{{\theta }_{1}}{{d}_{2}}+{{K}_{2}}{{\theta }_{2}}{{d}_{1}}}{{{K}_{2}}{{d}_{1}}+{{K}_{2}}{{d}_{2}}}\]You need to login to perform this action.
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