question_answer

Two slabs are of the thicknesses \[{{d}_{1}}\]and\[{{d}_{2}}\]. Their thermal conductivities are \[{{K}_{1}}\] and \[{{K}_{2}}\]respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures \[{{\theta }_{1}}\]and\[{{\theta }_{2}}.\] Assume \[{{\theta }_{1}}>{{\theta }_{2}}.\].The temperature 6 of their common junction is

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\[{{H}_{1}}={{H}_{2}}\] \[\therefore \] \[\frac{{{K}_{1}}({{\theta }_{1}}-\theta )A}{{{d}_{1}}}=\frac{{{K}_{2}}(\theta -{{\theta }_{2}})A}{{{d}_{2}}}\] \[\Rightarrow \] \[\theta =\frac{{{K}_{1}}{{\theta }_{1}}{{d}_{2}}+{{K}_{2}}{{\theta }_{2}}{{d}_{1}}}{{{K}_{2}}{{d}_{1}}+{{K}_{2}}{{d}_{2}}}\]