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 \[\theta \] 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