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Two walls of thicknesses d1 and d2 and thermal conductivities k1 and k­2 are in contact. In the steady state, if the temperatures at the outer surfaces are \[{{T}_{1}}\] and \[{{T}_{2}}\], the temperature at the common wall is [MP PMT 1990; CBSE PMT 1999]

A)            \[\frac{{{k}_{1}}{{T}_{1}}{{d}_{2}}+{{k}_{2}}{{T}_{2}}{{d}_{1}}}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}\]        

B)            \[\frac{{{k}_{1}}{{T}_{1}}+{{k}_{2}}{{d}_{2}}}{{{d}_{1}}+{{d}_{2}}}\]

C)            \[\left( \frac{{{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}}{{{T}_{1}}+{{T}_{2}}} \right){{T}_{1}}{{T}_{2}}\]  

D)            \[\frac{{{k}_{1}}{{d}_{1}}{{T}_{1}}+{{k}_{2}}{{d}_{2}}{{T}_{2}}}{{{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}}\]

Correct Answer: A

Solution :

                   In series both walls have same rate of heat flow. Therefore \[\frac{dQ}{dt}=\frac{{{K}_{1}}A({{T}_{1}}-\theta )}{{{d}_{1}}}=\frac{{{K}_{2}}A(\theta -{{T}_{2}})}{{{d}_{2}}}\]                    \[\Rightarrow {{K}_{1}}{{d}_{2}}({{T}_{1}}-\theta )={{K}_{2}}{{d}_{1}}(\theta -{{T}_{2}})\]            \[\Rightarrow \theta =\frac{{{K}_{1}}{{d}_{2}}{{T}_{1}}+{{K}_{2}}{{d}_{1}}{{T}_{2}}}{{{K}_{1}}{{d}_{2}}+{{K}_{2}}{{d}_{1}}}\]