question_answer

           Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are \[{{K}_{1}}\,\,\text{and}\,\,{{K}_{2}}\]. The temperature of the free ends of the first and second rods are maintained at \[{{\theta }_{1}}\,\,\text{and }{{\theta }_{2}}\] respectively. The temperature of the common junction is   [MP PET 2003]

A)            \[\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]               

B)            \[\frac{{{K}_{2}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}({{\theta }_{1}}+{{\theta }_{2}})\]

C)            \[\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]      

D)            \[\frac{{{K}_{2}}{{\theta }_{1}}+{{K}_{1}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]

Correct Answer: C

Solution :

                   At steady state, rate of heat flow for both blocks will be same i.e., \[\frac{{{K}_{1}}A({{\theta }_{1}}-\theta )}{{{l}_{1}}}=\frac{{{K}_{2}}A(\theta -{{\theta }_{2}})}{{{l}_{2}}}\](given \[{{l}_{1}}={{l}_{2}}\]) \[\Rightarrow {{K}_{1}}A({{\theta }_{1}}-\theta )={{K}_{2}}A(\theta -{{\theta }_{2}})\]  \[\Rightarrow \theta =\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]