question_answer

A, B and C are the three identical conductors but made from different materials. They are kept in contact as shown. Their thermal conductivities are K, 2K and \[\frac{K}{2}\]. The free end of A is at \[{{100}^{o}}C\] and the free end of C is at \[{{0}^{o}}C\]. During steady state, the temperature of the junction of A and B is nearly.....\[^{o}C\].

A)  \[71\]    

B)  \[29\]    

C)  \[63\]    

D)  \[37\]

Correct Answer: A

Solution :

: Here,  \[{{K}_{A}}=K,{{K}_{B}}=2K,\,{{K}_{C}}=\frac{K}{2}\] Let L be the length and A be area of cross-section of each conductor respectively. Let \[{{T}_{1}}\] and \[{{T}_{2}}\] be temperature of A-B and B-C junctions respectively. At the steady state, \[{{H}_{A}}={{H}_{B}}={{H}_{C}}\] \[\frac{{{K}_{A}}A(100-{{T}_{1}})}{L}=\frac{{{K}_{B}}A({{T}_{1}}-{{T}_{2}})}{L}=\frac{{{K}_{C}}A({{T}_{2}}-0)}{L}\]\[\frac{KA(100-{{T}_{1}})}{L}=\frac{2KA({{T}_{1}}-{{T}_{2}})}{L}=\frac{\frac{K}{2}A({{T}_{2}}-0)}{L}\] \[\therefore \] \[(100-{{T}_{1}})=2({{T}_{1}}-{{T}_{2}})\] or \[100-{{T}_{1}}=2{{T}_{1}}-2{{T}_{2}}\] \[2{{T}_{2}}=3{{T}_{1}}-100\] \[{{T}_{2}}=\frac{3}{2}{{T}_{1}}-50\]          ??..(i) and \[2({{T}_{1}}-{{T}_{2}})=\frac{1}{2}{{T}_{2}}\] \[2\left( {{T}_{1}}-\left( \frac{3}{2}{{T}_{1}}-50 \right) \right)=\frac{1}{2}\left( \frac{3}{2}{{T}_{1}}-50 \right)\](Using (i)) \[2\left( {{T}_{1}}-\left( \frac{3}{2}{{T}_{1}}-50 \right) \right)=\frac{1}{2}\left( \frac{3}{2}{{T}_{1}}-50 \right)\] \[-{{T}_{1}}+100=\frac{3{{T}_{1}}-100}{4}\] \[-4{{T}_{1}}+400=3{{T}_{1}}-100\] \[7{{T}_{1}}=500\] \[{{T}_{1}}=\frac{{{500}^{o}}C}{7}={{71}^{o}}C\]