question_answer

A wall has two layers A and B made of different materials. The thickness of both the layers is the same. The thermal conductivity of A and B are KA and KB such that KA = 3KB. The temperature across the wall is 20°C. In thermal equilibrium                                                                       [CPMT 1998]

A)            The temperature difference across \[A=15{}^\circ C\]

B)            The temperature difference across \[A=5{}^\circ C\]

C)            The temperature difference across A is 10°C

D)            The rate of transfer of heat through A is more than that through B.

Correct Answer: B

Solution :

                   In series rate of flow of heat is same Þ \[\frac{{{K}_{A}}A({{\theta }_{1}}-\theta )}{l}=\frac{{{K}_{B}}A(\theta -{{\theta }_{2}})}{l}\]                    Þ \[3{{K}_{B}}({{\theta }_{1}}-\theta )={{K}_{B}}(\theta -{{\theta }_{2}})\]                    Þ \[3({{\theta }_{1}}-\theta )=(\theta -{{\theta }_{2}})\]                    Þ \[3{{\theta }_{1}}-3\theta =\theta -{{\theta }_{2}}\]Þ \[4{{\theta }_{1}}-4\theta ={{\theta }_{1}}-{{\theta }_{2}}\]                    Þ \[4({{\theta }_{1}}-\theta )=({{\theta }_{1}}-{{\theta }_{2}})\]            Þ \[4({{\theta }_{1}}-\theta )=20\]Þ \[({{\theta }_{1}}-\theta )=5{}^\circ C\]