question_answer

A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is \[{{36}^{o}}C\]. Then the difference of temperature at the two surfaces of A will be [IIT 1980; CPMT 1991; BHU 1997; MP PET 1996, 99; DPMT 2000]

A)            \[{{6}^{o}}C\]                      

B)            \[{{12}^{o}}C\]

C)            \[{{18}^{o}}C\]                    

D)            \[{{24}^{o}}C\]

Correct Answer: B

Solution :

                   Suppose thickness of each wall is x then \[{{\left( \frac{Q}{t} \right)}_{combination}}={{\left( \frac{Q}{t} \right)}_{A}}\]Þ \[\frac{{{K}_{S}}A({{\theta }_{1}}-{{\theta }_{2}})}{2x}=\frac{2KA({{\theta }_{1}}-\theta )}{x}\]                    \[\because \] \[{{K}_{S}}=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}K\] and \[({{\theta }_{1}}-{{\theta }_{2}})=36{}^\circ \] Þ \[\frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA({{\theta }_{1}}-\theta )}{x}\] Hence temperature difference across wall A is       \[({{\theta }_{1}}-\theta )={{12}^{o}}C\]