question_answer

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are \['{{\theta }_{2}}'\]and \['{{\theta }_{1}}'\] respectively, \[({{\theta }_{2}}>{{\theta }_{1}}).\]The temperature at the interface is:-[JEE Main 9-4-2019 Afternoon]

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

B)      \[\frac{{{\theta }_{1}}}{10}+\frac{9{{\theta }_{2}}}{10}\]

C) \[\frac{{{\theta }_{1}}}{3}+\frac{2{{\theta }_{2}}}{3}\]           

D)      \[\frac{{{\theta }_{1}}}{6}+\frac{5{{\theta }_{2}}}{6}\]

Correct Answer: B

Solution :

[b]

Let the temperature of interface be\[''\theta ''\] \[{{i}_{1}}={{i}_{2}}\] {Steady state conduction}

\[\frac{3KA({{\theta }_{2}}-\theta )}{d}=\frac{KA(\theta -{{\theta }_{1}})}{3d}\]

\[\theta =\frac{9\theta {{ & }_{2}}}{10}+\frac{\theta {{ & }_{1}}}{10}\]