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 $\left( {\theta _2} > {\theta _1} \right)$ . The temperature at the interface is
JEE MAIN 2019, Medium
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At steady state:
${\left( {\frac{{\Delta q}}{{\Delta t}}} \right)_1} = {\left( {\frac{{\Delta q}}{{\Delta t}}} \right)_2}$
$\frac{{3mkA\left( {{\theta _2} - \theta } \right)}}{d} = \frac{{kA\left( {\theta - {\theta _1}} \right)}}{{3d}}$
$ \Rightarrow \,\,\theta = \frac{{{\theta _1} + 9{\theta _2}}}{{10}}$
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