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

821-1466

  • [JEE MAIN 2019]
  • 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

    $\frac{\theta _1}{6} + \frac{5\theta _2}{6}$

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