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 :

  • Option 1)

    \frac{\theta_{1}}{10}+\frac{9\theta _{2}}{10}

  • Option 2)

    \frac{\theta_{2}+\theta_{1}}{2}

  • Option 3)

    \frac{\theta_{1}}{6}+\frac{5\theta _{2}}{6}

  • Option 4)

    \frac{\theta_{1}}{3}+\frac{2\theta _{2}}{3}

Answers (1)

\Rightarrow Conservation \:\:of \:\:heat

\frac{T-\Theta_{2}}{d/3kA}+\frac{T-\Theta_{1}}{3d/kA}=0

\Rightarrow37-3\theta_{2}-\frac{T}{3}-\frac{\theta_{1}}{3}=0

\Rightarrow97-9\theta_{2}+T-\theta_{1}=0

\Rightarrow T=\frac{9}{10}\theta _{2}+ \frac{\theta_{1}}{10}


Option 1)

\frac{\theta_{1}}{10}+\frac{9\theta _{2}}{10}

Option 2)

\frac{\theta_{2}+\theta_{1}}{2}

Option 3)

\frac{\theta_{1}}{6}+\frac{5\theta _{2}}{6}

Option 4)

\frac{\theta_{1}}{3}+\frac{2\theta _{2}}{3}

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