Two slabs are of the thicknesses $d_{1}$ and $d_{2}$. Their thermal conductivities are $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures $\theta_{1}$ and $\theta_{2}$. Assume $\theta_{1}>\theta_{2}$. The temperature $\theta$ of their common junction is

• A

  $\frac{\mathrm{K}_{1} \theta_{1}+\mathrm{K}_{2} \theta_{2}}{\theta_{1}+\theta_{2}}$
• B

  $\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{1}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{2}}{\mathrm{~K}_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \mathrm{~d}_{1}}$
• Correct answer
  C

  $\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \mathrm{~d}_{1}}$
• D

  $\frac{\mathrm{K}_{1} \theta_{1}+\mathrm{K}_{2} \theta_{2}}{\mathrm{~K}_{1}+\mathrm{K}_{2}}$

Solution

For first slab

Heat current, $\mathrm{H}_{1}=\frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}}$ For second slab, Heat current, $\mathrm{H}_{2}=\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}}$ As slabs are in series $$ \begin{aligned} \mathrm{H}_{1} &=\mathrm{H}_{2} \\ & \therefore & \frac{\mathrm{K}_{1}\left(\theta_{1}-\theta\right) \mathrm{A}}{\mathrm{d}_{1}} &=\frac{\mathrm{K}_{2}\left(\theta-\theta_{2}\right) \mathrm{A}}{\mathrm{d}_{2}} \\ \Rightarrow & \theta &=\frac{\mathrm{K}_{1} \theta_{1} \mathrm{~d}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{~d}_{1}}{\mathrm{~K}_{2} \mathrm{~d}_{1}+\mathrm{K}_{1} \mathrm{~d}_{2}} \end{aligned} $$

Hence, the correct answer is option: 'C'.