JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{\theta _1}{10} + \frac{9\theta _2}{10}\)
Step-by-step Solution
Detailed explanation
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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