JEE Mains · Physics · STD 11 - 10.2 transmission of heat
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
- 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 \(\frac{\theta _1}{6} + \frac{5\theta _2}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{\theta _1}{10} + \frac{9\theta _2}{10}\)
Step-by-step Solution
Detailed explanation
At steady state: \({\left( {\frac{{\Delta q}}{{\Delta t}}} \right)_1} = {\left( {\frac{{\Delta q}}{{\Delta t}}} \right)_2}\) \(\frac{{3mkA\left( {{\theta _2} - \theta } \right)}}{d} = \frac{{kA\left( {\theta - {\theta _1}} \right)}}{{3d}}\)…
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