JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2025
- Medium
- Numerical
Three conductors of same length having thermal conductivity \(\mathrm{k}_1, \mathrm{k}_2\) and \(\mathrm{k}_3\) are connected as shown in figure.
Area of cross sections of \(1^{\text {st }}\) and \(2^{\text {nd }}\) conductor are same and for \(3^{\text {rd }}\) conductor it is double of the \(1^{\text {st }}\) conductor. The temperatures are given in the figure. In steady state condition, the value of \(\theta\) is _______ \({ }^{\circ} \mathrm{C}\).
(Given : \(\mathrm{k}_1=60 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, \mathrm{k}_2=120 \mathrm{Js}^{-1}\mathrm{~m}^{-1}\) \( \mathrm{~K}^{-1}, \mathrm{k}_3=135 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 40
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{R}_1=\frac{2 \mathrm{~L}}{\mathrm{~K}_1 \mathrm{~A}} \\ & \mathrm{R}_2=\frac{2 \mathrm{~L}}{\mathrm{~K}_2 \mathrm{~A}} \\ & \mathrm{R}_3=\frac{\mathrm{L}}{\mathrm{K}_3 \mathrm{~A}} \\ & \frac{\theta-100}{\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}}+\frac{\theta-0}{\mathrm{R}_3}=0 \\ & \theta=40\end{aligned}\)
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