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QJEEM25Ph3306: JEE Main 2025 Physics - Class 11 JEEM25Ph3306 Solution: Three conductors of same length having thermal conductivity and …

Three conductors of same length having thermal conductivity k1,k2k_1, k_2 and k3k_3 are connected as shown in figure.

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Heat and Thermodynamics Question 18 English

Area of cross sections of 1st 1^{\text {st }} and 2nd 2^{\text {nd }} conductor are same and for 3rd 3^{\text {rd }} conductor it is double of the 1st 1^{\text {st }} conductor. The temperatures are given in the figure. In steady state condition, the value of θ\theta is _________ ∘C{ }^{\circ} \mathrm{C} . (Given : k1=60Js−1 m−1 K−1,k2=120Js−1 m−1 K−1,k3=135Js−1 m−1 K−1\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} )

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JEE Main 2025 (Online) 22nd January Morning Shift Physics - Heat and Thermodynamics Question 18 English Explanation

amp;R1=2 L K1 Aamp;R2=2 L K2 Aamp;R3=L K3 Aamp;θ−100R1R2amp;R1+R2amp;θ=40\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}{\mathrm{R}_1 \mathrm{R}_2} \\& \mathrm{R}_1+\mathrm{R}_2 \\& \theta=40\end{aligned}

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