Physics · Heat Transfer

JEE Main 2025 — 22 January, Morning Shift — Question 63

Three conductions of same length having thermal conductivity k1,k2\mathrm{k}_{1}, \mathrm{k}_{2} and k3\mathrm{k}_{3} are connected as shown in figure. 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 \qquad ∘C{ }^{\circ} \mathrm{C}. (Given : k1=60Js−1 m−1 K−1,k2=120Js−1 m−1 K−1,k3=\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}= 135Js−1 m−1 K−1135 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1} )

Question figure

Answer: 40

Numerical answer — enter this value.

Step-by-step solution

Let each conductor length be LL. For conductors 11 and 22 in parallel, each has area AA, so R1=Lk1AR_1 = \frac{L}{k_1 A} and R2=Lk2AR_2 = \frac{L}{k_2 A}. For conductor 33, area is 2A2A, so R3=Lk3⋅2AR_3 = \frac{L}{k_3 \cdot 2A}. In steady state, heat current through the parallel combination equals heat current through conductor 33:

100−θRparallel=θ−0R3\frac{100 - \theta}{R_{\text{parallel}}} = \frac{\theta - 0}{R_3}

where 1Rparallel=1R1+1R2=AL(k1+k2)\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{A}{L}(k_1 + k_2). Thus:

(100−θ)AL(k1+k2)=θk3⋅2AL(100 - \theta) \frac{A}{L}(k_1 + k_2) = \theta \frac{k_3 \cdot 2A}{L}

Cancel AL\frac{A}{L}:

(100−θ)(60+120)=θ⋅270(100 - \theta)(60 + 120) = \theta \cdot 270 180(100−θ)=270θ180(100 - \theta) = 270\theta 18000=450θ  ⟹  θ=40∘C18000 = 450\theta \implies \theta = 40^\circ\text{C}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Heat Transfer
Topic
Introductory Heat transfer by Conduction
Three conductions of same length having thermal conductivity k 1 , k… | JEE Main 2025 PYQ with Solution · DhiX AI