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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2025 · Paper 1 · Physics · Thermal Properties of Matter

Two identical plates P and Q , radiating as perfect black bodies,...

Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures (TP\mathrm{T}_{\mathrm{P}}) and (TQ\mathrm{T}_{\mathrm{Q}}), respectively, with (TQ<TP\mathrm{T}_{\mathrm{Q}}\lt \mathrm{T}_{\mathrm{P}}), as shown in Fig. 1. The radiated power transferred per unit area from P to Q is W0W_0. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is WSW_S, then the ratio (W0WS\frac{W_0}{W_S}) is ________. Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures

Correct answer

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Step-by-step solution

Initially : W0=σ\mathrm{W}_0=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4\right)<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/gukJjWT8BR1VAuvNQmQ44Pfz8ATGhjTwjTkWpx8s9GI.original.fullsize.png"><br>Finally:<br>Puttingheatcurrentsequalinsteadystate:<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeeadvanced/ZhcSG5RXdEWK44ZEGtOtHHEGidzL4RmTN4wgNVgd0vw.original.fullsize.png"><br><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/gukJjWT8BR1VAuvNQmQ44Pfz8ATGhjTwjTkWpx8s9GI.original.fullsize.png"><br>Finally :<br>Putting heat currents equal in steady state : <img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/ZhcSG5RXdEWK44ZEGtOtHHEGidzL4RmTN4wgNVgd0vw.original.fullsize.png"><br>σ(TP4−T14)=σ(T14−T24)σ(T14−T24)=σ(T24−TQ4)\begin{aligned} & \sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)=\sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right) \\ & \sigma\left(\mathrm{T}_1^4-\mathrm{T}_2^4\right)=\sigma\left(\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\right)\end{aligned}Adding:Adding :\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4=\mathrm{T}_2^4-\mathrm{T}_{\mathrm{Q}}^4\Rightarrow \mathrm{T}_1^4+\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4andand\quad \Rightarrow \mathrm{T}_1^4-\mathrm{T}_2^4=\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4Adding:Adding :\mathrm{T}_1^4=\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}SoSo\mathrm{W}_{\mathrm{S}}=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_1^4\right)=σ=\sigma\left(\mathrm{T}_{\mathrm{P}}^4-\left(\frac{2 \mathrm{~T}_{\mathrm{P}}^4+\mathrm{T}_{\mathrm{Q}}^4}{3}\right)\right)=\sigma\left(\frac{\mathrm{T}_{\mathrm{P}}^4-\mathrm{T}_{\mathrm{Q}}^4}{3}\right)hencehence\frac{\mathrm{W}_{\mathrm{S}}}{\mathrm{W}_0}=3
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