Physics · Heat Transfer

JEE Advanced 2025 — Paper 1 — Question 8

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}}<\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 PP to QQ (Fig. 2) in the steady state is WSW_{S}, then the ratio W0 Ws\frac{\mathrm{W}_{0}}{\mathrm{~W}_{\mathrm{s}}} is \qquad Fig. 1 Fig. 2

figure

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Initially : W0=σ(TP4−TQ4)\mathrm{W}_{0}=\sigma\left(\mathrm{T}_{\mathrm{P}}^{4}-\mathrm{T}_{\mathrm{Q}}^{4}\right)

figure

Finally : Putting heat currents equal in steady state : σ(TP4−T14)=σ(T14−T24)\sigma\left(T_{P}^{4}-T_{1}^{4}\right)=\sigma\left(T_{1}^{4}-T_{2}^{4}\right)

σ(T14−T24)=σ(T24−TQ4)\sigma\left(\mathrm{T}_{1}^{4}-\mathrm{T}_{2}^{4}\right)=\sigma\left(\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{Q}}^{4}\right)

Adding : TP4−T14=T24−TQ4\mathrm{T}_{\mathrm{P}}^{4}-\mathrm{T}_{1}^{4}=\mathrm{T}_{2}^{4}-\mathrm{T}_{\mathrm{Q}}^{4}

⇒T14+T24=TP4+TQ4\Rightarrow \mathrm{T}_{1}^{4}+\mathrm{T}_{2}^{4}=\mathrm{T}_{\mathrm{P}}^{4}+\mathrm{T}_{\mathrm{Q}}^{4} and

⇒T14−T24=TP4−T14\Rightarrow \mathrm{T}_{1}^{4}-\mathrm{T}_{2}^{4}=\mathrm{T}_{\mathrm{P}}^{4}-\mathrm{T}_{1}^{4}

Adding : T14=2 TP4+TQ43\mathrm{T}_{1}^{4}=\frac{2 \mathrm{~T}_{\mathrm{P}}^{4}+\mathrm{T}_{\mathrm{Q}}^{4}}{3}

So WS=σ(TP4−T14)\mathrm{W}_{\mathrm{S}}=\sigma\left(\mathrm{T}_{\mathrm{P}}^{4}-\mathrm{T}_{1}^{4}\right)

=σ(TP4−(2 TP4+TQ43))=σ(TP4−TQ43)=\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)

hence WSW0=3\frac{W_{S}}{W_{0}}=3

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 1
Subject
Physics
Chapter
Heat Transfer
Topic
Convection and Radiation
Two identical plates P and Q , radiating as perfect black bodies, are… | JEE Advanced 2025 PYQ with Solution · DhiX AI