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JEE Advanced 2025 · Paper 1, Physics question 10

Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures TPT_P and TQT_Q, respectively, with TQ<TPT_Q < T_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\dfrac{W_0}{W_S} is _____.

Fig. 1: two parallel vertical plates, P (temperature T_P) on the left and Q (T_Q) on the right, with an arrow W₀ from P to Q. Fig. 2: the same plates P (T_P) and Q (T_Q) with two more identical parallel plates placed between them (four plates in all); an arrow W_S from P to the next plate.
Answer and worked solution

Official answer: 3

Between two black plates, W=σ(T14−T24)W=\sigma(T_1^4-T_2^4).

With two plates inserted there are three gaps in series. In steady state each gap carries the same flux WSW_S, so the temperature differences add:
TP4−TQ4=3WSσ.T_P^4-T_Q^4=\frac{3W_S}{\sigma}.

Since W0=σ(TP4−TQ4)W_0=\sigma(T_P^4-T_Q^4), we get W0=3WSW_0=3W_S and
W0WS=3.\frac{W_0}{W_S}=3.

Topics:Radiation and Stefan's law
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