GATE 2026 ME – Question 54
Two rectangular surfaces both having 1 m$^2$ area are placed perpendicular to each other with a common edge. One surface is hot, having a temperature of 1000 K and emissivity of 0.4, while the other is insulated and in radiant balance with a large surrounding room at 300 K. If the fraction of radiation leaving the hot surface which reaches the cold surface is 0.2, then the equivalent overall resistance for the radiation heat loss from the hot surface is ________ m$^{-2}$ (*rounded off to 2 decimal places*).
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Correct answer: 2.53 to 2.55
Explanation
Draw the radiation network. The surface resistance of the hot surface is $\frac{1 - \varepsilon}{\varepsilon A} = \frac{0.6}{0.4} = 1.5$ m$^{-2}$. From the hot surface to the room directly (view factor 0.8) the resistance is $\frac{1}{0.8} = 1.25$. The other path goes through the insulated surface: $\frac{1}{0.2} = 5$ to reach it, and then $\frac{1}{0.8} = 1.25$ from it to the room, which is a total of 6.25. These two paths are in parallel: $\left(\frac{1}{1.25} + \frac{1}{6.25}\right)^{-1} = 1.04$. The total resistance is $1.5 + 1.04 = 2.54$ m$^{-2}$.