Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer…
Physics · JEE Advanced · NTA Exams — Thermodynamics
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power of the same rate. The wavelength lB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 mm. If the temperature of A is 5802 K
- the temperature of B is 1934 K
- lB = 1.5 mm
- the temperature of B is 11604 K
- the temperature of B is 2901 K.
Answer
(A) the temperature of B is 1934 K
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