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Two bodies A and B ahave 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelenth corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K :
  • a)
    the temperature of B is 1934 K
  • b)
  • c)
    the temperature of B is 11604 K
  • d)
    the temperature of B is 2901 K
Correct answer is option 'A,B'. Can you explain this answer?
Verified Answer
Two bodies A and B ahave thermal emissivities of 0.01 and 0.81 respect...
 Energy emitted per second by body
Energy emitted per second by body
Given that power radiated are equal  
According to Wein's displacement law
Since temperature of A is more therefore
Also according to Wein's displacement law
On solving (i) and (ii), we get
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Two bodies A and B ahave 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelenth corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K :a)the temperature of B is 1934 Kb)c)the temperature of B is 11604 Kd)the temperature of B is 2901 KCorrect answer is option 'A,B'. Can you explain this answer?
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Two bodies A and B ahave 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelenth corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K :a)the temperature of B is 1934 Kb)c)the temperature of B is 11604 Kd)the temperature of B is 2901 KCorrect answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two bodies A and B ahave 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelenth corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K :a)the temperature of B is 1934 Kb)c)the temperature of B is 11604 Kd)the temperature of B is 2901 KCorrect answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two bodies A and B ahave 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelenth corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K :a)the temperature of B is 1934 Kb)c)the temperature of B is 11604 Kd)the temperature of B is 2901 KCorrect answer is option 'A,B'. Can you explain this answer?.
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