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Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant s = 5.7 × 10–8 Wm–2K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to
  • a)
    330K
  • b)
    660K
  • c)
    990K
  • d)
    1550K
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Parallel rays of light of intensity I = 912 Wm–2 are incident on...
In steady state Energy lost = Energy gained
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Parallel rays of light of intensity I = 912 Wm–2 are incident on...
It seems like there is a missing piece of information in your question. The intensity of light is typically measured in watts per square meter (W/m^2), but you only provided the value of intensity as 912 Wm.

For parallel rays of light, the intensity would typically be given as an intensity per unit area. So, if you provide the area over which the intensity is measured, we can calculate the actual value of the intensity.
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Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant s = 5.7 × 10–8 Wm–2K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toa)330Kb)660Kc)990Kd)1550KCorrect answer is option 'A'. Can you explain this answer?
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Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant s = 5.7 × 10–8 Wm–2K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toa)330Kb)660Kc)990Kd)1550KCorrect answer is option 'A'. 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 Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant s = 5.7 × 10–8 Wm–2K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toa)330Kb)660Kc)990Kd)1550KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant s = 5.7 × 10–8 Wm–2K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close toa)330Kb)660Kc)990Kd)1550KCorrect answer is option 'A'. Can you explain this answer?.
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