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JEE Advanced · Physics · 12. Thermal Properties

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 σ=5.7×10-8 Wm-2 K-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

  1. A 330 K
  2. B 660 K
  3. C 990 K
  4. D 1550 K
Verified Solution

Answer & Solution

Correct Answer

(A) 330 K

Step-by-step Solution

Detailed explanation

Rate of radiation energy lost by the sphere = Rate of radiation energy incident on it
σ×4πr2T4-3004=912×πr2
T=11×102330 K
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