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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2014 · Paper 2 · Physics · Thermal Properties of Matter

Parallel rays of light of intensity I = 912W m ^ - 2 are incident...

Parallel rays of light of intensity I=912Wm−2I = 912W{m}^{- 2} are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ=5.7×10−8Wm−2K−4\sigma = 5.7 \times {10}^{-8}W{m}^{- 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

Options

  1. A330 K
  2. B660 K
  3. C990 K
  4. D1550 K

Correct answer

A. 330 K

Step-by-step solution

Rate of radiation energy lost by the sphere = Rate of radiation energy incident on it
⇒σ×4πr2[T4−(300)4]=912×πr2⇒ \sigma \times 4\pi{r}^{2}\left[{T}^{4} - {\left(300\right)}^{4}\right] = 912 \times \pi{r}^{2}
⇒T=11×102≈330K⇒ T = \sqrt{11} \times {10}^{2} \approx 330K
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