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JEE Advanced 2014 · Paper 2, Physics question 6

Parallel rays of light of intensity I=912 W m−2I=912\ \mathrm{W\,m^{-2}} are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ=5.7×10−8 W m−2 K−4\sigma=5.7\times10^{-8}\ \mathrm{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

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Answer and worked solution

Official answer: A

Energy balance. The sphere intercepts the beam over its cross-section πr2\pi r^2 and exchanges radiation over its surface 4πr24\pi r^2:
Iπr2=σ4πr2(T4−T04)  ⟹  T4=T04+I4σ.I\pi r^2=\sigma4\pi r^2\left(T^4-T_0^4\right)\implies T^4=T_0^4+\frac I{4\sigma}.

Numbers.
T4=8.1×109+9124×5.7×10−8=8.1×109+4.0×109=12.1×109.T^4=8.1\times10^9+\frac{912}{4\times5.7\times10^{-8}}=8.1\times10^9+4.0\times10^9=12.1\times10^9.
T≈332T\approx332 K.

(A)

Topics:Radiation and Stefan's law
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