Parallel rays of light of intensity l - 912 Wm-2 are incident on a spherical black body kept in surroundings of temperature 300 K- Take Stefan-Boltzmann constant - - 5-7- 10-5 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-

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Standard XII

Physics

YDSE Problems I

Parallel rays...

Question

Parallel rays of light of intensity $l = 912 W m^{- 2}$ are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant $σ = 5.7 \times 10^{- 5} W m^{- 2} K^{- 4}$ 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:

330 K

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660 K

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990 K

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1550 K

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Solution

The correct option is A 330 K
Rate of radiation energy incident on it=Rate of radiation energy lost by the sphere:
Using Stefan Boltzmann's law:
$912 π r^{2} = σ 4 π r^{2} (T^{4} - 300^{4})$
where, $σ = 5.7 \times 10^{- 5}$
$T \approx 330 K$ i.e. option A

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The correct option is A 330 KRate of radiation energy incident on it=Rate of radiation energy lost by the sphere:Using Stefan Boltzmann's law:912πr2=σ4πr2(T4−3004)where, σ=5.7×10−5T≈330K i.e. option A

The correct option is A 330 K
Rate of radiation energy incident on it=Rate of radiation energy lost by the sphere:
Using Stefan Boltzmann's law:
$912 π r^{2} = σ 4 π r^{2} (T^{4} - 300^{4})$
where, $σ = 5.7 \times 10^{- 5}$
$T \approx 330 K$ i.e. option A