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Standard XII

Physics

Stefan-Boltzman's Law

The ratio of ...

Question

The ratio of energies of emitted radiation by a black body at $600 K$ and $900 K$ , when the surrounding temperature is $300 K$ , is:

\frac{5}{16}

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\frac{7}{16}

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\frac{3}{16}

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\frac{9}{16}

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Solution

The correct option is B $\frac{3}{16}$

$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4} - T_{s}^{4}}{T_{2}^{4} - T_{s}^{4}}$

$= \frac{(600)^{4} - (300)^{4}}{(900)^{4} - (300)^{4}} = \frac{3}{16}$

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The ratio of energies of emitted radiation by a black body at 600 K and 900 K, when the surrounding temperature is 300 K, is:

The correct option is B 316 E1E2=T41−T4sT42−T4s =(600)4−(300)4(900)4−(300)4=316

The ratio of energies of emitted radiation by a black body at $600 K$ and $900 K$ , when the surrounding temperature is $300 K$ , is:

The correct option is B $\frac{3}{16}$

$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4} - T_{s}^{4}}{T_{2}^{4} - T_{s}^{4}}$

$= \frac{(600)^{4} - (300)^{4}}{(900)^{4} - (300)^{4}} = \frac{3}{16}$