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The rate of emission of radiation of a black body at temperature $27 ° C$ is $E_{1}$ . If its temperature is increased to $327 ° C$ the rate of emission of radiation is $E_{2}$ . The relation between $E_{1} a n d E_{2}$ is

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$E_{2} = 4 E_{1}$

$E_{2} = 16 E_{1}$

$E_{2} = 24 E_{1}$

$E_{2} = 8 E_{1}$

answer is B.

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Detailed Solution

$∵ E \propto T^{4}$

$∴ \frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4}$

$\Rightarrow \frac{E_{1}}{E_{2}} = {(\frac{37 + 273}{347 + 273})}^{4} = {(\frac{310}{620})}^{4} = {(\frac{1}{2})}^{4} = \frac{1}{16}$

$∴ E_{2} = 16 E_{1}$

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