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(D) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $E = \sigma A T^4$.Since the surface area $A$ and the Stefan-

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$1/16$

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(D) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $E = \sigma A T^4$.Since the surface area $A$ and the Stefan-Boltzmann constant $\sigma$ are constant,the power radiated is directly proportional to the fourth power of the absolute temperature: $E \propto T^4$.Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = T/2$.The ratio of the final power $E_2$ to the initial power $E_1$ is given by:$E_2 / E_1 = (T_2 / T_1)^4 = ( (T/2) / T )^4 = (1/2)^4 = 1/16$.Therefore,the amount of radiation emitted reduces to $1/16$ of its initial value.

The temperature of a black body reduces to half of its initial value. By what fraction will the amount of radiation emitted reduce?

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