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(D) According to the Stefan-Boltzmann law,the rate of radiation $E$ from a black body is given by $E = A \sigma T^4$,where $A$ is the surface area,$\s

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$9 E$

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(D) According to the Stefan-Boltzmann law,the rate of radiation $E$ from a black body is given by $E = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.For a sphere,the surface area $A = 4 \pi R^2$.Thus,$E = (4 \pi R^2) \sigma T^4$,which implies $E \propto R^2 T^4$.Let the initial state be $E_1 = E$,$R_1 = R$,and $T_1 = T$.Let the final state be $E_2$,$R_2 = R/3$,and $T_2 = 3T$.Taking the ratio: $\frac{E_2}{E_1} = \left( \frac{R_2}{R_1} ight)^2 \left( \frac{T_2}{T_1} ight)^4$.Substituting the values: $\frac{E_2}{E} = \left( \frac{R/3}{R} ight)^2 \left( \frac{3T}{T} ight)^4$.$\frac{E_2}{E} = \left( \frac{1}{3} ight)^2 	imes (3)^4 = \frac{1}{9} 	imes 81 = 9$.Therefore,$E_2 = 9E$.

$A$ black sphere has radius $R$ whose rate of radiation is $E$ at temperature $T$. If the radius is made $R/3$ and the temperature $3T$,the rate of radiation will be:

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