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(B) According to the Stefan-Boltzmann Law,the power radiated by a black body is given by $P = \sigma A T^4$,where $A = 4 \pi r^2$ is the surface area

Vedclass · Accepted Answer

$E$

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(B) According to the Stefan-Boltzmann Law,the power radiated by a black body is given by $P = \sigma A T^4$,where $A = 4 \pi r^2$ is the surface area of the sphere.Thus,$P \propto r^2 T^4$.Let $r_1 = 0.2 \ m$,$T_1 = 800 \ K$,and $P_1 = E$.Let $r_2 = 0.8 \ m$,$T_2 = 400 \ K$,and $P_2$ be the energy radiated by the bigger body.Taking the ratio: $\frac{P_2}{P_1} = \left( \frac{r_2}{r_1} ight)^2 \left( \frac{T_2}{T_1} ight)^4$.Substituting the values: $\frac{P_2}{E} = \left( \frac{0.8}{0.2} ight)^2 \left( \frac{400}{800} ight)^4$.$\frac{P_2}{E} = (4)^2 	imes (\frac{1}{2})^4 = 16 	imes \frac{1}{16} = 1$.Therefore,$P_2 = E$.

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