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(C) According to the Stefan-Boltzmann law,the power radiated by a sphere is given by $P = A \varepsilon \sigma T^4$,where $A = 4\pi r^2$ is the surfac

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Equal in both cases

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(C) According to the Stefan-Boltzmann law,the power radiated by a sphere is given by $P = A \varepsilon \sigma T^4$,where $A = 4\pi r^2$ is the surface area,$\varepsilon$ is the emissivity,and $\sigma$ is the Stefan-Boltzmann constant.Since both spheres are made of the same material,$\varepsilon$ is the same for both.Thus,$P \propto r^2 T^4$.For the first sphere $(P_1)$: $P_1 \propto (1)^2 	imes (4000)^4 = 1 	imes (4000)^4$.For the second sphere $(P_2)$: $P_2 \propto (4)^2 	imes (2000)^4 = 16 	imes (2000)^4$.Calculating the ratio: $\frac{P_1}{P_2} = \frac{1^2}{4^2} 	imes \left( \frac{4000}{2000} ight)^4 = \frac{1}{16} 	imes (2)^4 = \frac{1}{16} 	imes 16 = 1$.Therefore,$P_1 = P_2$,which means the energy radiated per second is equal in both cases.

Two spheres of the same material have radii $1 \ m$ and $4 \ m$ and temperatures $4000 \ K$ and $2000 \ K$ respectively. The energy radiated per second by the first sphere is

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