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(D) According to the Stefan-Boltzmann law,the power $P$ radiated by a sphere of radius $r$ at temperature $T$ is given by $P = A \sigma T^4$,where $A

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$\frac{1}{81}$

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(D) According to the Stefan-Boltzmann law,the power $P$ radiated by a sphere of radius $r$ at temperature $T$ is given by $P = A \sigma T^4$,where $A = 4 \pi r^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.Since the spheres are made of the same material,the emissivity is the same.Thus,$P \propto r^2 T^4$.The ratio of the emitted energy per second is $\frac{P_1}{P_2} = \left(\frac{r_1}{r_2}ight)^2 	imes \left(\frac{T_1}{T_2}ight)^4$.Given $\frac{r_1}{r_2} = \frac{1}{4}$ and $\frac{T_1}{T_2} = \frac{2}{3}$.Substituting these values: $\frac{P_1}{P_2} = \left(\frac{1}{4}ight)^2 	imes \left(\frac{2}{3}ight)^4$.$\frac{P_1}{P_2} = \frac{1}{16} 	imes \frac{16}{81} = \frac{1}{81}$.

The ratio of the radii of two spheres made of the same material is $1:4$ and the ratio of their temperatures is $2:3$. The ratio of the energy emitted by the spheres per second will be:

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