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(A) According to the Stefan-Boltzmann law,the energy radiated per second $(P)$ by a sphere of radius $r$ and temperature $T$ is given by $P = \sigma A

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$1:1$

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(A) According to the Stefan-Boltzmann law,the energy radiated per second $(P)$ by a sphere of radius $r$ and temperature $T$ is given by $P = \sigma A T^4$,where $A = 4 \pi r^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.Thus,$P = \sigma (4 \pi r^2) T^4$.The ratio of the energy radiated by the two spheres is $\frac{P_1}{P_2} = \frac{\sigma (4 \pi r_1^2) T_1^4}{\sigma (4 \pi r_2^2) T_2^4} = \left( \frac{r_1}{r_2} ight)^2 \left( \frac{T_1}{T_2} ight)^4$.Given $r_1 = 1 \; m$,$r_2 = 4 \; m$,$T_1 = 4000 \; K$,and $T_2 = 2000 \; K$.Substituting these values: $\frac{P_1}{P_2} = \left( \frac{1}{4} ight)^2 \left( \frac{4000}{2000} ight)^4 = \left( \frac{1}{16} ight) 	imes (2)^4 = \frac{16}{16} = 1$.Therefore,the ratio is $1:1$.

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

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