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(A) According to the Stefan-Boltzmann law,the energy radiated per second (power) $P$ by a black body is given by $P = \sigma A T^4$,where $\sigma$ is

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$1:1$

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