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(A) The energy radiated per second by a sphere is given by the Stefan-Boltzmann law: $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area of th

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$1 : 1$

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