Two spherical black bodies of radii r_1 and r | vedclass

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(B) The power radiated by a black body is given by Stefan-Boltzmann Law: $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area of the sphere.For

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$\frac{1}{\sqrt{2}}\left(\frac{T_2}{T_1}\right)^2$

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(B) The power radiated by a black body is given by Stefan-Boltzmann Law: $P = \sigma A T^4$,where $A = 4\pi r^2$ is the surface area of the sphere.For the first black body: $P_1 = \sigma (4\pi r_1^2) T_1^4$.For the second black body: $P_2 = \sigma (4\pi r_2^2) T_2^4$.The ratio of the powers is given as $\frac{P_1}{P_2} = \frac{1}{2}$.Substituting the expressions: $\frac{1}{2} = \frac{\sigma 4\pi r_1^2 T_1^4}{\sigma 4\pi r_2^2 T_2^4} = \frac{r_1^2 T_1^4}{r_2^2 T_2^4}$.Rearranging for the ratio of radii: $\frac{r_1^2}{r_2^2} = \frac{1}{2} \left(\frac{T_2}{T_1}\right)^4$.Taking the square root on both sides: $\frac{r_1}{r_2} = \frac{1}{\sqrt{2}} \left(\frac{T_2}{T_1}\right)^2$.

Two spherical black bodies of radii $r_1$ and $r_2$ at temperatures $T_1$ and $T_2$ respectively radiate power in the ratio $1:2$. Then $r_1:r_2$ is:

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