Two spherical black bodies of radii r_1 and r | vedclass

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Question

(A) According to the Stefan-Boltzmann law,the power $P$ radiated by a black body of surface area $A$ and temperature $T$ is given by $P = A \sigma T^4

Vedclass · Accepted Answer

$(\frac{T_2}{T_1})^2$

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(A) According to the Stefan-Boltzmann law,the power $P$ radiated by a black body of surface area $A$ and temperature $T$ is given by $P = A \sigma T^4$.Since the bodies are spherical,the surface area $A = 4 \pi r^2$.Thus,$P = 4 \pi r^2 \sigma T^4$.Given that both bodies radiate the same power,we have $P_1 = P_2$.$4 \pi r_1^2 \sigma T_1^4 = 4 \pi r_2^2 \sigma T_2^4$.Simplifying this,we get $r_1^2 T_1^4 = r_2^2 T_2^4$.Rearranging for the ratio $\frac{r_1}{r_2}$,we get $(\frac{r_1}{r_2})^2 = (\frac{T_2}{T_1})^4$.Taking the square root on both sides,we get $\frac{r_1}{r_2} = (\frac{T_2}{T_1})^2$.

Two spherical black bodies of radii $r_1$ and $r_2$ and with surface temperatures $T_1$ and $T_2$ respectively radiate the same power. Then the ratio of $r_1$ and $r_2$ will be:

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