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(D) 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 = \sigma A T^4

Vedclass · Accepted Answer

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

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

Two spherical black bodies have radii $r_1$ and $r_2$. Their surface temperatures are $T_1$ and $T_2$. If they radiate the same power,then $\frac{r_2}{r_1}$ is:

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