Two spherical black bodies have radii R_1 and | vedclass

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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

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

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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 black 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 $\frac{R_1}{R_2}$,we have $\frac{R_1^2}{R_2^2} = \frac{T_2^4}{T_1^4}$.Taking the square root of both sides,we get $\frac{R_1}{R_2} = \sqrt{\frac{T_2^4}{T_1^4}} = \left(\frac{T_2}{T_1}\right)^2$.

Two spherical black bodies have radii $R_1$ and $R_2$. Their surface temperatures are $T_1 \ K$ and $T_2 \ K$ respectively. If they radiate the same power,the ratio $\frac{R_1}{R_2}$ is

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