Two spherical black bodies of radii R_1 and R | vedclass

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(C) According to the Stefan-Boltzmann law, the power radiated by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann c

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) According to the Stefan-Boltzmann law, the power radiated by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature.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 $R_1^2 T_1^4 = R_2^2 T_2^4$.Taking the square root of both sides, we get $R_1 T_1^2 = R_2 T_2^2$.Rearranging to find the ratio $R_1/R_2$, we get $\frac{R_1}{R_2} = (\frac{T_2}{T_1})^2$.

Two spherical black bodies of radii $R_1$ and $R_2$ having surface temperatures $T_1$ and $T_2$ respectively, radiate the same power. The ratio of $R_1$ to $R_2$ will be

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