Two spherical black bodies of radii R_1 and R | vedclass

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(B) According to the Stefan-Boltzmann law,the power $P$ radiated by a spherical black body of radius $R$ and temperature $T$ is given by $P = \sigma A

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$(T_2/T_1)^2$

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(B) According to the Stefan-Boltzmann law,the power $P$ radiated by a spherical black body of radius $R$ and temperature $T$ is given by $P = \sigma A T^4$,where $A = 4 \pi R^2$ is the surface area and $\sigma$ is the Stefan-Boltzmann constant.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 to find the ratio $R_1/R_2$:$(R_1/R_2)^2 = (T_2/T_1)^4$.Taking the square root on both sides:$R_1/R_2 = (T_2/T_1)^2$.

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. The ratio of $R_1$ to $R_2$ will be

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