Two bodies A and B at temperatures T_1 K and | vedclass

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(B) According to the Stefan-Boltzmann law,the heat radiated per unit area per unit time is given by $E = e \sigma T^4$,where $e$ is the emissivity,$\s

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$3^{1/4}: 1$

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(B) According to the Stefan-Boltzmann law,the heat radiated per unit area per unit time is given by $E = e \sigma T^4$,where $e$ is the emissivity,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.Given that the heat radiated per unit area per unit time is the same for both bodies,we have $e_1 \sigma T_1^4 = e_2 \sigma T_2^4$.Since $\sigma$ is a constant,this simplifies to $e_1 T_1^4 = e_2 T_2^4$.Given the ratio of emissivities $e_1 : e_2 = 1 : 3$,we have $\frac{e_1}{e_2} = \frac{1}{3}$.Rearranging the equation for the ratio of temperatures: $\left(\frac{T_1}{T_2}\right)^4 = \frac{e_2}{e_1} = \frac{3}{1}$.Taking the fourth root on both sides,we get $\frac{T_1}{T_2} = \left(\frac{3}{1}\right)^{1/4} = 3^{1/4} : 1$.

Two bodies $A$ and $B$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. Their emissivities are in the ratio $1: 3$. If they radiate the same amount of heat per unit area per unit time,then the ratio of their temperatures $(T_1: T_2)$ is

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