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

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(D) According to Stefan-Boltzmann Law,the heat radiated per unit area per unit time (emissive power) is given by $E = \epsilon \sigma T^4$,where $\eps

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$0.5$

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(D) According to Stefan-Boltzmann Law,the heat radiated per unit area per unit time (emissive power) is given by $E = \epsilon \sigma T^4$,where $\epsilon$ is the emissivity,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.Given that the two bodies radiate the same amount of heat per unit area per unit time,we have $E_A = E_B$.Therefore,$\epsilon_A \sigma T_1^4 = \epsilon_B \sigma T_2^4$.Given the ratio of emissivities $\frac{\epsilon_A}{\epsilon_B} = \frac{16}{1}$.Substituting this into the equation: $16 \sigma T_1^4 = 1 \sigma T_2^4$.This simplifies to $16 T_1^4 = T_2^4$.Taking the fourth root on both sides: $2 T_1 = T_2$,which implies $T_1 = 0.5 T_2$.Comparing this with $T_1 = x T_2$,we get $x = 0.5$.

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 $16:1$. At $T_1 = x T_2$,they radiate the same amount of heat per unit area per unit time. The value of $x$ is

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