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At $273^{\circ} \mathrm{C}$, the emissive power of a perfectly black body is $R$. It emissive power at $0^{\circ} \mathrm{Cis}$
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The correct answer is:
$\frac{R}{16}$
From Stefan's law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature $(T)$ of the body $E=\sigma T^4$
Given, $T_1=273^{\circ} \mathrm{C}=273+273 \mathrm{~K}=564 \mathrm{~K}$
$T_2=0^0 \mathrm{C}=273 \mathrm{~K}$
$\begin{aligned}
& \therefore \frac{E_1}{E_2}=\frac{T_1^4}{T_2^4} \\
& \Rightarrow E_2=\frac{T_2^4}{T_1^4} E_1=\frac{(273)^4}{(564)^4} R=\frac{R}{16}
\end{aligned}$
Given, $T_1=273^{\circ} \mathrm{C}=273+273 \mathrm{~K}=564 \mathrm{~K}$
$T_2=0^0 \mathrm{C}=273 \mathrm{~K}$
$\begin{aligned}
& \therefore \frac{E_1}{E_2}=\frac{T_1^4}{T_2^4} \\
& \Rightarrow E_2=\frac{T_2^4}{T_1^4} E_1=\frac{(273)^4}{(564)^4} R=\frac{R}{16}
\end{aligned}$
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