A) 1/4
B) 1/2
C) 1/16
D) \[\frac{{{473}^{4}}-{{300}^{4}}}{{{673}^{4}}-{{300}^{4}}}\]
Correct Answer: D
Solution :
If temperature of surrounding is considered, then net loss of energy of a body by radiation \[Q=A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})t\Rightarrow Q\propto ({{T}^{4}}-T_{0}^{4})\] \[\Rightarrow \] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{T_{1}^{4}-T_{0}^{4}}{T_{2}^{4}-T_{0}^{4}}\] \[=\frac{{{(273+200)}^{4}}-{{(273+27)}^{4}}}{{{(273+400)}^{4}}-{{(273+27)}^{4}}}\] \[=\frac{{{(473)}^{4}}-{{(300)}^{4}}}{{{(673)}^{4}}-{{(300)}^{4}}}\]
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