A) 0.054
B) 0.0034
C) 1
D) 2
Correct Answer: C
Solution :
Total energy radiated from a body \[Q=A\varepsilon \sigma {{T}^{4}}t\] \[\Rightarrow \] \[Q\propto A{{T}^{4}}\propto {{r}^{2}}{{T}^{4}}\] \[(\because A=4\pi {{r}^{2}})\] \[\Rightarrow \] \[\frac{{{Q}_{P}}}{{{Q}_{Q}}}={{\left( \frac{{{r}_{P}}}{{{r}_{Q}}} \right)}^{2}}{{\left( \frac{{{T}_{P}}}{{{T}_{Q}}} \right)}^{4}}\] \[={{\left( \frac{8}{2} \right)}^{2}}{{\left[ \frac{273+127}{273+527} \right]}^{4}}=1\]
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