A) 0.054
B) 0.0034
C) 1
D) 2
Correct Answer: A
Solution :
Total energy radiated from a body \[Q=A{{\varepsilon }_{0}}{{T}^{4}}t\] \[\Rightarrow \] \[Q\propto A{{T}^{4}}\propto {{r}^{2}}{{T}^{4}}\] \[(\because A=4\pi {{r}^{2}})\] \[\therefore \] \[\frac{{{Q}_{p}}}{{{Q}_{Q}}}={{\left( \frac{{{r}_{p}}}{{{r}_{Q}}} \right)}^{2}}{{\left( \frac{{{T}_{p}}}{{{T}_{Q}}} \right)}^{4}}\] \[={{\left( \frac{8}{2} \right)}^{2}}{{\left( \frac{273+127}{273+527} \right)}^{4}}=1\]
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