A) 400 K
B) 4000 K
C) 80000 K
D) 40000 K
Correct Answer: C
Solution :
Energy radiated from a body \[Q=Ae\sigma {{T}^{4}}t\] \[\Rightarrow \] \[\frac{{{Q}_{2}}}{{{Q}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{Q}_{2}}}{{{Q}_{1}}} \right)}^{1/4}}={{\left( \frac{4.32\times {{10}^{6}}}{2.7\times {{10}^{-3}}} \right)}^{1/4}}\] \[\Rightarrow \] \[\frac{{{T}_{2}}}{400}={{\left( \frac{16\times 27}{27}\times {{10}^{8}} \right)}^{1/4}}=2\times {{10}^{2}}\] \[\Rightarrow \] \[{{T}_{2}}=-200\times 400=80000\,K\]
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