A) 1 : 3
B) 1 : 81
C) 1 : 27
D) 1 : 9
Correct Answer: B
Solution :
Energy per second \[P\left( =\frac{Q}{t} \right)\propto {{T}^{4}}\] \[\frac{{{P}_{1}}}{{{P}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}={{\left( \frac{273-73}{273+327} \right)}^{4}}={{\left( \frac{200}{600} \right)}^{4}}=\frac{1}{81}\]
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