A) 16P
B) 4P
C) 2P
D) P
Correct Answer: B
Solution :
Energy received per second i.e., power \[P\propto ({{T}^{4}}-T_{0}^{4})\] Þ \[P\propto {{T}^{4}}\] \[(\because \,{{T}_{0}}<<T)\] Also energy received per sec (p)\[\propto \frac{1}{{{d}^{2}}}\] (inverse square law) Þ \[P\propto \frac{{{T}^{4}}}{{{d}^{2}}}\] Þ \[\frac{{{P}_{1}}}{{{P}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}\times {{\left( \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{2}}\] Þ \[\frac{P}{{{P}_{2}}}={{\left( \frac{T}{2T} \right)}^{2}}\times {{\left( \frac{2d}{d} \right)}^{2}}=\frac{1}{4}\] Þ\[{{P}_{2}}=4P.\]
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