A) 1 : 4
B) 16 : 1
C) 1 : 16
D) 1 : 1
Correct Answer: D
Solution :
\[P=\sigma A{{t}^{4}}\] or \[P\propto A{{T}^{4}}\] or \[P\propto {{r}^{2}}{{T}^{4}}\] \[(\because A=4\pi {{r}^{2}})\] \[\frac{{{P}_{A}}}{{{P}_{B}}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{2}}{{\left( \frac{{{T}_{A}}}{{{T}_{B}}} \right)}^{4}}\] \[\frac{{{P}_{A}}}{{{P}_{B}}}={{\left( \frac{r}{4r} \right)}^{2}}{{\left( \frac{T}{T/2} \right)}^{4}}\] Or \[\frac{{{P}_{A}}}{{{P}_{B}}}=\frac{1}{{{(4)}^{2}}}\times {{(2)}^{4}}\] Or \[\frac{{{P}_{A}}}{{{P}_{B}}}=\frac{1}{16}\times 16\] \[\frac{{{P}_{A}}}{{{P}_{B}}}=\frac{1}{1}\]
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