A) 6
B) 7
C) 8
D) 9
Correct Answer: D
Solution :
The given, \[{{r}_{1}}=9\,cm\] \[{{r}_{2}}=27\,cm\] \[{{\lambda }_{p}}=300\,nm\] \[{{\lambda }_{Q}}=900\,nm\] So, the ratio of the rate of energy radiated \[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}{{\left( \frac{{{\lambda }_{Q}}}{{{\lambda }_{p}}} \right)}^{4}}\] \[=\frac{9\times 9}{27\times 27}\times {{\left( \frac{900}{300} \right)}^{4}}\] \[=\frac{9}{3\times 27}\times \frac{9\times 9\times 9\times 9}{3\times 3\times 3\times 3}\] \[=\frac{9}{1}=9:1\]or = 9You need to login to perform this action.
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