A) 2 : 1
B) 4 : 1
C) 8 : 1
D) 16 : 1
Correct Answer: D
Solution :
We know that, \[{{r}_{n}}\propto {{n}^{2}}\] Hence, area \[\Rightarrow \] \[{{A}_{n}}\propto {{n}^{4}}\] Hence, \[\frac{{{A}_{2}}}{{{A}_{1}}}-\frac{n_{2}^{4}}{n_{1}^{4}}=\frac{{{(2)}^{4}}}{{{(1)}^{4}}}\] \[\Rightarrow \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{16}{1}\]
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