A) \[\frac{r}{2}\]
B) 2r
C) 4r
D) \[\frac{r}{8}\]
E) \[\frac{r}{4}\]
Correct Answer: E
Solution :
\[p\propto \frac{1}{\sqrt{r}}\] \[r\propto \frac{1}{{{p}^{2}}}\] Hence, \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)}^{2}}\] \[={{\left( \frac{p}{2p} \right)}^{2}}=\frac{1}{4}\] \[{{r}_{1}}=\frac{{{r}_{2}}}{4}\]
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