A) 1
B) 2
C) 0
D) \[R/2\]
Correct Answer: C
Solution :
Given, \[=\sqrt{\frac{2GM}{R}\left( 1-\frac{1}{\sqrt{5}} \right)}\] \[E\propto \frac{1}{{{r}^{2}}}\] (say) \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] \[=\frac{{{(4)}^{2}}}{{{(3)}^{2}}}=\frac{16}{9}\] And \[\frac{{{E}_{2}}}{{{E}_{1}}}-1=\frac{16}{9}-1=\frac{7}{9}\] \[\frac{{{E}_{2}}-{{E}_{1}}}{{{E}_{1}}}=\frac{7}{9}\times 100=\frac{700}{9}\] \[=77.77%(increase)\] \[4\mu F\] \[10\mu F\] \[8\mu F\] \[120\mu F\] \[\omega \]
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