A) \[6v\]
B) \[\frac{3v}{4}\]
C) \[12v\]
D) \[\frac{3v}{2}\]
Correct Answer: A
Solution :
The velocity v of a satellite varies inversely as the square root of the orbital radius r i.e., \[v\propto \frac{1}{\sqrt{r}}\] So, \[\frac{{{v}_{x}}}{{{v}_{y}}}=\sqrt{\frac{{{r}_{B}}}{{{r}_{A}}}}\] Given, \[{{v}_{x}}=3v\] and \[\frac{{{r}_{B}}}{{{r}_{A}}}=\frac{R}{4R}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{3v}{{{v}_{y}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\] Or \[{{v}_{y}}=2\times 3v=6v\]
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