A) \[+\frac{k}{r}\]
B) \[-\frac{k}{2r}\]
C) \[-\frac{\text{ 2K}}{r}\]
D) \[+\frac{\text{ K}}{{{r}^{2}}}\]
Correct Answer: B
Solution :
Potential energy,\[U=-\delta F.dr=-\delta \frac{K}{{{r}^{2}}}dr.=-\frac{K}{r}\] Kinetic energy\[,\]\[K=\frac{1}{2}\]. \[m{{v}^{2}}=\frac{K}{2r}\left( \operatorname{since}\frac{m{{v}^{2}}}{r}=\frac{K}{{{r}^{2}}}\Rightarrow {{v}^{2}}=\frac{K}{mr} \right)\] \[\therefore \]Total energy\[=-\frac{K}{r}+\frac{K}{2r}=-\frac{K}{2r}\]You need to login to perform this action.
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