A) \[\frac{K}{2r}\]
B) \[-\frac{K}{2r}\]
C) \[-\frac{K}{r}\]
D) \[\frac{K}{r}\]
Correct Answer: B
Solution :
Here \[\frac{m{{v}^{2}}}{r}=\frac{K}{{{r}^{2}}}\] \ K.E.\[=\frac{1}{2}m{{v}^{2}}=\frac{K}{2r}\] \[U=-\int_{\infty }^{r}{F.dr}=-\int_{\infty }^{r}{\left( -\frac{K}{{{r}^{2}}} \right)}\,dr=-\frac{K}{r}\] Total energy \[E=\text{K}\text{.E}\text{.}+\text{P}\text{.E}\text{.}=\frac{K}{2r}-\frac{K}{r}=-\frac{K}{2r}\]
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