A) \[\frac{K}{2r}\]
B) \[-\frac{K}{2r}\]
C) \[-\frac{K}{r}\]
D) \[\frac{K}{r}\]
Correct Answer: B
Solution :
Here \[\frac{m{{\nu }^{2}}}{r}\,\,=\,\,\frac{K}{{{r}^{2}}}\,\,\,\Rightarrow \,\,K.E.=\,\,\frac{1}{2}\,m{{\nu }^{2}}=\frac{K}{2r}\] \[U=-\int_{\infty }^{r}{F.dr=-\int_{\infty }^{r}{\left( -\frac{K}{{{r}^{2}}} \right)}\,dr=-\frac{K}{r}}\] Total energy \[\operatorname{E}=K.E. + P.E.=\frac{K}{2r}-\frac{K}{r}=-\frac{K}{2r}\]You need to login to perform this action.
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