A)
B)
C)
D)
Correct Answer: A
Solution :
\[\frac{m{{v}^{2}}}{r}=\frac{{{k}^{2}}}{{{r}^{2}}}\] \[\therefore \] \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\frac{{{k}^{2}}}{r}\] \[=\] kinetic energy of the particle Potential energy of the particle\[=-\frac{k}{r}\] \[\therefore \]Total energy\[E=K.E.+P.E.\] \[=\frac{k}{2r}-\frac{k}{r}=-\frac{k}{2r}\] \[\therefore \] \[|E|\,\,=\frac{k}{2r}\]
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