A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r^{2}}$ . Its total energy is:-

\frac{k}{2 a^{2}}

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z e r o

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- \frac{3 k}{2 a^{2}}

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- \frac{k}{4 a^{2}}

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Solution

The correct option is B $z e r o$
$\begin{matrix} F = \frac{\partial u}{\partial e} = \frac{K}{r^{3}} sin c e i t i s p e r f o r min g c i r c u l a r m o t i o n F = \frac{m v^{2}}{r} = \frac{K}{r^{3}} m v^{2} = \frac{K}{r^{2}} \Rightarrow K . E . = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} T o t a l e n e r g y = P . E + K . E = - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = z e r o \end{matrix}$
Hence,
option $(B)$ is correct answer.

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