A particle is moving in a circular path of radius a under the action of an attractive potential `U=-(k)/(2r^(2))`. Its total energy is :

Options

A. `-(K)/(4a^(2))`
B. `(K)/(2a^(2))`
C. Zero
D. `-(3k)/(2a^(2))`

Answer

C

Explanation

`F=(delu)/(delr)htr=(K)/(r^(3))impliesmv^(2)=(K)/(r^(2))`
`therefore K.E.=(1)/(2)mv^(2)=(K)/(2r^(2))`
Total energy =P.E+K.E.
`=-(K)/2r^(2)+(K)/(2r^(2))=zero (because P.E.=-(K)/(2r^(2))given)`