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

Options

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

Answer

C

Explanation

`U = k/(2r^(2)) " " ` ….(1)
` :. (dU)/(dr) = k/2 (-2r^(-3) (dr)/(dr))`
` :. F = k/(r^(3))`
Now particle is moving in circular path of radius a instead of radius r , the centripetal force
`(mv^(2))/a = k/(a^(3)) " " [ :. r =a]`
` :. 1/2 mv^(2) =k/(2a^(2))`
` :. k = k/(2a^(2))" "...(2)`
` :. "Total energy " = U+k`
` = - k/(2a^(2) +k/(2a^(2))`
` [ :. " from equ .(1) and (2) r =1"] `
= 0