A small particle of mass `m`, moves in such a way that the potential energy `U = ar^(3)`, where a is position constant and `r` is the distance of the particle from the origin. Assuming Rutherford's model of circular orbits, then relation between `K.E` and `P.E` of the particle is :

Options

A. `K.E = (3)/(2) U`
B. `K.E = U`
C. `K.E = 3U`
D. `K.E = (U)/(2)`

Answer

A

Explanation

`F=|(dU)/(dr)| = 3ar_(2)(mv^(2))/(r)=3ar_(2)`
`mv_(2)=3ar_(3)`
`(1)/(2)mv_(2)=(3)/(2) ar_(3) K.E = (3)/(2) U`.