JEE Main 2018Session, Shift 1PhysicsMediumWork, Energy and Power70× since 2002A particle is moving in a circular path of radius aaa under the action of an attractive potential U=−k2r2U = - {k \over {2{r^2}}}U=−2r2k Its total energy is:AOption A. −32ka2- {3 \over 2}{k \over {{a^2}}}−23a2kBOption B. ZeroCOption C. −k4a2- {k \over {4{a^2}}}−4a2kDOption D. k2a2{k \over {2{a^2}}}2a2kCheckSkip · show solution
