Q. $A\ particle\ of\ mass\ m\ is\ moving\ in\ a\ horizontal\ circle\ of\ radius\ r,\ under\ a\ centripetal\ force$ $= \frac{k}{r^{2}'}\ where\ k\ is\ a\ constant.$
A) The potential energy of the particle is zero
B) $The\ potential\ energy\ of\ the\ particle\ is\frac{k}{r}$
C) $The\ total\ energy\ of\ the\ particle\ is - \frac{k}{2r}$
D) $The\ Kinetic\ energy\ of\ the\ particle\ is - \frac{k}{r}$
Explanation
(a) For horizontal planes potential energy remains constant equal to zero, if we assumes surface to be the zero level.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.