Q 11-05-199JEE MainJEE Main 2018 (8 Apr)Medium
A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = -\dfrac{k}{2r^2}$. Its total energy is:
Answer: (D) Zero
Force magnitude: $F = \left|\dfrac{dU}{dr}\right| = \dfrac{k}{r^3}$ (attractive). For circular motion of radius $a$:
$$\frac{mv^2}{a} = \frac{k}{a^3} \;\Rightarrow\; K = \frac12mv^2 = \frac{k}{2a^2}$$
$$E = K + U = \frac{k}{2a^2} - \frac{k}{2a^2} = 0$$
Solution by Sreeraj P, M.Sc Physics