Q 11-07-199JEE MainJEE Main 2018 (8 Apr)Easy
A particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the $n^{\text{th}}$ power of $R$. If the period of rotation of the particle is $T$, then:
Answer: (D) $T \propto R^{\frac{n+1}{2}}$
The central force provides the centripetal force:
$$m\omega^2R = \frac{k}{R^n} \;\Rightarrow\; \omega^2 \propto R^{-(n+1)}$$
$$T = \frac{2\pi}{\omega} \propto R^{\frac{n+1}{2}}$$
(For gravity, $n = 2$ gives Kepler's $T\propto R^{3/2}$.)
Solution by Sreeraj P, M.Sc Physics