Q 11-04-168JEE MainJEE Main 2021 (26 Feb, Shift 1)Easy
A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^3$. Its time period of revolution will be given by:
Answer: (D) $T \propto R^2$
The central force provides the centripetal force:
$$\frac{mv^2}{R} = \frac{k}{R^3} \Rightarrow v^2 \propto \frac{1}{R^2} \Rightarrow v \propto \frac{1}{R}$$
Time period $T = \dfrac{2\pi R}{v} \propto R \cdot R = R^2$.
Solution by Sreeraj P, M.Sc Physics