A particle moves at a constant speed along the circumference of a circle with a radius \(R,\) subject to a central fictitious force \(F\) that is inversely proportional to \(R^3.\) Its time period of revolution will be given by:
Hint: \(F=mw^2R\)
Step: Find the time period of the revolution.
Given: The central fictitious force \(F\) that is inversely proportional to \(R^3\) i.e.,
\(F\propto \frac{1}{R^3}\Rightarrow F= \frac{K}{R^3}\)
\(\Rightarrow \frac{K}{R^3}= m\omega^2 R\Rightarrow \omega^2 = \frac{K}{mR^4}\)
\(\Rightarrow \left(\frac{2\pi}{T}\right)^2= \frac{K}{mR^4}~~~\left[\omega = \frac{2\pi}{T}\right]\)
\(\Rightarrow T^2 \propto R^4\)
\(\Rightarrow T\propto R^2\)
Therefore, the time period of the revolution is proportional to \(R^2.\)
Hence, option (1) is the correct answer.