A particle is moving with uniform speed along the circumference of a circle of radius $\mathrm{R}$ under the action of a central fictitious force $\mathrm{F}$ which is inversely proportional to $\mathrm{R}^{3}$. Its time period of revolution will be given by:
Correct Option: , 2
(2)
$F \propto \frac{1}{R^{3}}$
$F=\frac{K}{R^{3}}$
$\frac{m v^{2}}{R}=\frac{K}{R^{3}}$
$m(\omega R)^{2}=\frac{K}{R^{2}}$
$m \omega^{2} R^{2}=\frac{K}{R^{2}}$
$\omega^{2}=\frac{K}{m}\left(\frac{1}{R^{4}}\right)$
$\left(\begin{array}{l}\frac{2 \pi}{T} \\ 2\end{array}\right)^{2} \frac{1}{R^{4}}$
$\frac{4 \pi^{2}}{T^{2}} \propto \frac{1}{R^{4}}$
$T \propto R^{2}$
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