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:

A particle moves in a circle of radius $$R$$ under a central force that is inversely proportional to $$R^3$$. We can write $$F = \frac{k}{R^3}$$, where $$k$$ is a positive constant.

For uniform circular motion, the centripetal force required is $$F_c = \frac{mv^2}{R}$$, where $$m$$ is the mass and $$v$$ is the speed of the particle. Setting the applied force equal to the centripetal force: $$\frac{mv^2}{R} = \frac{k}{R^3}$$.

Solving for $$v^2$$: $$v^2 = \frac{k}{mR^2}$$. Taking the square root: $$v = \sqrt{\frac{k}{m}} \cdot \frac{1}{R}$$, so $$v \propto \frac{1}{R}$$.

The time period of one complete revolution is the circumference divided by the speed: $$T = \frac{2\pi R}{v}$$. Substituting $$v = \sqrt{\frac{k}{m}} \cdot \frac{1}{R}$$:

$$T = \frac{2\pi R}{\sqrt{k/m} \cdot (1/R)} = \frac{2\pi R^2}{\sqrt{k/m}} = 2\pi\sqrt{\frac{m}{k}} \cdot R^2$$.

Therefore $$T \propto R^2$$.

Hence the correct answer is Option 4: $$T \propto R^2$$.