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(A) The centripetal force required for circular motion is provided by the central force $F$. Given $F \propto \frac{1}{R^{3}}$,we can write $F = \frac

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$T \propto R^{2}$

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(A) The centripetal force required for circular motion is provided by the central force $F$. Given $F \propto \frac{1}{R^{3}}$,we can write $F = \frac{K}{R^{3}}$,where $K$ is a constant.For a particle of mass $m$ moving with angular velocity $\omega$ in a circle of radius $R$,the centripetal force is $F = m \omega^{2} R$.Equating the two expressions: $\frac{K}{R^{3}} = m \omega^{2} R$.Rearranging for $\omega^{2}$: $\omega^{2} = \frac{K}{m R^{4}}$.Since the time period $T = \frac{2 \pi}{\omega}$,we have $\omega = \frac{2 \pi}{T}$.Substituting this into the equation: $(\frac{2 \pi}{T})^{2} = \frac{K}{m R^{4}}$.This simplifies to $\frac{4 \pi^{2}}{T^{2}} = \frac{K}{m R^{4}}$,which implies $T^{2} \propto R^{4}$.Taking the square root of both sides,we get $T \propto R^{2}$.

$A$ 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

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