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(C) In an inverse square field,the force $F$ acting on a body of mass $m$ moving in a circular orbit of radius $r$ is given by $F = \frac{k}{r^2}$,whe

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$T^2 \propto r^3$

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(C) In an inverse square field,the force $F$ acting on a body of mass $m$ moving in a circular orbit of radius $r$ is given by $F = \frac{k}{r^2}$,where $k$ is a constant.For circular motion,this force provides the necessary centripetal force: $\frac{k}{r^2} = \frac{mv^2}{r}$.Simplifying this,we get $v^2 = \frac{k}{mr}$,which implies $v = \sqrt{\frac{k}{mr}}$.The time period $T$ for one revolution is $T = \frac{2\pi r}{v}$.Substituting the expression for $v$: $T = \frac{2\pi r}{\sqrt{\frac{k}{mr}}} = 2\pi \sqrt{\frac{m}{k}} \cdot r^{3/2}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{m}{k} r^3$.Since $4, \pi, m,$ and $k$ are constants,we have $T^2 \propto r^3$.

If a body describes a circular motion under an inverse square field,the time taken to complete one revolution $T$ is related to the radius of the circular orbit $r$ as:

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