Q67 / 90 · KVPY 2014 (SB/SX) · Physics

A body moves in a circular orbit of radius $\mathrm{R}$ under the action of a central force. Potential due to the central force is given by $V(r)=\operatorname{kr}(k$ is a positive constant). period of revolution of the body is proportional to :

  • Correct answer
    A
    $R^{1 / 2}$
  • B
    $R^{-1 / 2}$
  • C
    $R^{-3 / 2}$
  • D
    $\mathrm{R}^{-5 / 2}$

Solution

$V(r)=k r$ $U(r)=m k r \quad \omega^{2}=\frac{k}{r}$ $\begin{aligned} F=-\frac{d u}{d r}=-m k \quad \omega=\sqrt{\frac{k}{r}} \\ & T=\frac{2 \pi \sqrt{r}}{\sqrt{k}} \end{aligned}$

Hence, the correct answer is option: 'A'.

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