A sphere of radius ' $r$ ' is Kept on a concave minor of radius of curvature ' $R$ ' The arrangement is kept on a horizontal table If the sphere is displaced from its equilibrium position and left then it executes S.H.M. The period of oscillation will be ( $\mathrm{g}=$ acceleration due to gravity.)

• A

  $2 \pi[(\mathrm{R} / \mathrm{gr})]^{\frac{1}{2}}$
• Correct answer
  B

  $2 \pi[(\mathrm{R}-\mathrm{r}) / \mathrm{g}]^{\frac{1}{2}}$
• C

  $2 \pi[(\mathrm{R}-\mathrm{r}) 1.4 / \mathrm{g}]^{\frac{1}{2}}$
• D

  $2 \pi[(\mathrm{Rr}) / \mathrm{g}]^{\frac{1}{2}}$

Solution

Consider the figure below: Restoring torque about axis of rotation $\mathrm{O}$ is given by, $\begin{aligned} & \tau=-\mathrm{mgr}_{\perp}=-\mathrm{mg}(\mathrm{R}-\mathrm{r}) \sin \theta \\ & =\mathrm{m}(\mathrm{R}-\mathrm{r})^2 \alpha \\ & \Rightarrow \alpha=-\left(\frac{\mathrm{g}}{(\mathrm{R}-\mathrm{r})}\right) \theta=-\omega^2 \theta\end{aligned}$ The angular frequency can be written as: $\omega^2=\left(\frac{\mathrm{g}}{\mathrm{R}-\mathrm{r}}\right)$ $\therefore \mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{(\mathrm{R}-\mathrm{r})}{\mathrm{g}}}$

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