(c) Minimum angular velocity $\omega_{\min} = \sqrt{g/R}$ $\therefore T_{\max} = \frac{2\pi}{\omega_{\min}} = 2\pi\sqrt{\frac{R}{g}} = 2\pi\sqrt{\frac{2}{10}} \cong 3s$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.