(b) Using the relation $\frac{{mv}^{2}}{r} = \mu R,\ \ R = mg$ $\frac{{mv}^{2}}{r} = \mu mg \Longrightarrow v^{2} = \mu rg$ $v^{2} = 0.6 \times 150 \times 10$ $v = 30{\ ms}^{- 1}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.