Q 11-04-198JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25\ \text{cm}$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $(g = 10\ \text{m/s}^2)$
Answer: (D) $0.6$
Static friction provides the centripetal force, so $\mu mg \ge m\omega^2r$.
$\omega = 2\pi\times3.5 = 7\pi \approx 22\ \text{rad s}^{-1}$:
$$\mu \ge \frac{\omega^2 r}{g} = \frac{(22)^2\times0.0125}{10} \approx 0.60$$
The coefficient of friction is about $0.6$.
Solution by Sreeraj P, M.Sc Physics