Q 11-04-059JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
A $0.5$ kg mass is in contact against the inner wall of a cylindrical drum of radius $4$ m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is $5$ rad/s. The coefficient of friction between the drum's inner wall surface and mass is ______. (Take $g=10\ \text{m/s}^2$)
Answer: (A) $0.1$
The wall's normal force provides the centripetal force: $N=m\omega^2r$.
Friction must hold the weight: $\mu N\ge mg$. At the minimum speed, $\mu m\omega^2r=mg$.
$$\mu=\frac{g}{\omega^2r}=\frac{10}{5^2\times4}=0.1$$
Solution by Sreeraj P, M.Sc Physics