A 0.5 kg mass is in contact against the inner wall of a cylindrical… | JEE Main 2026 (2 Apr, Shift 2)

Q1JEE Main 2026 · 2 APR · SHIFT 2Easy

Level Road & Rotating Platform

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 $g = 10\text{ m/s}^2$ )

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Answer: (a)

Wall's normal force gives the centripetal force: $N = m\omega^2 r$ .
Friction holds the weight: $\mu N \ge mg \Rightarrow \mu = \frac{g}{\omega_{min}^2 r}$ .
$\mu = \frac{10}{25 \times 4} = 0.1$
So the answer is (a) 0.1.

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