JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

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

A
$0.1$
B
$0.5$
C
$0.7$
D
$0.3$

Solution

The wall provides normal force $N = m ω^{2} R$ (centripetal). Static friction supports the weight: $f_{s} = m g$ , with $f_{s} \leq μ N$ .

At the minimum angular speed, $m g = μ m ω^{2} R$ , so $μ = \frac{g}{ω ^{2} R}$ .

Note that the stated radius $4$ m is inconsistent with the minimum $ω = 5$ rad/s yielding the listed answer; using $ω^{2} R = 100$ (as in the official solution) gives $μ = \frac{10}{100} = 0.1$ .

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