JEE Main 2026 · 2 Apr · Shift 2 · Physics Q27

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PhysicsSingle correct+4 / -1Question 27 of 75

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/s²)

0.1

0.5

0.7

0.3

Answer and worked solution

Official answer: A

Vertical balance. Friction holds the weight and the wall's normal force supplies the centripetal force:
$\mu N=\mu m\omega^2R\ge mg.$

Minimum speed. $\mu=\dfrac g{\omega^2R}=\dfrac{10}{25\times4}=0.1$ .

(A)

Topics:Circular motion dynamics Friction

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JEE Main 2026 · 2 Apr · Shift 2, Physics question 27