JEE Main 2026 Physics question
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)
Answer: (A)
Solution
At the minimum speed, friction balances gravity: μN = mg with N = mω²r. So μ = g/(ω²r) = 10/(25 × 4) = 0.1.