Learn

Friday, 9 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2026, Shift 2, Physics Q27

Question 27 of 75 in this shift, Physics question 2 of 25, Section A.

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=10g = 10 m/s2^2)
  1. (1)0.10.1Official answer
  2. (2)0.50.5
  3. (3)0.70.7
  4. (4)0.30.3

Official answer

Option 1

NTA final key.