JEE Main201815 Apr 2018Evening ShiftPhysicsLaws of MotionActual
A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25 ~cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is ( g =10 ~m / s ^2 )
Options
- A0.5
- B0.7
- C0.3
- D0.6
Correct answer
D. 0.6
Step-by-step solution
Using, m g= m v^2 r =m r ^2 =2 n=2 3.5=7 rad / sec Radius, r=1.25 ~cm =1.25 10⁻² ~m Coefficient of friction, = ? m g= m(r )^2 r ( v=r ) gathered = r ^2 g = 1.25 10⁻² (7 22 7 )^2 10 = 1.25 10⁻² 22^2 10 =0.6 gathered