99 Percentile Qs Bank for JEE MainPhysicsRotational Motion
A disc of radius 0.4 meter and mass 1 ~kg rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is 10 rad s ⁻² . The tangential force applied to the rim of the disc is
Options
- A2 ~N
- B3 ~N
- C4 ~N
- D5 ~N
Correct answer
A. 2 ~N
Step-by-step solution
R =0.4 ~m , M =1 ~kg , =10 rad / s ^2 I = MR ^2 2 = 1 (0.4)^2 2 =0.08 ~kg ~m ^2 Torque, = I =0.08 10=0.8 ~N - m Also, = FR or F= R = 0.8 0.4 =2 ~N