NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
Four identical uniform rods of mass m = 6 kg each are welded their ends to form a square and then welded to a uniform ring having mass m = 4 kg and radius R = 1 m . The system is allowed to roll down a rough incline of inclination θ = 3 0 ∘ . ( where g is in m s - 2 ) The force of friction between the system and the plane is
Options
- A6 g downwards
- B6 g  upwards
- C3 5 g 6 upwards
- D3 5 g 6   downwards
Correct answer
C. 3 5 g 6 upwards
Step-by-step solution
I = M R 2 2 1 2 + M R 2 2 × 4 + mR 2 = 20 kg m 2 4 M + m g sin θ - F = 4 M + m a FR = I a R Solving, a = 7 g 2 4 ∴ F = 4 M + m g 2 - a = 4 M + m 12 g - 7 g 2 4 = 7 5 g 6 F = 35 g 6 upwards