The correct option is
D 12√3Given :- Weight of thin hoop = 500 N
Radius of hoop ( R ) = 1 m
To Find :- Minimum coefficient of friction in equilibrium ( μmin )
Solution :- According to diagram , N=MgCos30∘
∴ f=μmgCos30∘ (∵f=μN)
Now , taking equilibrium along the plane , we get
MgSin30∘ - μmgCos30∘ = τ −−−−− (i)
We know , In rotational equilibrium about centre of mass
τ×R = f×R ⟹ τ = f ⟹ τ = μmgCos30∘
Applying in eq. (i) we get ,
MgSin30∘ = 2μmgCos30∘
∴ μmin = 14 × 2√3 ⟹ 12√3––––––
Hence , Option D(12√3) is correct.––––––––––––––––––––––––––––––