Correct Answer - A::B::C::D
`N_(1)=mg-f_(2).....(1), N_(2)=f_(1).....(2)`
Now taking moment about `A`
`N_(2)xxlsin theta+f_(2)xxlcos theta=mgl/2xxcos theta......(3)`
Take moment about point `B`
`N_(1)xxIcos theta=mgxxl/2 cos theta+f_(1)xxl sin theta....(4)`
`tau_(0)`(of normals)=
`N_(1)l cos theta+N_(2)xxl sin theta=(mg)/2`
`l cos theta+(mg)/2 cot theta=2mglcos theta//2`
`B,C &D`are easily explain