A smooth circular wire hoop of radius R is fixed so that it rotates with constant angular speed ω about its vertical diameter, with ω^2 = 2g/R. A…

Let the bead sit at angle $\theta$ from the lowest point. In the frame of the hoop it is in equilibrium under gravity, the centrifugal force $m\omega^2 R\sin\theta$ (horizontal, outward) and the hoop's reaction $N$ (along the radius, because the hoop is smooth and the bead is at rest relative to it, so there is no azimuthal component). Along the tangent: $mg\sin\theta = m\omega^2R\sin\theta\cos\theta$, so $\cos\theta = g/(\omega^2 R) = 1/2$ and $\theta = 60^\circ$ (the other root $\sin\theta=0$ is excluded). Along the radius, towards the centre: $N = mg\cos\theta + m\omega^2R\sin^2\theta = mg\cdot\tfrac12 + m\cdot\tfrac{2g}{R}\cdot R\cdot\tfrac34 = 2mg$. In general $N = mg/\cos\theta$. Option (A) results from $N=mg$ ignoring rotation, and (C), (D) from mis-resolving the centrifugal term. Hence $N = 2mg$.