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A thin circular loop of radius R rotates about its vertical diameter with angular speed $ω = \sqrt{\frac{2 g}{R}} .$ If a bead on the wire loop remains at rest relative to loop, the angle made by the radius vector of bead with the downward
vertical is (Neglect friction)

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$90^{\circ}$

$45^{\circ}$

$60^{\circ}$

$30^{\circ}$

answer is C.

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Detailed Solution

$Ncos θ = mg$

$Nsin θ = m ω^{2} r where \sin θ = \frac{r}{R}$

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