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Q.

A small ring P is threaded on a smooth wire bent in the form of a circle of radius a and center O. The wire is rotating with constant angular speed ω about a vertical diameter XY, while the ring remains at rest relative to the wire at a distance a/2 from XY. Then ω2 is equal to

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a

2ga

b

g2a

c

2ga3

d

g32a

answer is C.

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Detailed Solution

Ncos⁡θ=mg   ................(i)Nsin⁡θ=mω2a2 ............(ii)From Eqs. (ii)/(i), tan⁡θ=ω2a2g⇒ω2=2gtan⁡θa Now, sin⁡θ=a2×1a=12 or θ=30∘ or  ω2=2g3a
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A small ring P is threaded on a smooth wire bent in the form of a circle of radius a and center O. The wire is rotating with constant angular speed ω about a vertical diameter XY, while the ring remains at rest relative to the wire at a distance a/2 from XY. Then ω2 is equal to