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A ring is suspended at a point on its ri...

A ring is suspended at a point on its rim and it behaves as a second's pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be (g = `pi^(2))`

Text Solution

Verified by Experts

Time period of second pendulum T = 2 cm.
`T = 2pi sqrt((I)/(Mgd))`
Moment of inertia with respect to axis O
`I = MR ^(2) + MR^(2) = 2MR^(2)`
the distance between centre of mass and the axis O
d= R
`2= 2pisqrt((2MR^(2))/(MgR))implies R =0.5 m`
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