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Question

A ring of mass 7.2 kg and of radius 25 cm is experiencing a tension of 880 N while making rotation about its own axis is such a way that each part of the ring has uniform speed. The frequency of rotation of the ring (in rev/min) is (assume surfaces of contact are frictionless)

A
10 rev/min
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B
60 rev/min
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C
520 rev/min
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D
529 rev/min
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Solution

The correct option is D 529 rev/min
Tension in a ring of mass m rotating with angular velocity ω about its axis is given by
T=mRω22π

Since, ω=2πf
T=mR(2πf)22π
f=T2πmR

Given: T=880 N;m=7.2 kg;R=25 cm=0.25 m

We get f=8802π×7.2×0.25=880×72×22×7.2×0.25
=8.82 rev/s=529 rev/min

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