Question

# A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is

A
2m
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B
4m
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C
1.5m
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D
3m
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Solution

## The correct option is B 1.5m Time period of a physical pendulum is given by T=2π√IsupportmglcmIn case of the given ringIsupport=(mR22+mR2)=32mR2lcm=RSo we have,Tring=2π  ⎷(32mR2)mgR=2π√3R2gand we know that the time period of a simple pendulum is given byTp=2π√lgfor equal time period Tring=Tp⇒3R2g=lg⇒l=3R2=1.5mhence the equivalent length of the simple pendulum is 1.5m

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