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A man standing at the centre of a rotating table, with his hands stretched outwards. The table is rotating at the rate of 30 rev/minute. If the man brings his hands towards his chest and thereby reduces his moment of inertia to `(3)/(5)` times of its original moment of inertia, then the number of revolutions performed by the rotating table per minute will be

A

25

B

30

C

40

D

50

Text Solution

Verified by Experts

The correct Answer is:
D
`I_(1)omega_(1)=I_(2)omega_(2)`
`therefore I_(1)xx2pin_(1)=(3)/(5)I_(1)xx2pin_(2)`
`therefore n_(2)=(5)/(3)n_(1)=(5)/(3)xx30="50 rev/min."`
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