If acycle wheel of radius 0.4 m completes one revolution in one second , then the acceleration of a point on the rim is in one second , then the acceleration of a point on the rim is
A
`0.8 pi ^(2) m//s^(2)`
B
`1.2pi ^(2) m//s^(2)`
C
`1.6pi^(2)m//s^(2)`
D
`0.4 pi m//s^(2)`
Text Solution
Verified by Experts
The correct Answer is:
C
`a=(v^(2))/(r ) = omega^(2) r = 4pi ^(2) n^(2) r = 4pi ^(2) xx 1xx0.4 = 1.6 pi ^(2) m//s^(2)`
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