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A thin uniform circular ring is rolling ...

A thin uniform circular ring is rolling down an inclined plane of inclination `30^(@)` without slipping. Its linear acceleration along the inclined plane will be

A

g/2

B

g/3

C

g/4

D

2g/3

Text Solution

Verified by Experts

The correct Answer is:
C
`a= (g sin theta)/(1 + (k^(2))/(R^(2))) = (g sin 30^(@))/(1+1)= (g)/(4)` [As `(k^(2))/(R^(2))=1 and theta= 30^(@)`]
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