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A ring starts to roll down the inclined ...

A ring starts to roll down the inclined plane of height h without slipping. The velocity with which it reaches the ground is

A

`sqrt((10gh)/(7))`

B

`sqrt((4gh)/(7))`

C

`sqrt((4gh)/(3))`

D

`sqrt(gh)`

Text Solution

Verified by Experts

The correct Answer is:
D
For a ring `k^(2) = r^(2)` then
`v^(2) = (2gh)/(1 + k^(2)//r^(2)) = (2gh)/(2) = gh`
`v = sqrt(gh)`
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