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A ring is rolling down an inclined plane...

A ring is rolling down an inclined plane which is inclined at an angle of `30^(@)` to the horizontal. Then the speed of ring after travelling a distance of 10 m is :-

A

`10 ms^(-1)`

B

`sqrt(50) ms^(-1)`

C

`sqrt(30)ms^(-1)`

D

`2 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`v=sqrt((2 gs sin theta )/(1+(K^(2))/(R^(2))))=sqrt(50)`
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