A uniform ring rolls down an inclined plane without slipping. If it reaches the bottom of a speed of `2m//s`, then calculate the height of the inclined plane (use `g=10m//s^(2)`)

Explanation

`h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))`
`M.I.` of ring `=MR^(2)`, `K=R`
`h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))=(v^(2))/(2g)xx2=(v^(2))/(g)=(2xx2)/(10)=0.4m`
Hint : `h=(v^(2))/(2g)(1+(k^(2))/(R^(2)))`