A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]A. `5 cm`B. `0.5 cm`C. `sqrt10 cm`D. `sqrt5 m`

A solid cylinder rolls down an inclined plane of height `3m` and reach

1.	A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]A. `5 cm`B. `0.5 cm`C. `sqrt10 cm`D. `sqrt5 m`
Answer» `v=sqrt((2gh)/(1+k^(2)//R^(2)))=sqrt((2gh)/(1+(1//2)))=sqrt((4gh)/(3))=sqrt((4xx10xx3)/(3))` `2sqrt10` `omega=(v)/(R )implies R=(v)/(omega)=(2sqrt10)/(2sqrt2)=sqrt5m`

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A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]A. `5 cm`B. `0.5 cm`C. `sqrt10 cm`D. `sqrt5 m`

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