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)`]

Options

A. 5 cm
B. 0.5 cm
C. `sqrt(10) cm`
D. `sqrt5m`

Answer

D

Explanation

The relation between linear velocity (v) and angular velocity `(omega)` is `v=romega rArr r=(v)/(omega)`
From equation of motion, `v^(2)=u^(2)+2gh`
When `u=0, h=3mm, g=10ms^(-2)`
`omega=2sqrt2"rad s"^(-1)`
`v=sqrt(2xx10xx3)=sqrt(60)ms^(-1)`
`therefore" "r=(sqrt(60))/(2sqrt2)=(sqrt(30)xx2)/(2sqrt2)=(sqrt((30)/(4)))=sqrt5m`