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

Answer

A

Explanation

When the solid cylinder rolls down the smooth inclined plane, its velocity is given by
`v^(2)=(2gh)/(1+(I)/(MR^(2)))`
But for the solid cylinder, `I=(MR^(2))/(2) and v=romega`
`therefore r^(2)omega^(2)=(2gh)/(1+(1)/(2)(MR^(2))/(MR^(2)))=(4)/(3)gh`
`therefore r^(2)=(4)/(3)ghxx(1)/(omega^(2))=(4)/(3)xx10xx3xx(1)/(4xx2)`
`therefore r^(2)=5 " "therefore r=sqrt5m`