Correct Answer - A
We know that the kinetic energy, `K = (1)/(2)mv^(2)=(1)/(2)I omega^(2)`
where, `m = 27 kg` (mass of the body)
`omega = 3 "rads"^(-1)` (angular velocity)
`I = 3 "kg-m"^(2)` (moment of inertia)
`v = ? or mv^(2)=I omega^(2)`
`rArr v^(2)=(I omega^(2))/(m)rArrv^(2)=(3xx3^(2))/(27)rArrv^(2)=(27)/(27)=1`
`rArr` Velocity of body, `v = sqrt(1) = 1 ms^(-1)`.