Let the angular velocity of ring be `omega`, when a particle is at the lowest position.
Taking `PQ` as reference level, appliying mechanical energy conservation between two positions
`mg(R+(R)/(4))+mg(R/4)= -mg(R+(R)/(4))-mg(R/4)+(1)/(2)I_(PQ)omega^(2)`
`3mgR=(1)/(2)I_(PQ)omega^(2)`
`I_(PQ)=m(R+(R)/(4))^(2)+[(1)/(4)mR^(2)+m((R)/(4))^(2)]=(15)/(8)mR^(2)`
`3mgR=(1)/(2)xx(15)/(8)mR^(2)omega^(2)implies omega=sqrt((16g)/(5R))`
Linear speed of particle is
`v=(R+(R)/(4))omega=(5R)/(4)sqrt((16g)/(5R))`
`=sqrt(5gR)`