Let the speed of the block be v when it descends thrugha height h. Sol is the speed of the string and hence of a particle at the rim of the wheel. The angular velocity of the wheel is `v/r` and it skinetic energy at this instant is `1/2I(v/r)^2`. Using the prnciple of conservation of energy the gragvitastioN/Al potential energy lost by the block must be equal to the kinetic energy gained by the block and the wheel. Thus,
`mgh-1/2mv^2+1/2Iv^2/r^2`
`or v=[(2mgh)/(m+I/r^2)]^(1/2)`