Let speed of cylinder upon reaching the bottom of the inclined plane be v.
Here, gain in kinetic energy=loss in potential energy
So, 21Iω2+21mv2=mglsinθ
⇒21(2mr2)ω2+21mv2=mglsin30∘
For pure rolling, v=rω
⇒21(2mr2)(rv)2+21mv2=mglsin30∘
⇒41mv2+21mv2=mglsin30∘⇒43mv2=2mgl⇒v=64gl
Putting the values, we have
⇒v=64×10×0.6=2ms−1