The cone’s slant angle θ is given by cosθ=0.20/1=0.2. sinθ=√1-0.04=√0.96.
The outward force on the ball is mv²r horizontally and mg vertically. If T is the tension, Tcosθ=mv²/r and Tsinθ=mg. So T=0.500×9.8/√0.96=0.5N approx.
So since Tcosθ=mv²/r, 0.500×0.98/√0.96×0.2=0.500v²/0.20; v²=(0.98/√0.96)/0.500=2, and v=1.41 m/s. The angular speed is v/r=√2/0.2=7.07 revs per sec=424.3 rpm approx.