Answer:
K1/K2=4
Explanation:
The kinetic energy of a rotating sphere is given by:

The moment of inertia of a solid sphere is given by

The initial kinetic energy is therefore


The final kinetic energy is given by

Therefore the relation K1/K2 if R2 = 0.5R1

The text says nothing about the final angular velocity just the collapse of the collapse of the radius
