A same torque is applied to a disc and a ring of equal mass and radii then

Step-by-step solution

Rotational dynamics analysis shows that for a constant applied torque , the angular acceleration relates to moment of inertia I through = I . For identical mass and radius, a disc has I_ disc = 1 2 MR^2 while a ring has I_ ring = MR^2 , establishing I_ ring = 2I_ disc . With constant, I remains equal for both objects. Since I_ ring > I_ disc , it follows that _ disc > _ ring . From the inertia ratio, _ disc = 2 _ ring . Starting from rest, angular frequency evolves as = t . The disc's greater angular acceleration p