Answer
The correct answer is:
$(c) ~\frac{K_{rot}'}{K_{tr}'} \gt \frac{K_{rot}}{K_{tr}}$
Work Step by Step
Since the hole was drilled along the axis, the new shape has less mass located near the axis.
For a cylinder, the rotational inertia is $\frac{1}{2}MR^2$
The new shape will have a rotational inertia $I'$ such that $I' \gt \frac{1}{2}MR^2$
Since the expression for the rotational inertia increases, a higher fraction of the energy at the bottom of the incline will be in the form of rotational kinetic energy.
The correct answer is:
$(c) ~\frac{K_{rot}'}{K_{tr}'} \gt \frac{K_{rot}}{K_{tr}}$
