Answer
The correct answer is:
$(a) ~v' \lt v$
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. Therefore, the translational velocity will be smaller.
The correct answer is:
$(a) ~v' \lt v$
