= Solution
{c}
For a uniform solid <cylinder>, set $a=0$ in the <moment of inertia> formula, giving
$$
\ddot x_{\mathrm{solid}}=\frac23g\sin\varphi.
$$
For the hollow <cylinder>, $a>0$ makes the denominator larger:
$$
\ddot x_{\mathrm{hollow}}=\frac{2b^2}{3b^2+a^2}g\sin\varphi
<\frac23g\sin\varphi
$$
on a downward incline with $\sin\varphi>0$. Thus \b[the solid <cylinder> accelerates faster], assuming <rolling without slipping> for both. More of the hollow <cylinder>'s <mass> lies far from the rotation axis, increasing its <moment of inertia> relative to $Mb^2$.
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