A solid disc and a thin ring, having the same mass and same radius, are released from rest at the top of the same incline. Both roll without slipping. Which reaches the bottom first?

**Key formula.** $a = \dfrac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)}$. Whichever body has smaller $I_{\text{cm}}/(MR^2)$ has the larger acceleration. **Apply.** - Disc: $I_{\text{cm}} = \dfrac{1}{2}MR^2 \Rightarrow$ ratio $= \dfrac{1}{2}$, $a_{\text{disc}} = \dfrac{g\sin\theta}{1.5} = \dfrac{2}{3}g\sin\theta$. - Ring: $I_{\text{cm}} = MR^2 \Rightarrow$ ratio $= 1$, $a_{\text{ring}} = \dfrac{g\sin\theta}{2} = \dfrac{1}{2}g\sin\theta$. The disc has higher acceleration → reaches the bottom first. **Why option D is wrong.** The ratio of accelerations is independent of $\theta$ — both scale the same way with $\sin\theta$. The ordering never flips.