1. For a body rolling down an incline without slipping, the linear acceleration is

$a = \frac{g\sin\theta}{1+\frac{I}{mR^2}}$

where $I$ is the moment of inertia about its center.

2. The body that reaches first will have the largest acceleration. So we compare $\dfrac{I}{mR^2}$ for each body.

3. Moments of inertia:

• Ring: $I = mR^2 \Rightarrow \frac{I}{mR^2}=1$ $a = \frac{g\sin\theta}{1+1}=\frac{g\sin\theta}{2}$

• Disc: $I = \frac{1}{2}mR^2 \Rightarrow \frac{I}{mR^2}=\frac{1}{2}$ $a = \frac{g\sin\theta}{1+\frac{1}{2}}=\frac{2g\sin\theta}{3}$

• Solid cylinder: $I = \frac{1}{2}mR^2 \Rightarrow \frac{I}{mR^2}=\frac{1}{2}$ $a = \frac{2g\sin\theta}{3}$

• Solid sphere: $I = \frac{2}{5}mR^2 \Rightarrow \frac{I}{mR^2}=\frac{2}{5}$ $a = \frac{g\sin\theta}{1+\frac{2}{5}}=\frac{5g\sin\theta}{7}$

4. Compare the accelerations:

$\frac{g\sin\theta}{2}, \quad \frac{2g\sin\theta}{3}, \quad \frac{2g\sin\theta}{3}, \quad \frac{5g\sin\theta}{7}$

Since

$\frac{5}{7} > \frac{2}{3} > \frac{1}{2}$

the solid sphere has the greatest acceleration.

5. Therefore, the body that reaches the bottom first is:

$\boxed{4}$