1. Use conservation of energy

All three bodies start from rest and roll down the same incline without slipping, so the loss of gravitational potential energy becomes translational plus rotational kinetic energy:

$mgh = \frac12 mv^2 + \frac12 I\omega^2$

Since rolling is without slipping,

$v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R}$

So,

$mgh = \frac12 mv^2 + \frac12 I\frac{v^2}{R^2}$

$mgh = \frac12 mv^2\left(1 + \frac{I}{mR^2}\right)$

Hence,

$v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}$

Thus, for the same height $h$, the body with smaller $\dfrac{I}{mR^2}$ will have greater speed at the bottom.

2. Write moment of inertia for each body

• Ring: $I = mR^2$ $\frac{I}{mR^2} = 1$

• Solid cylinder: $I = \frac12 mR^2$ $\frac{I}{mR^2} = \frac12$

• Solid sphere: $I = \frac{2}{5}mR^2$ $\frac{I}{mR^2} = \frac{2}{5}$

3. Compare speeds

Using

$v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}$

we get:

• Ring: $v_{\text{ring}}^2 = \frac{2gh}{2} = gh$

• Cylinder: $v_{\text{cyl}}^2 = \frac{2gh}{1+1/2} = \frac{2gh}{3/2} = \frac{4gh}{3}$

• Sphere: $v_{\text{sphere}}^2 = \frac{2gh}{1+2/5} = \frac{2gh}{7/5} = \frac{10gh}{7}$

Therefore,

$v_{\text{sphere}} > v_{\text{cyl}} > v_{\text{ring}}$

So the sphere has the greatest velocity and the ring has the least velocity at the bottom.

4. Check options

• A: False
• B: False
• C: True
• D: False

5. Final answer

The correct option is C.