Concept:
For an object sliding down a smooth inclined plane and entering a loop-the-loop to complete a vertical circle, the conservation of energy dictates the minimum height. To barely complete the circle, the velocity at the top of the loop must be \( v_{top} = \sqrt{gr} \).
Step 1: State the energy conservation condition.
Using conservation of energy from the release height \(h\) to the top of the circle \(2r\):
$$ mgh = mg(2r) + \frac{1}{2}m v_{top}^2 $$
$$ mgh = 2mgr + \frac{1}{2}m(gr) = 2.5mgr $$
$$ h = 2.5r = \frac{5}{2}r $$
Step 2: Convert diameter to radius.
Diameter \( D = 20 \) cm, so radius \( r = 10 \) cm \(= 0.1\) m.
Step 3: Calculate height.
$$ h = 2.5 \times 0.1 \text{ m} = 0.25 \text{ m} $$
$$\boxed{0.25 \text{ m}}$$