The C.G. of a right circular cone lies on its axis of symmetry at a height of A. $$\frac{{\text{h}}}{2}$$ B. $$\frac{{\text{h}}}{3}$$ C. $$\frac{{\text{h}}}{4}$$ D. $$\frac{{\text{h}}}{5}$$

$$rac{{ ext{h}}}{2}$$
$$rac{{ ext{h}}}{3}$$
$$rac{{ ext{h}}}{4}$$
$$rac{{ ext{h}}}{5}$$

The correct answer is $\boxed{\frac{{\text{h}}}{2}}$.

The center of gravity (CG) of a right circular cone is located at a height of $\frac{{\text{h}}}{2}$ from the base, where $h$ is the height of the cone. This is because the mass of the cone is evenly distributed around the axis of symmetry, so the CG must lie on this axis.

Option A is incorrect because it is the height of the cone. Option B is incorrect because it is the radius of the base of the cone. Option C is incorrect because it is the radius of the cone. Option D is incorrect because it is the diameter of the base of the cone.

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