2.27. A cone is inscribed in a sphere, the generatrix of which is equal to the diameter of the base. Find the ratio of the total surface area of the cone to the surface area of the sphere.
2.27. Let us consider the axial section of the cone passing through the center of the sphere. Since the diameter of the base of the cone is equal to the generatrix, we will obtain an equilateral triangle inscribed in a circle in the section (Fig. 2.23). Let the radius of the sphere be \( R \): then \( AB = R \sqrt{3}, AD = \frac{R \sqrt{3}}{2} \). Denote the total surface area of the cone by \( S_{1} \), and the surface area of the sphere by \( S_{2} \). We have

\[
S_{1} = \pi \frac{R \sqrt{3}}{2} \cdot R \sqrt{3} + \pi \left( \frac{R \sqrt{3}}{2} \right)^{2} = \frac{9}{4} \pi R^{2}, \quad S_{2} = 4 \pi R^{2}
\]

from which \( S_{1} : S_{2} = 9 : 16 \).

Answer: \( 9 : 16 \).