2. The maximum volume of a cone inscribed in a sphere compared to the volume of the sphere is

保留了源文本的换行和格式。
2. Let the radius of the sphere be $R$, the height of the inscribed cone be $h$, and the base radius be $r$.
Using any plane passing through the center of the sphere and the vertex of the inscribed cone to cut, the resulting section is as shown in the figure.
It is easy to know that $r^{2}=h(2 R-h)$.
$$
\begin{array}{c}
\therefore V_{\text {cone}}=\frac{1}{3} \pi r^{2} h \\
=\frac{1}{3} \pi h^{2}(2 R-h) \\
=\frac{1}{3} \pi \cdot \frac{1}{2}[h \cdot h \cdot(4 R-2 h)] .
\end{array}
$$
$\because h+h+(4 R-2 h)=4 R$ is a constant, \# $h+1+h^{-1}$
$$
\text { ( } \left.V_{\text {cone}}\right)_{\max }=\frac{1}{3} \pi \cdot\left(\frac{4}{3} R\right)^{2} \cdot\left(2 R-\frac{4}{3} R\right)=\frac{32}{81} \pi R^{3} \text {, }
$$

Also,
$$
V_{\text {sphere}}=\frac{4}{3} \pi R^{3},
$$
$\therefore \quad\left(V_{\text {cone}}\right)_{\max }: V_{\text {sphere}}=\frac{32}{81}: \frac{4}{3}=8: 27$