Solution.

From $\triangle S O A: A O=R=\sqrt{25-16}=3$ cm (Fig. 11.74). The volume of the desired hemisphere is $V=\frac{2}{3} \pi r^{3}$. From $\triangle S B O: r^{2}+(5-x)^{2}=16$. From $\triangle A B O: x^{2}+r^{2}=9$. Solving the system $\left\{\begin{array}{l}r^{2}+(5-x)^{2}=16, \\ r^{2}+x^{2}=9,\end{array}\right.$ we get $x=\frac{9}{5}$. Then $r=\sqrt{9-x^{2}}=$ $=\frac{12}{5}$ (cm). Finally, we obtain $V=\frac{2}{3} \pi \frac{12^{3}}{5^{3}}=\frac{1152}{125} \pi$ (cm $\left.{ }^{3}\right)$.

Answer: $\frac{1152}{125} \pi \mathrm{cm}^{3}$.
11.082. The height of the cone and its slant height are 4 and 5 cm, respectively. Find the volume of a hemisphere inscribed in the cone, with its base lying on the base of the cone.