A right circular cone's surface area is $10 \mathrm{dm}^{2}$ larger than the surface area of the sphere inscribed in the cone. What is the difference in volume between the cone and the sphere?
Let $R$ denote the radius of the base circle of the cone, and $r$ the radius of the inscribed sphere. The relationship between the two radii is: $r=\frac{R}{\sqrt{3}}$. The height of the cone is $m=R \sqrt{3}$.

# 1993-04-171-1.eps 

The surface area of the equilateral cone is $A_{k}=3 \pi R^{2}$, and the surface area of the inscribed sphere is $A_{g}=4 \pi r^{2}=\frac{4 \pi}{3} R^{2}$. The difference in surface areas is $\frac{5 \pi}{3} R^{2}$, which we know is $10 \mathrm{dm}^{2}$. From this, it follows that $R=\sqrt{\frac{6}{\pi}} \mathrm{dm}$. The difference in volumes is:

$$
\begin{aligned}
V_{k}-V_{g} & =\frac{\pi \sqrt{3}}{3} R^{3}-\frac{4 \pi}{9 \sqrt{3}} R^{3}=\frac{5 \pi \sqrt{3}}{27} R^{3}= \\
& =\frac{10}{3} \sqrt{\frac{2}{\pi}} \mathrm{dm}^{3} \approx 2.6596 \mathrm{dm}^{3}
\end{aligned}
$$