11.181. The total surface area of the cone is $\pi S$ sq. units. The lateral surface of the cone, when unfolded onto a plane, forms a sector with an angle of $60^{\circ}$. Determine the volume of the cone.
Solution.

Let $R$ be the radius of the base of the cone, $l$ its slant height, and $H$ its height. The lateral surface of the cone unfolds into a sector of a circle with radius $l$ and a central angle of $60^{\circ}$, the length of the arc of which is $2 \pi R$. Then $2 \pi R = \frac{1}{6} \cdot 2 \pi l ; l = 6 R . H = \sqrt{l^{2} - R^{2}} = R \sqrt{35}$. The total surface area of the cone $\pi S = \pi R l + \pi R^{2}$. Therefore, $S = R \cdot 6 R + R^{2} = 7 R^{2}, R = \sqrt{\frac{S}{7}}$, $H = \sqrt{\frac{S}{7}} \cdot \sqrt{35} = \sqrt{5 S}$. The volume of the cone $V = \frac{1}{3} \pi R^{2} H = \frac{1}{3} \pi \frac{S}{7} \sqrt{5 S} = \frac{\pi S \sqrt{5 S}}{21}$. Answer: $\frac{\pi S \sqrt{5 S}}{21}$.