91Ó°ÊÓ

Since all the spheres have the same mass, the center of mass of the system must be equally distant from each sphere's center. Place the spheres in an equilateral triangle configuration with the centers being vertices. Each side of the triangle has a length of \(20\, \mathrm{cm}\).

Let's first determine the center of mass for only two spheres. The two spheres' centers will be at points A and B, being \(20\, \mathrm{cm}\) apart. The center of mass of two identical spheres is the midpoint of the segment joining their centers. Let's call the midpoint M. We find that M is 10 cm away from each sphere's center, and the distance between the spheres will be \(10+10=20\, \mathrm{cm}\)

Now, let's add the third sphere, with its center at point C. We need to find the center of mass of the system such that it is also equidistant from C. At this point, we can consider the masses of spheres A, B, and C to be located at their centers. Remember that the center of mass M of spheres A and B was 10 cm from each sphere. Draw a circle with a 10 cm radius around point C. M must be on that circle. Since the system is symmetric, M will also be on the bisector of angle \(\angle ACB\). Let's call the center of mass of the entire system G, which is the intersection of the circle and the bisector of the angle.

First, note that the triangle formed by A, B, and C is equilateral. Therefore, \(\angle ACB = 60°\). Knowing that the bisector of the angle intersects side AB at a 90° angle at point D, and the distance AD will be half of the side length, which is 10 cm. Now, we know that G is the intersection of the circle centered at point C with a 10 cm radius and the bisector of angle ACB. We can find the altitude CG in triangle AGC using the Pythagorean theorem in the right triangle ADC: \[AC^2 = AD^2 + DC^2 \Rightarrow DC^2 = AC^2 - AD^2.\] We know AC is the sum of the radii, so: \[DC^2 = (20\, \mathrm{cm})^2 - (10\, \mathrm{cm})^2 = 300\, \mathrm{cm}^2.\] Now, using the right triangle AGC: \[GC^2 = AC^2 - AG^2 \Rightarrow GC = \sqrt{AC^2 - AG^2}.\] We know the center of mass G is equidistant from all three sphere centers, so \(AG = BG = CG= x\). We can substitute \(AC = x\): \[GC = \sqrt{x^2 - AG^2} = \sqrt{x^2 - x^2 + 300\, \mathrm{cm}^2} = \sqrt{300\, \mathrm{cm}^2}.\] The distance between the center of mass and the horizontal surface (point G and the plane) is the altitude \(\sqrt{300\, \mathrm{cm}^2} = 10\sqrt{3}\, \mathrm{cm}\). So, the correct answer is: (E) None of these