10. (5th "Hope Cup" Invitational Competition Question) The radius of the upper base of a frustum is 5 cm, the radius of the lower base is 10 cm, and the slant height $AB$ is 20 cm (where $B$ is on the circumference of the lower base). A rope is pulled from the midpoint $M$ of the slant height $AB$, wraps around the lateral surface of the frustum to point $B$. When the rope is the shortest, the shortest distance between a point on the rope and a point on the circumference of the upper base of the frustum is ( ).
A. 6 cm
B. 5 cm
C. 4 cm
D. 3 cm
10. C. Reason: The lateral development of the frustum is shown in the figure. $AB=20$ cm, $\widetilde{AC}=10\pi$ cm, $\overparen{BD}=20\pi$ cm. Let the center of the sector be $O$, and let $OA=x$ cm, $\angle AOC=\theta$, then $\overparen{AC}=x\theta, \overparen{BD}=(x+20)\theta$. Solving the system of equations $\left\{\begin{array}{l}x\theta=10\pi, \\ (x+20)\theta=20\pi \text{. }\end{array}\right.$ yields $x=20, \theta=\frac{\pi}{2}$. At this point, the shortest distance from point $O$ to $BM$ is the altitude $ON$ from the hypotenuse $BM$ in the right triangle $BOM$, $ON=\frac{OB \cdot OM}{BM}=$ 24 cm, the shortest distance from any point on $\overparen{AC}$ to $BM$ is $24-20=4$ cm.