Correct Answer: 5 : 3

Solution : Given: The radius of the base of a hollow cone = 8 cm
Height = 15 cm
Now consider the problem in two dimensions.
Area of the triangle = $\frac{1}{2}\times$ base $\times$ height.
The inradius of the circle = $\frac{A}{s}$, where $A$ is the area of the triangle and $s$ is the semi–perimeter of the triangle.

Using Pythagoras theorem, $(CB)^2=(BD)^2+(CD)^2$
$⇒(CB)^2=(8)^2+(15)^2$
$⇒(CB)^2=64+225$
$⇒(CB)^2=289$
$⇒(CB)=17$ cm
Now, inradius = $\frac{A}{s}$
$A =\frac{1}{2}\times 15\times 16$
$= 120$ cm$^2$
$⇒s=\frac{17+16+17}{2}$
$⇒s= 25$ cm
Inradius = $\frac{120}{25}=\frac{24}{5}$
The ratio of the radius of the base of a cone to the radius of a sphere is $8:\frac{24}{5}=5:3$
Hence, the correct answer is 5 : 3.