A hollow right circular cone has a base radius of 8 cm and a height of 15 cm. A solid sphere of the largest possible radius is placed inside the cone so that it touches the base and the slant surface. What is the ratio of the radius of the base of the cone to the radius of the sphere?

Introduction / Context:
This problem involves a sphere inscribed in a right circular cone. The largest sphere that can fit inside a hollow cone touches the base and the lateral surface, and its cross section along the axis matches the incircle of a corresponding isosceles triangle. The question is asking for the ratio between the cone base radius and the radius of this inscribed sphere, which is a classic use of triangle inradius formulas in three dimensional geometry.

Given Data / Assumptions:

Base radius of the cone R = 8 cm.
Height of the cone h = 15 cm.
The cone is right circular, so its axis is perpendicular to the base.
The sphere is the largest that fits inside, so it is tangent to the base and the lateral surface.
We work with an axial cross section that forms an isosceles triangle with base 16 cm and height 15 cm.

Concept / Approach:
When you slice the cone through its axis, you obtain an isosceles triangle whose base is the diameter of the cone and whose height is the cone height. The sphere becomes a circle in this cross section, tangent to the base and the equal sides. This circle is the incircle of the triangle. Therefore, the radius of the sphere is equal to the inradius r of the triangle. The inradius of a triangle is given by r = A / s, where A is the area of the triangle and s is its semiperimeter. Once r is found, we can form the ratio R : r.

Step-by-Step Solution:
In the cross section, the triangle has base b = 2R = 16 cm and height h = 15 cm. The equal sides form a right triangle with legs 8 cm and 15 cm, so each slant side length is sqrt(8^2 + 15^2) = sqrt(64 + 225) = sqrt(289) = 17 cm. Therefore, the triangle sides are 16 cm, 17 cm and 17 cm. Area A = 1/2 * base * height = 1/2 * 16 * 15 = 120 square centimetres. Semiperimeter s = (16 + 17 + 17) / 2 = 50 / 2 = 25 cm. Inradius r = A / s = 120 / 25 = 4.8 cm. The cone base radius R is 8 cm. The ratio R : r = 8 : 4.8. Divide both terms by 1.6 to simplify: 8 / 1.6 = 5, 4.8 / 1.6 = 3, so R : r = 5 : 3.

Verification / Alternative check:
We can check the result by using another scale factor. The ratio 8 : 4.8 simplifies to 80 : 48, and dividing by 16 gives 5 : 3 again. Also, the triangle sides 16, 17, 17 form a valid scalene isosceles triangle, and the inradius computed from A = 120 and s = 25 is consistent. The numeric values are neat, which matches the style of the multiple choice options provided in aptitude exams.

Why Other Options Are Wrong:
4 : 1 would require the sphere radius to be 2 cm, which does not satisfy the inradius formula for this triangle.
2 : 1 would mean the sphere radius is 4 cm, which is too large and would not fit while remaining tangent to both the base and slant sides.
7 : 3 does not match the exact ratio calculated using the inradius relation and arises from no correct geometric argument in this case.
3 : 2 suggests a sphere radius larger than half the cone radius, which is impossible for the given cone dimensions because the inscribed sphere must sit closer to the vertex.

Common Pitfalls:
A frequent mistake is to treat the problem as if a sphere is inscribed in a cylinder instead of a cone, leading to a trivial radius equal to the base radius. Another error is to use the height of the cone directly as the radius of the sphere, ignoring the need to satisfy tangency conditions to both sides of the cone and the base. Some learners also confuse the inradius with the circumradius of a triangle and apply the wrong formula, which produces incorrect values.

Final Answer:
The ratio of the radius of the base of the cone to the radius of the largest inscribed sphere is 5 : 3.