A hollow right circular cone rests on a sphere as shown in the figure. The height of the cone is $4$ metres and the radius of the base is $1$ metre. The volume of the sphere is same as that of the cone. What is the distance between the centre of the sphere and the vertex of the cone?
Since the volume of the sphere is same as the volume of the cone I found that the radius of the sphere is also $1$ metre and hence half of the sphere will be inside the cone i.e. the distance between the centre of the sphere and the vertex of the cone is same as the height of the cone which is given to be $4$ metres. Am I right?
Best Answer
In that case, at height h from the center of the sphere to the vertex of the cone, radius of the circle in the sphere will be -
$$r_s=\sqrt{1^2-h^2}$$ Now, $$\text {Also the radius of the cone at height h from the base, } r_c = \frac{4-h}{4}$$ Let's now assume h = 0.2.
$$\text {We get, } r_s > r_c$$
It should have been the other way round. That tells you that the cone will touch sphere at a circle above the equator.
Now to solve it, wherever cone will touch the sphere, the line from center of the sphere to the point of contact will be perpendicular to the edge of the cone. Let's say distance from vertex of the cone to the point of contact is distance a. Then,
$$\frac{1}{a} = \frac{1}{4}$$ $$\text {So, } a = 4$$ $$\text {Distance from center of the sphere to the vertex of the cone } = \sqrt{1^2+a^2} = \sqrt{17}$$