An inverted rt circular cone is full of water. What is the radius of a ball that will displace the most water?

Question

Give your answer in terms of the height of the cone, h, and the angle of the vertex, a.

Answer 1

I am not very sure as I had done such questions in around 1979-81. Displaced water (if fall is not restrained) = density*g* volume of ball

it will be most when volume of ball is most or radius of ball is maximum

If you allow different radii balls to go in cone, all will get trapped at some point along the depth depending on the lower frustum of sphere (below dia line) fitting in the circular loop of cone.

If the ball traps, then hydrostaic pressure will not be genuine because walls will also neutralise its weight. Thus, falling object will be RESTRAINED.

As the cone is given, both height (h) and radius [R=h tan (a/2)] are numerically fixed.

Thus, let us get rid of cone itself and prepare equivalent cylinder
volume of cone = (1/3) Pi * R^2 (h)
= Pi * {R/sqrt (3)}^2 (h)

= volume of cylinder of same height but [R/sqrt (3)] radius

So this equivalent cylinder will have
height = h and radius = [h * tan(a/2)] / {sqrt (3)]

Now this cylinder will always diaplace most water when a ball of radius = its cross-sectional radius

so Answer is [h * tan(a/2)] / {sqrt (3)] = 0.577 *[h * tan(a/2)]

Answer 2

Let us take any general inverted right circular cone, but one which is symmetrical from all sides, & geometrically observe what its sides & angles would be. A diagram can’t be displayed here, otherwise I would have drawn it, but with a little application of geometrical manipulation, you can see that the vertex of a RIGHT circular cone is always 60 degrees. And moreover, a right circular cone, as far as I remember, is defined as a cone whose base angle is 60 degrees. So, the vertex of a right circular cone is 60 degrees anyway. Now, for a ball to displace most water, its volume should be the maximum that can be incorporated inside the cone, and this will happen only when the radius of the ball is the maximum possible one. Now, let us apply some geometry to the figure and draw a suitable ball inside the cone & join its centre with the centre of the base of the cone & also with one of the base vertices. By application of simple geometry & trigonometry, we can find that the length of the segment whose end points are (1) the point where the radius of the spherical ball intersects a side of the cone, and (2) the vertex of the cone considered in the previous line of this discussion, is h/sqrt3. By the application of simple trigonometry, we can now find that the radius of the spherical ball is h/3.

[Note: Even if our cone is not a right circular cone, & instead is any general but symmetric cone, the radius, from simple geometry & trigonometry, R = (h/sqrt3)*tan(a/2). In this value of R, if we put a=60 degrees, we get, R= (h/sqrt3)*tan30 =(h/sqrt3)*(1/sqrt3) =h/3. In fact, this is what I had done in the statements where I referred to the use of “simple geometry & trigonometry”.]

Answer 3

The problem does not state specifically whether the ball is contained completely within the cone, but I will assume so. If this assumption is incorrect, please see the discussion linked to below.

Although I cannot claim an exhaustive solution to this problem, I reduced it to solving "What is the radius of a circle with maximal area inscribed within an isosceles triangle with height h and vertex angle a?"

In other words, I reduced your question to a 2-dimensional problem, with the 'isosceles triangle' taken as the cross-section of the given right circular cone.

Step 1. Draw an isosceles triangle and mark vertex angle 'a' and height 'h'.

Step 2. Inscribe a circle that is tangent to the base of the triangle as well as both its sides. Note that a radius drawn to each point of tangency automatically forms a right angle there.

Step 3. Denote one-half of the triangle's base as 'x' and conclude that this length is equal to (h)(tan(a/2)).

Step 4. Draw a line segment from the center of the inscribed circle, to one of the triangle's base angles. I assert that this line segment will bisect the base angle, giving each 'half' of this angle the measure (180-a)/4. Likewise, I assert that we have thus formed two congruent smaller right triangles.

Step 5. Note that the following holds within each smaller right triangle: tan ((180-a)/4) = r/x

Step 6. Substitute x = (h) tan (a/2) into the equation from Step 5.

Step 7. Solve for r: r = h tan (a/2) tan [(180-a)/4]

I hope this is correct and helps in your analysis!

For a similar discussion wherein the sphere is permitted to extend outside the cone, see the link below: