Question

The smaller cone is replaced with another cone of equal radii, but a height twice as big.

Answer 1

Let's make this simple. Let's have the small cone have a radius 1 and the height 1. This would make the bigger cone have a radius of 1 and the height of 2.

With this information, lets get the volume of both cones. The formula is this:

$V = \dfrac{(\pi r^2 h)}{3}$

Plug in numbers:
Small cone: $V = \dfrac{(\pi 1^2 \times 1)}{3}$
Big cone: $V = \dfrac{(\pi 1^2 \times 2)}{3}$

The small cone has a volume of $\dfrac{\pi}{3}$
The big cone has a volume of $\dfrac{2 \pi}{3}$

Now, you want to find how many small cones you need to have the same total volume of the big cone.

$\dfrac{2\pi}{3} - \dfrac{\pi}{3} = \dfrac{\pi}{3}$

You have the difference of pi over 3 comparing the big cone to the small one. You realize that the small cone has the same volume of that. Therefore, you need 2 small cones to have the same total volume as the larger cone