Hello
My question is about the new water level when an object is partially submerged. As the water level raises, it swallows a different amount of volume of the ball and so changes at different rates, from which I am assuming calculus is involved. Aside from that I am unsure of how to begin answering this question, I hope someone can help me.
Thanks.
EDIT: I've thought about it for a while and I have an idea.
The height right now is 1, so if we consider half the sphere (which is what would be submerged if the water didn't rise) we see that we are essentially adding half of $$4/3*pi*1^3$$
This raises the water level by a certain known amount, which now submerges a little more of the sphere. This raises the water level by a very small amount, submerging a tiny fraction of the sphere. As the volume being submerged is getting smaller maybe I can try taking some kind of limit.
Suppose the new water level is at $ h $, then, from the balance of volumes
$ 25 + V_1 = 25 h $
Where $V_1$ is the volume of the sphere between levels of $0$ an $h$,
$\begin{equation} \begin{split} V_1 &= \dfrac{r^3}{3} (2 \pi) (1 - \dfrac{r - h}{r} ) - \dfrac{\pi}{3}(r - h) (r^2 - (r - h)^2 ) \\ &= \dfrac{\pi}{3} ( 2 r^2 h - (r - h) (2 r h - h^2) )\\ &= \dfrac{\pi}{3}(2 r^2 h - 2 r^2 h + r h^2 + 2 r h^2 - h^3 ) \\ &= \dfrac{\pi}{3}(3 r h^2 - h^3 ) \\ &= \dfrac{\pi}{3} h^2 ( 3 r - h) \\ \end{split} \end{equation} $
Therefore, since $r = 1$, the equation we want to solve is
$ \dfrac{\pi}{3} h^2 (3 - h ) = 25 (h - 1) $
The solution of which is $ h = 1.095774 $