Confusion regarding volume of fluid displaced by a partially immersed body Say I have a cylindrical apparatus partially filled with a water column, the height of the column being $h$. Now I have a solid cylinder of radius smaller than the apparatus and height $H>h$ but less than height of apparatus.

If I immerse the red cylinder in the water, it will displace some water. But if I cut the cylinder in pieces and stack them on top of each other perfectly in the water, the displaced water will be the same. Keeping that idea in mind,
I cut the cylinder to height $h$ and immersed it in water.

Some water got displaced(shown in lighter blue) which is equal to the volume of immersed cylinder.
Now I put the piece of cylinder of height equal to the rise in water level, and do it again.
 
Now, if I put the remaining piece of cylinder in,

So the final water level turned out to be higher than the cylinder height? But this does not make sense as this means that the displacement is going to be higher than the cylinder height irrespective of the radius of the cylinder? But how is that possible? Can someone please explain the flaw in my logic?
 A: Short answer: the total height of all the small cylinders stacked together will always be less than $H$, and will never even get close to $H$.
Long answer: The volume of a cylinder is is its height times its cross sectional area. Let $A$ be the cross sectional area of the cylindrical apparatus containing water to height $h=h$ and let $a$ be the cross sectional area of the solid red cylinder. Necessarily, $a<A$. If there is not enough water to rise above the cylinder when it is immersed, then the actual volume of water is less than the volume of the apparatus up to height $H$ minus the volume of the cylinder: $$(A-a)H>Ah \Leftrightarrow H>h\frac{A}{A-a}.$$
If the red cylinder is cut to height $h$ and immersed, the volume of displaced water is $ah$. The displaced water raises the water level by the amount $(a/A)h$. Therefore, the next small cylinder is cut to height $(a/A)h$. This cylinder has volume $(a^2/A)h$ and immersing it raises the water level by a height of $(a/A)^2h$, which is the height of the third cylinder. We can continue calculating in this pattern to conclude that placing the $n$th cylinder must have height $(a/A)^{n-1}h$. After  stacking the first $N$ cylinders, we calculate their total height as
$$H_N=\sum_{n=1}^N (a/A)^{n-1} h=h\frac{1-(a/A)^N}{1-a/A}.$$
Multiply top and bottom by $A$ to get
$$H_N=h\frac{A-A(a/A)^N}{A-a}<h\frac{A}{A-a}<H.$$
Therefore, $H_N$ never reaches, or even approaches $H$. So no matter how many fully immersed cylinders you stack on top of each other, you will never reach the height of the original cylinder.
A: If $R$ is the radius of the container holding water and $r$ is the radius of the cylinder being dipped into water, equating volumes of water before you dip the cylinder into the water and after,
$ \pi R^2 h = \pi R^2 h_{\text{new}} - \pi r^2 h_{\text{new}}~$ (assuming $H \gt h_{\text{new}})$
$ \displaystyle h_{\text {new}} = \frac{R^2 h}{R^2 - r^2}$
Now in your second approach, you are starting with a cylinder of height $h$ and after you dip the cylinder into water fully,
$ \displaystyle \pi R^2 h = \pi R^2 h_1 - \pi r^2 h \implies h_1 = \left(1 +  \frac{r^2}{R^2}\right)h$
Next you add on top a cylinder of height $h_1$ and again equating volumes before and after, we get
$ \displaystyle h_2 = \left(1 + \frac{r^4}{R^4} + \frac{r^2}{R^2}\right)h$
Note that the level of water rise after dipping each cylinder is given by, $ \displaystyle h_1 - h = \frac{r^2h}{R^2}, h_2 - h_1 = \frac{r^4h}{R^4}$
So we can see that with every subsequent cylinder, the level of rise in water forms an infinite geometric sequence,
$ \displaystyle h_{\text{new}} = h \left ( 1 + \frac{r^2}{R^2} + \frac{r^4}{R^4} + ....\right) = \frac{R^2 h}{R^2 - r^2}$
This is same as the first answer when we dipped the cylinder without breaking it into pieces.
