How does one determine when a variable is dependent or independent? So I was going to through a fairly basic problem which I have posted below. 
The first step is to use similarity of triangles and solve it(pic also posted below). We use similarity of triangles for the other cone as well(skipping q few steps)
Lastly, we arrive at an equation in which 'R²' cancels out on both sides and it is written that the volume of water in the cone is independent of 'R'. How does one arrive at that conclusion just because the R² terms cancels out ?


 A: What these calculations revealed is that any cone that is $1+\sqrt{85}$ centimeters tall could be filled with water to meet the conditions of this problem.  You could choose any radius for the base that you liked, and as long as you filled it up with water to $2$ cm from the top, when you sealed it and flipped it over there would be $8$ cm of air at the top.  
You discovered this independence when you started solving an equation with two variables and one of them cancelled out like the $R$ did in this equation.  The amount of water that goes into the cone was dependent on the radius of the base of the cone, but it was not necessary to consider it when determining the height of the cone.
A: 
The volume of a cone:
$$V_{cone}=\frac13S_{base}h=\frac13\cdot \pi R^2\cdot h.$$

I assume you understand how the RHS formula is derived:
$\hspace{2cm}$
$$\begin{align}Left:& \begin{cases}V_{water}=V_{cone}-V_{empty \ \ cone}=\frac{\pi R^2}{3}\cdot h-\frac{\pi r_1^2}{3}\cdot 8=\color{red}{\frac13\cdot \pi R^2}\cdot \color{blue}{(h-\frac{512}{h^2})}\\ \frac{r_1}{8}=\frac{R}{h} \Rightarrow r_1=\frac{8R}{h}\end{cases}\\
Right:& \begin{cases}V_{water}=\frac{\pi r_2^2}{3}\cdot (h-2)=\color{red}{\frac13 \cdot \pi R^2}\cdot \color{blue}{\frac{(h-2)^3}{h^2}}\\ \frac{r_2}{h-2}=\frac{R}{h} \Rightarrow r_2=\frac{R(h-2)}{h}\end{cases} \end{align}$$
To fit the water: For the left cone, decrease height $h$ to $\color{blue}{h-\frac{512}{h^2}}$, while holding the base area ($\color{red}{\pi R^2}$) unchanged. For the right cone, decrease height $h$ to $\color{blue}{\frac{(h-2)^3}{h^2}}$, while holding the base area ($\color{red}{\pi R^2}$) unchanged. Since the water volumes are equal, the heights of the resulting cones must be equal. Hence, only the heights are effected, while the radius $R$ is independent.
