Steel Ball Drop

Let $V_c$ be the volume of the cylinder, $V_w$ be the volume of the water and $V_s$ be the volume of the sphere.

Then,

$V_c = V_w+V_s$

$\pi r^2(h+4)=\pi r^2h+\frac{4}{3}\pi r^3$

$\pi r^2(h+4)=\pi r^2(h+\frac{4}{3}r)$ $\implies$ $\pi r^2$ cancels out

$h+4=h+\frac{4}{3}r$ $\implies$ $h$ cancels out

$3=r$

but: $h=2r$ $\implies$ $r=\frac{h}{2}$

therefore,

$3=\frac{h}{2}$

$6=h$

Finally, the height of the cylinder is

$h+4=6+4=$ $\boxed{\large\color{#D61F06}10~cm}$ $\color{#69047E}answer$

The volume of the steel ball equals to the cylindrical volume raised.

Thus, $\dfrac{4}{3}\pi(r^3) =\pi(r^2)h = 4\pi(r^2)$ .

Hence, $\dfrac{r}{3}=1$ . $r=3$ .

Finally, the glass's height = $2\times 3 + 4 =\boxed{10}$ .

Let $r$ be the radius of the cylinder and sphere. Then the initial height of water is $h=d=2r$ . The volume of rise of water is equal to the volume of the sphere. So we have

$\pi r^2(4)=\dfrac{4}{3} \pi r^3$

$r=3$

So the height of the the glass is $h+4=2r+4=2(3)+4=6+4=$ $\boxed{10}$ .