Water flowing from a sphere

Invert the universe so that the water is now flowing out through top. Now, at a certain instant, the surface of water(blue) is at distance $x$ from the center.

At this instant, mass of water body is given by:

$mw=\int _{ x }^{ 1 }{ \rho \pi (1-{ t }^{ 2 })dt }$

where $\rho$ is the density of liquid.

therefore, center of mass of water body is given by

$x_{1}=\frac { \int _{ x }^{ 1 }{ \rho \pi t(1-{ t }^{ 2 })dt } }{ mw}$

this means that center of mass of whole system is:

$K=\frac { \int _{ x }^{ 1 }{ \rho \pi t(1-{ t }^{ 2 })dt } }{ \int _{ x }^{ 1 }{ \rho \pi (1-{ t }^{ 2 })dt } +m }$

where $m$ is mass of container.

now, put $f\left( t \right) =\rho \pi (1-{ t }^{ 2 })$ , then the equation reduces to:

$K=\frac { \int _{ x }^{ 1 }{ tf\left( t \right) dt } }{ \int _{ x }^{ 1 }{ f\left( t \right)dt } +m }$

now putting the condition $\frac { dK }{ dx } =0$ , we get that:

$x=\frac { \int _{ x }^{ 1 }{ tf\left( t \right) dt } }{ \int _{ x }^{ 1 }{ f\left( t \right)dt } +m }$

$\Rightarrow x=K$ (when $K$ is maximum)

Now, integrating and cross multiplying, we get:

$x^4-6x^2+24x-3=0$

solving this equation yields the required value of $x$ , which is $\boxed{{12}^{\frac{1}{4}}-{9}^{\frac{1}{4}} metre}$

thus, answer is: $12+9+4=\boxed{25}$