Question 1102126
<br>
You didn't specify whether the 12m is the radius or the diameter of the top of the tank.  By the way you state the problem, I assume it is the full width (diameter) of the top of the tank.<br>
Then the radius of the top of the tank is 6m; since the depth of the tank is 24m, the radius is 1/4 of the depth.  As the tank is being filled, the volume of water forms a cone that is similar to the whole tank; so at any time the radius of the surface of the water in the tank is 1/4 the depth of the water.<br>
So the volume of water when the depth is h is
{{{V = (1/3)(pi)(r^2)h = (1/3)(pi)((h/4)^2)h = (1/48)(pi)h^3}}}<br>
{{{dV/dt = (1/16)(pi)h^2(dh/dt)}}}<br>
{{{dh/dt = (16/((pi)h^2))*(dV/dt)}}}<br>
We know dV/dt is 4; when the depth h is 4, dh/dt is
{{{((16/((pi)4^2))*4) = 4/pi}}}