the rate at which the sand is leaking, is the same rate at which the volume of the cone is increasing, so the rate of the sand leak is then the same as dv/dt of the cone.
now, we know the height is always the same length as the radius, thus h = r.
[tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\qquad \boxed{h=r}\qquad V=\cfrac{\pi h^2 h}{3}\implies V=\cfrac{\pi h^3}{3}
\\\\\\
\cfrac{dV}{dt}=\cfrac{\pi }{3}\cdot 3h^2\cdot \cfrac{dh}{dt}\implies \cfrac{dV}{dt}=\pi h^2\cfrac{dh}{dt}\qquad
\begin{cases}
h=10\\
\frac{dh}{dt}=6
\end{cases}
\\\\\\
\cfrac{dV}{dt}=\pi \cdot 10^2\cdot 6\implies \cfrac{dV}{dt}=600\pi ~\frac{in^3}{min}[/tex]
