Water is being poured into the inverted conical vessel at the rate of 1.5 cubic meter per minute. Its depth is always equal to twice its radius. The level of water is rising at the rate of $\frac{3}{8\pi }$meter per minute when its depth is (in meters)

Suppose that $r$ is radius of base of cone and $h$ be the height of the cone. 

Given that the rate of change in volume in water is $1.5 \mathrm{cubic} \mathrm{meter} \mathrm{per} \mathrm{minute}$

Given that depth is always equal to twice its radius

$h=2r$ and $\frac{dV}{dt}=1.5$

The level of raising the rate is $\frac{dh}{dt}=\frac{3}{8\mathrm{\pi }}$

Consider the volume of cone and then differentiate 

$V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Substitute $h=2r$

$\begin{array}{rcl}V& =& \frac{1}{3}\mathrm{\pi }\left(\frac{{\mathrm{h}}^2}{4}\right)\mathrm{h}\\ & =& \frac{1}{12}{\mathrm{\pi h}}^3\end{array}$

Differenitate both sides 

$\begin{array}{rcl}\frac{dV}{dt}& =& \frac{1}{4}{\mathrm{\pi h}}^2\frac{\mathrm{dh}}{\mathrm{dt}}\\ 1.5& =& \frac{1}{4}{\mathrm{\pi h}}^2\left(\frac{3}{8\mathrm{\pi }}\right)\\ {\mathrm{h}}^2& =& 16\\ \mathrm{h}& =& 4 \mathrm{meters}\end{array}$