$\text{ An inverted cone is }10 \mathrm{cm}\text{ in diameter and }10 \mathrm{cm}\text{ deep. Water }$$\text{ is poured into it at the rate of }4{ \mathrm{cm}}^3/\min \text{ . When the depth of }$$\text{ water level is }6 \mathrm{cm}\text{ , then it is rising at the rate }$

$\begin{array}{c}\text{Let y be the level of water at time t, }\\ \text{x the be radius of the surface and }\\ \text{V be the volume of water.}\\ \text{We know that the volume of the cone is }\\ =\frac{1}{3}\mathrm{\pi }{\left(\mathrm{radius}\right)}^2\times \text{ height }\\ \therefore \text{V}=\frac{1}{3}\pi {\text{x}}^2\text{y}.\text{ Let }\angle \text{BAD}=\alpha \\ \therefore \text{V}=\frac{1}{3}\pi {\text{x}}^2\text{y}.\text{ Let }\angle \text{BAD}=\alpha \\ \Rightarrow \tan \alpha =\frac{BD}{AD}=\frac{5}{10}=\frac{1}{2}.\\ \tan \alpha =\frac{\text{MR}}{\text{AR}}=\frac{\text{x}}{\text{y}};\Rightarrow \frac{1}{2}=\frac{\text{x}}{\text{y}};\therefore \text{x}=\frac{\text{y}}{2}\\ \therefore \text{V}=\frac{1}{3}\pi {\text{x}}^2\text{y}=\frac{1}{3}\pi {\left(\frac{\text{y}}{2}\right)}^2\cdot \text{y}=\frac{\pi }{12}{\text{y}}^3\\ \frac{\text{dV}}{\text{dt}}=4\text{cub}.\text{cm}/\text{min}\\ \frac{\text{dV}}{\text{dt}}=\frac{\pi }{12}\cdot 3{\text{y}}^2\cdot \frac{\text{dy}}{\text{dt}};\therefore 4=\frac{\pi }{4}{\text{y}}^2\cdot \frac{\text{d}y}{\text{dt}};\therefore \frac{\text{dy}}{\text{dt}}=\frac{16}{\pi {\text{y}}^2}\\ \text{ When y}=6 \text{cm},\frac{\text{dy}}{\text{dt}}=\frac{16}{\pi 6^2}=\frac{4}{9\pi }\text{cub}.\text{cm}/\text{min}\end{array}$