A right circular cone with radius R=4 and height H=8 contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Then the time after which the cone is empty.
 

Let the semi vertical angle of the cone be  

$\mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{R}}{\mathrm{H}}\right)$
Let height of the liquid at time ‘t’ be ‘h’ from the base BC and radius r.
Volume of liquid at time ‘t’ = V =$=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$
$=\frac{1}{3}{\mathrm{\pi r}}^3\cot \mathrm{\theta }$
S = Surface area in contact with air at time 
${ }^{\text{'}}{\mathrm{t}}^{\text{'}}={\mathrm{\pi r}}^2$
Given that $-\frac{\mathrm{dV}}{\mathrm{dt}}\propto \mathrm{S}$
$\begin{array}{l}\Rightarrow -\frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{kS}={\mathrm{k\pi r}}^2\\ \Rightarrow \frac{\cot \mathrm{\theta }}{3}\mathrm{\pi }3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}=-{\mathrm{k\pi r}}^2\\ \Rightarrow \cot \mathrm{\theta }{\int }_{\mathrm{R}}^0 \mathrm{dr}=-\mathrm{k}{\int }_0^{\mathrm{\top }} \mathrm{dt}\text{ (where T is the required time) }\\ \Rightarrow \frac{\mathrm{H}}{\mathrm{R}}\mathrm{R}=\mathrm{kT}\Rightarrow \mathrm{T}=\frac{\mathrm{H}}{\mathrm{k}}=2\end{array}$