A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant k is positive). Suppose that r(t) is the radius of the liquid cone at time t. The time after which the cone is empty is

Explanation

Let at time t ,r and h be the radius and height of cone of water respectively. Thus, at time t, surface and area of liquid in contact with air `=pir^(2)`

Now according to question `-(dV)/(dt) propto pir^(2)` Therefore ["-" sign shows that V decreases with time]
Or `(dV)/(dt) = -kpir^(2)`.............(1)
But from the figure, we get [Similar `Delta`s]
`r/h=R/h` or `h=(rH)/R`
or `1/3d/(dt)[r^(2)(rH)/R]=-kr^(2)` (from 1)
or `(dr)/(dt) =-(kR)/H`
or `r=(-kR)/(H)t+C` (interating)
Now, at t=0, r=R
`therefore R=0+C` or `C=R`
`therefore r=(-kRt)/(H)+R`
Now, let the time at which cone is empty be T. Then at T,r=0 (no liquid is left).
`therefore 0=(-kRT)/(H)+R` or `T=H//k`