Question

A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius be $3mm$ and $1hour$ later it reduces to $2mm$, find an expression for the radius of the rain drop at any time $t$.

Answer 1

Let $V$ and $S$ be volume and surface area of the rain drop whose radius at time $t$ is $r$

Given

$\frac{dV}{dt}\propto -S$

$\frac{dV}{dt}=-kS........\left(1\right)$

where, k is the proportionality constant

Now,

$V=\frac{4}{3}\pi r^2$ and $S=4\pi r^2$

From (1), we have

$\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-k\left(4\pi r^2\right)$

$\therefore \frac{4}{3}r\left(3r^2\frac{dr}{dt}\right)=-k\left(4\pi r^2\right)$

$\Rightarrow 4\pi r^2\frac{dr}{dt}=-k\left(4\pi r^2\right)$

$\Rightarrow \frac{dr}{dt}=-k$

Thus, the expression for the radius of the rain at any instant of time is

$\frac{dr}{dt}=-k$