A spherical rain drop evaporates at a rate proportional to its surface area at any instant $t$. The

Question

A spherical rain drop evaporates at a rate proportional to its surface area at any instant $t$. The rate of change of the radius of the rain drop is

Professor PC Thomas Classes · Answer

Let $r$ be the radius of the spherical rain drop since spherical rain drop evaporates at a rate proportional to its surface area.\ i.e., $-\frac{d V}{d t} \propto s$\ (- sign represents when $t$ increases, $V$ decreases)\ where $s=$ surface area $=4 \pi r^{2}$\ and $V=$ volume $=\frac{4}{3} \pi r^{3}$\ $\Rightarrow-\frac{12}{3} \pi r^{2} \frac{d r}{d t}=K 	imes 4 \pi r^{2}$\ where $K$ is constant of proportionality.\ $\Rightarrow \frac{d r}{d t}=\frac{-K}{1}=-K=$ constant\