A spherical rain drop evaporates at a rate proportional to its surface area. If initially its radius is 3 mm and after 1 second it is reduced to 2 mm , then at any time t its radius is (where \(0 \leq \mathrm{t} \lt 3\) )

\(\begin{aligned}
& \frac{\mathrm{dv}}{\mathrm{dt}} \propto-\mathrm{s} \\
\therefore \quad & \frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{ks}, ...(i)\\
& \text { where } \mathrm{k}\gt0 \\
& \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \text { and } \mathrm{s}=4 \pi \mathrm{r}^2 \\
& \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}}
\end{aligned}\)
Equation (i) becomes
\(\begin{aligned}
& 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}\left(4 \pi \mathrm{r}^2\right) \\
& \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}
\end{aligned}\)
Integrating on both sides, we get
\(\begin{array}{ll}
& r=-k t+c...(ii) \\
& \text { When } t=0, r=3 \\
\therefore \quad & 3=-k(0)+c \Rightarrow c=3 \\
\therefore \quad & r=-k t+3 ...[From(ii)]\\
& \text { When } t=1, r=2 \\
\therefore \quad & 2=-k(1)+3 \Rightarrow k=1 \\
\therefore \quad & r=-t+3 \\
& \Rightarrow r=3-t
\end{array}\)