A spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3…
A spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3 \mathrm{~mm}$. and 1 hour later has been reduced to $2 \mathrm{~mm}$, then the expression of radius $\mathrm{r}$ of the raindrop at any time $t$ is (where $0 \leq t < 3$ )
$\mathrm{r}=\mathrm{t}+5$
$r=t-5$
$\mathrm{r}=3-\mathrm{t}$
$r=t+3$
Solution
We have $\frac{\mathrm{dv}}{\mathrm{dt}} \propto-\left(4 \pi \mathrm{r}^2\right)$
We know that $\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^2 \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}$
$\begin{aligned}
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \propto-\left(4 \pi \mathrm{r}^2\right) \\
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}=\left(-4 \mathrm{k} \pi \mathrm{r}^2\right) \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}
\end{aligned}$
