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 $0 \leq \mathrm{t} \lt 3$ )

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Aicharya · Accepted Answer

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

A spherical rain drop evaporates at a rate proportional to its surface area. If initially its radius is 3 mm…

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