MHT CET · Maths · Application of Derivatives
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 \(0 \leq t < 3\) )
- A \(\mathrm{r}=\mathrm{t}+5\)
- B \(r=t-5\)
- C \(\mathrm{r}=3-\mathrm{t}\)
- D \(r=t+3\)
Answer & Solution
Correct Answer
(C) \(\mathrm{r}=3-\mathrm{t}\)
Step-by-step Solution
Detailed explanation
We have \(\frac{\mathrm{dv}}{\mathrm{dt}} \propto-\left(4 \pi \mathrm{r}^2\right)\)
We know that \(\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^2 \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}\)
\(
\begin{aligned}
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \propto-\left(4 \pi \mathrm{r}^2\right) \\
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}=\left(-4 \mathrm{k} \pi \mathrm{r}^2\right) \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}
\end{aligned}
\)
We know that \(\mathrm{v}=\frac{4}{3} \pi \mathrm{r}^2 \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}\)
\(
\begin{aligned}
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \propto-\left(4 \pi \mathrm{r}^2\right) \\
& \therefore 4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}=\left(-4 \mathrm{k} \pi \mathrm{r}^2\right) \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=-\mathrm{k}
\end{aligned}
\)
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