MHT CET 2023 13th May Morning Shift Mathematics - Differential Equations Question 216 English Explanation

For the conical vessel, $$\mathrm{h}=9 \mathrm{~cm}, \mathrm{r}=5 \mathrm{~cm}$$

$$\therefore \quad$$ Full volume of the vessel,

$$\begin{aligned} \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\\\ & =\frac{1}{3} \pi \times 25 \times 9 \\\\ & =75 \pi \mathrm{~cm}^3 \end{aligned}$$

Now, $$\frac{\mathrm{h}}{\mathrm{r}}=\frac{9}{5}$$

$$\begin{aligned} & \therefore \quad r=\frac{5 h}{9} \\ & \therefore \quad A=\pi r^2=\pi \frac{25 h^2}{81} \end{aligned}$$

According to the given condition,

$$\begin{aligned} \frac{\mathrm{dh}}{\mathrm{dt}} & =\frac{\pi}{\mathrm{A}}=\pi \frac{81}{\pi 25 \mathrm{~h}^2}=\frac{81}{25 \mathrm{~h}^2} \\\\ \therefore \quad \mathrm{h}^2 \mathrm{dh} & =\frac{81}{25} \mathrm{dt} \end{aligned}$$

Integrating on both sides, we get

$$\begin{aligned} & \frac{h^3}{3}=\frac{81}{25} t+c_1 \\\\ \therefore \quad h^3 & =\frac{243}{25} t+c, \text { where } c=3 c_1 \end{aligned}$$

Naturally, $$\mathrm{h}=0$$, when $$\mathrm{t}=0$$ and hence, $$\mathrm{c}=0$$

$$\begin{aligned} \therefore \quad \mathrm{h}^3 & =\frac{243}{25} \mathrm{t} \quad \text{.... (i)}\\\\ \therefore \quad \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\\\ & =\frac{1}{3} \pi \frac{25 \mathrm{~h}^2}{81} \mathrm{~h} \\\\ & =\frac{25}{243} \pi \mathrm{h}^3 \\\\ & =\frac{25}{243} \pi \frac{243}{25} \mathrm{t} \quad \text{...[From (i)]}\\\\ \therefore \quad \mathrm{V} & =\pi \mathrm{t} \end{aligned}$$

But volume of vessel, $$\mathrm{V}=75 \pi$$

$$\begin{array}{ll} \therefore & \pi \mathrm{t}=75 \pi \\\\ \therefore & \mathrm{t}=75 \text { seconds. } \end{array}$$

Differential Equations

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