A right circular cone has height $9 \mathrm{~cm}$ and radius of base $5 \mathrm{~cm}$. It is inverted and…

1. $70 \mathrm{sec}$.
2. $75 \mathrm{sec}$.
3. $72 \mathrm{sec}$.
4. $77 \mathrm{sec}$.

Solution

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, $\frac{\mathrm{h}}{\mathrm{r}}=\frac{9}{5}$ $\begin{aligned} & \therefore \quad \mathrm{r}=\frac{5 \mathrm{~h}}{9} \\ & \therefore \quad \mathrm{A}=\pi \mathrm{r}^2=\pi \frac{25 \mathrm{~h}^2}{81} \end{aligned}$ According to the given condition, $\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{\pi}{\mathrm{A}}=\pi \frac{81}{\pi 25 \mathrm{~h}^2}=\frac{81}{25 \mathrm{~h}^2}$ $\therefore \quad h^2 \mathrm{dh}=\frac{81}{25} \mathrm{dt}$ Integrating on both sides, we get $\begin{aligned} & \frac{\mathrm{h}^3}{3}=\frac{81}{25} \mathrm{t}+\mathrm{c}_1 \\ \therefore \quad & \mathrm{h}^3=\frac{243}{25} \mathrm{t}+\mathrm{c}, \text { where } \mathrm{c}=3 \mathrm{c}_1 \end{aligned}$ Naturally, $\mathrm{h}=0$, when $\mathrm{t}=0$ and hence, $\mathrm{c}=0$ $\begin{aligned} \therefore \quad \mathrm{h}^3 & =\frac{243}{25} \mathrm{t} \\ \therefore \quad \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\ & =\frac{1}{3} \pi \frac{25 \mathrm{~h}^2}{81} \mathrm{~h} \\ & =\frac{25}{243} \pi \mathrm{h}^3 \\ & =\frac{25}{243} \pi \frac{243}{25} \mathrm{t} \\ \therefore \quad \mathrm{V} & =\pi \mathrm{t} \end{aligned}$ But volume of vessel, $\mathrm{V}=75 \pi$ $\begin{array}{ll} \therefore & \pi \mathrm{t}=75 \pi \\ \therefore & \mathrm{t}=75 \text { seconds. } \end{array}$

Asked in: MHT CET 2023 (13 May Shift 1)

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