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Water is being poured at the rate of $36 \mathrm{~m}^3 / \mathrm{min}$. into a cylindrical vessel, whose circular base is of radius $3 \mathrm{~m}$. then the water level in the cylinder is rising at the rate of
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Verified Answer
The correct answer is:
$\frac{4}{\pi} \mathrm{m} / \mathrm{min}$
Volume of cylinder $=\pi \mathrm{r}^2 \mathrm{~h}$
$$
\begin{aligned}
& \therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{dh}}{\mathrm{dt}} \\
& \therefore 36=\pi(3)^2 \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{4}{\pi} \mathrm{m} / \mathrm{min}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{dh}}{\mathrm{dt}} \\
& \therefore 36=\pi(3)^2 \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{4}{\pi} \mathrm{m} / \mathrm{min}
\end{aligned}
$$
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