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 meters. Then the water level in the cylinder is rising at the rate of

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{dV}}{\mathrm{dt}}=36 \mathrm{~m}^3 / \mathrm{min}, 	ext { radius }(\mathrm{r})=3 \mathrm{~m} \ & 	ext { Volume }(\mathrm{V})=\pi \mathrm{r}^2 \mathrm{~h} \ 	herefore \quad & \frac{\mathrm{dV}}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{dh}}{\mathrm{dt}} \ \Rightarrow & 36=\pi 	imes(3)^2 \frac{\mathrm{dh}}{\mathrm{dt}} \ \Rightarrow & \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{4}{\pi} \mathrm{~m} / \mathrm{min}\end{aligned}$

Water is being poured at the rate of $36 \mathrm{~m}^3 / \mathrm{min}$ into a cylindrical vessel, whose…

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