If the water is being poured at the rate $36 \mathrm{~m}^3 / \mathrm{sec}$ in cylindrical vessel of base radius $3 \mathrm{~m}$, then the rate at which water level is rising, is

Question

Aicharya · Accepted Answer

$\frac{\mathrm{d} v}{\mathrm{~d} t}=36 \mathrm{~m}^3 / \mathrm{sec}$ and $v=\pi r^2 \mathrm{~h}=\pi 	imes 3^2 \mathrm{~h}=9 \pi \mathrm{h}$ $\Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$ $\Rightarrow 36=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$ $\Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{4}{\pi} \mathrm{m} / \mathrm{sec}$

If the water is being poured at the rate $36 \mathrm{~m}^3 / \mathrm{sec}$ in cylindrical vessel of base…

Solution