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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
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The correct answer is:
$\frac{4}{\pi} \mathrm{m} / \mathrm{sec}$
$\frac{\mathrm{d} v}{\mathrm{~d} t}=36 \mathrm{~m}^3 / \mathrm{sec}$
and $v=\pi r^2 \mathrm{~h}=\pi \times 3^2 \mathrm{~h}=9 \pi \mathrm{h}$
$\Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$
$\Rightarrow 36=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$
$\Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{4}{\pi} \mathrm{m} / \mathrm{sec}$
and $v=\pi r^2 \mathrm{~h}=\pi \times 3^2 \mathrm{~h}=9 \pi \mathrm{h}$
$\Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$
$\Rightarrow 36=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}$
$\Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{4}{\pi} \mathrm{m} / \mathrm{sec}$
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