Water is running into an underground right circular conical reservoir, which is 10 m deep and radius of its base is 5 m. If the rate of change in the volume of water in the reservoir is $\frac{3}{2} π m^{3} / m i n$ ., then the rate (in m/min) at which water rises in it, when the water level is 4m, is:

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$\frac{1}{8}$

$\frac{3}{8}$

$\frac{3}{2}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

Let $h$ be the height of the water and $r$ the radius of the cone formed by the water at time $t .$

$\begin{array}{l} ΔOAB ~ ΔOCD \\ \frac{O A}{A B} = \frac{O C}{C D} \\ \Rightarrow \frac{10}{5} = \frac{h}{r} \Rightarrow r = \frac{h}{2} \end{array}$

Let V = volume of water, the

$V = \frac{1}{3} π r^{2} h$

$\begin{array}{l} = \frac{1}{3} π {(\frac{h}{2})}^{2} h = \frac{1}{12} π h^{3} \\ \frac{d V}{d t} = \frac{1}{4} π h^{2} \frac{d h}{d t} \\ \Rightarrow \frac{3}{2} π = \frac{1}{4} π h^{2} \frac{d h}{d t} \\ \Rightarrow \frac{d h}{d t} = \frac{6}{h^{2}} \Rightarrow \frac{d h}{d t} ∣ h = 4 = \frac{6}{16} = \frac{3}{8} . \end{array}$