A water tank has the shape of an inverted right circular cone whose semi vertical angle is
$\tan^{- 1} (\frac{1}{2})$ . Water is poured into it at a constant rate of $5 m^{3} / \min$ , then the rate $(i n m / \min)$ at which level of water is rising, at the instant when the depth of the water in the tank is $10 m$ is

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$\frac{1}{10 π}$

$\frac{1}{15 π}$

$\frac{1}{5 π}$

$\frac{2}{π}$

answer is B.

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Detailed Solution

$θ = \tan^{- 1} (\frac{1}{2}) \Rightarrow \tan θ = \frac{1}{2} \frac{r}{h} = \frac{1}{2} \frac{d V}{d t} = 5 m^{3} / \min V = \frac{1}{3} π r^{2} h$

$V = \frac{π}{3} \frac{h^{2}}{4} h = \frac{π h^{3}}{12} \frac{d v}{d t} = \frac{π}{12} (3 h^{2}) \frac{d h}{d t} 5 = \frac{π}{4} (100) \frac{d h}{d t} \frac{d h}{d t} = \frac{1}{5 π} m e t / \min$

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