# A water tank has a shape of an inverted right circular cone , whose semi-vertical angle is  $\tan ^{-1}\left ( \frac{1}{2} \right )$   . Water is poured into it at a constant rate of 5 cubic meter per minut . Then the rate  ( in m /min) , at which the level of water is rising at the instant when the depth of water in the tank is 10 m ; is : Option 1) $1/15\pi$ Option 2) $1/10\pi$ Option 3) $2/\pi$ Option 4) $1/5\pi$

$\\ \tan\theta =\frac{r}{h}=\frac{1}{2}\\\\\:2r=h$

$\\ v=1/3\:\pi\frac{h^{3}}{4}=\frac{\pi h^{3}}{12} \\\\\: \frac{dv}{dt} = \frac{\pi}{4} h^{2} \frac{dh}{dt}$

$\\5=\frac{\pi}{4}(10)^{2}\frac{dh}{dt}\\\\\:\frac{dh}{dt}=\frac{1}{5}\cdot\frac{1}{\pi}=\frac{1}{5\pi} \:m/min$

Option 1)

$1/15\pi$

Option 2)

$1/10\pi$

Option 3)

$2/\pi$

Option 4)

$1/5\pi$

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