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

 

Answers (1)

\\ \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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