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 cubic meter per minute. 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 10m; is:

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$1 / 10 π$

$1 / 15 π$

$1 / 5 π$

$2 / π$

answer is C.

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Detailed Solution

$tanθ = \frac{1}{2} \frac{dv}{dt} = 5 h = 10 tanθ = \frac{r}{h} = \frac{1}{2} r = \frac{h}{2} V = \frac{1}{3} {πr}^{2} h \Rightarrow V = \frac{1}{3} π \times \frac{h^{2}}{4} \times h = \frac{π}{12} h^{3} \frac{dv}{dt} = \frac{π}{4} h^{2} \frac{dh}{dt} \Rightarrow 5 = \frac{π}{4} \times 100 \times \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{5 π}$