A water tank has the shape of an inverted rig | vedclass

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(B) Let $h$ be the depth of water and $r$ be the radius of the water surface at time $t$.Given the semi-vertical angle $	heta = 	an^{-1}(1/2)$,we ha

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$\frac{1}{5\pi}$

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(B) Let $h$ be the depth of water and $r$ be the radius of the water surface at time $t$.Given the semi-vertical angle $	heta = 	an^{-1}(1/2)$,we have $	an 	heta = \frac{r}{h} = \frac{1}{2}$,which implies $r = \frac{h}{2}$.The volume $V$ of the water in the cone is given by $V = \frac{1}{3}\pi r^2 h$.Substituting $r = \frac{h}{2}$,we get $V = \frac{1}{3}\pi \left(\frac{h}{2}ight)^2 h = \frac{\pi h^3}{12}$.Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = \frac{\pi}{12} (3h^2) \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$.Given $\frac{dV}{dt} = 5 \ m^3/min$ and $h = 10 \ m$,we substitute these values:$5 = \frac{\pi (10)^2}{4} \frac{dh}{dt}$$5 = \frac{100\pi}{4} \frac{dh}{dt}$$5 = 25\pi \frac{dh}{dt}$$\frac{dh}{dt} = \frac{5}{25\pi} = \frac{1}{5\pi} \ m/min$.

$A$ water tank has the shape of an inverted right circular cone,whose semi-vertical angle is $\tan^{-1}(1/2)$. Water is poured in at a constant rate of $5 \ m^3/min$. 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:

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