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

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(A) Let $r$ be the radius and $h$ be the depth of the water at any time $t$. The semi-vertical angle $\alpha$ is given by $	an \alpha = 0.5 = \frac{1

Vedclass · Accepted Answer

$\frac{5}{4\pi} \ m/h$

Vedclass · Answer

(A) Let $r$ be the radius and $h$ be the depth of the water at any time $t$. The semi-vertical angle $\alpha$ is given by $	an \alpha = 0.5 = \frac{1}{2}$.From the geometry of the cone,$	an \alpha = \frac{r}{h}$,so $\frac{r}{h} = \frac{1}{2}$,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 with respect to time $t$,we get $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$.Given $\frac{dV}{dt} = 5 \ m^3/h$ and $h = 4 \ m$,we substitute these values:$5 = \frac{\pi (4)^2}{4} \cdot \frac{dh}{dt} = 4\pi \cdot \frac{dh}{dt}$.Therefore,$\frac{dh}{dt} = \frac{5}{4\pi} \ m/h$.

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