A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is $\tan^{-1}(0.5)$. Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4 \text{ m}$.

$\frac{35}{88} \text{ m/h}$

The correct answer is Option (2) → $\frac{35}{88} \text{ m/h}$ ##

Let $r, h$ and $\alpha$ be as in Fig. Then $\tan \alpha = \frac{r}{h}$.

So, $\alpha = \tan^{-1}\left(\frac{r}{h}\right)$

But $\alpha = \tan^{-1}(0.5)$ (given)

or $\frac{r}{h} = 0.5 ⇒r = \frac{h}{2}$

Let $V$ be the volume of the cone. Then

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$

Therefore, $\frac{dV}{dt} = \frac{d}{dh}\left(\frac{\pi h^3}{12}\right) \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} \quad \text{(by Chain Rule)}$

Now rate of change of volume, i.e., $\frac{dV}{dt} = 5 \text{ m}^3/\text{h}$ and $h = 4 \text{ m}$.

Therefore, $5 = \frac{\pi}{4} (4)^2 \frac{dh}{dt}$

$ \frac{dh}{dt} = \frac{5}{4\pi} = \frac{35}{88} \text{ m/h} \quad \left(\text{taking } \pi = \frac{22}{7}\right)$

Thus, the rate of change of water level is $\frac{35}{88} \text{ m/h}$.