Question

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 meter per hour. The rate, at which the level of the water is rising at the instant when the depth of water in tank is $4 \,m$, is

• A. $35\, m / h$
• B. $37 \,m / h$
• C. $88\, m / h$
• D. None of the above

Answer 1

Answer: (D)

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

So, $ \alpha=\tan ^{-1}\left(\frac{r}{h}\right)$.
But $ \alpha=\tan ^{-1}(0.5)$ (given)
$\Rightarrow \frac{r}{h}=0.5$ or $r=\frac{h}{2}$
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{d V}{d t}=\frac{d}{d h}\left(\frac{\pi h^3}{12}\right) \cdot \frac{d h}{d t}$
(by Chain rule)
$=\frac{\pi}{4} h^2 \frac{d h}{d t}$
Now, rate of change of volume, i.e., $\frac{d V}{d t}=5 m ^3 / h$ and $h=4 m$.
Therefore, $ 5=\frac{\pi}{4}(4)^2 \cdot \frac{d h}{d t}$
$\Rightarrow \frac{d h}{d t}=\frac{5}{4 \pi}=\frac{35}{88} m / h \left(\pi=\frac{22}{7}\right)$
Thus, the rate of change of water level is $\frac{35}{88} m / h$.