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 $4m$, is

• A. $35m/h$
• B. $37m/h$
• C. $88m/h$
• 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, $\frac{dV}{dt}=\frac{d}{dh}\left(\frac{\pi h^3}{12}\right)\cdot \frac{dh}{dt}$
(by Chain rule)
$=\frac{\pi }{4}{h}^2\frac{dh}{dt}$
Now, rate of change of volume, i.e., $\frac{dV}{dt}=5m^3/h$ and $h=4m$.
Therefore, $5=\frac{\pi }{4}(4{)}^2\cdot \frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{5}{4\pi }=\frac{35}{88}m/h\left(\pi =\frac{22}{7}\right)$
Thus, the rate of change of water level is $\frac{35}{88}m/h$.