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(power-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 m.​

Answer 1

$\underline{\huge{Answer:-}}$

Step-by-step explanation:

$\mathcal{Let \: r, h \: and \: \alpha . \: \:Then \: tan \: \alpha = \frac{r}{h} }$

$\implies\mathcal{ \alpha = {tan}^{ - 1}( \frac{r}{h}) } \\ \implies \mathcal{ \alpha = {tan}^{ - 1} (0.5) \: \: (given)} \\ \implies\mathcal{ \frac{r}{h} = 0.5} \\ \implies\mathcal{r = \frac{h}{2} }$

Let V be the volume of the cone. Then

$\implies\mathcal{V = \frac{1}{3}\pi {r}^{2}h = \frac{1}{3}\pi( \frac{h}{2})^{2}h = \frac{\pi \: {h}^{3} }{12}} \\ \\ \implies \mathcal{ \frac{dV}{dt} = \frac{d}{dh}( \frac{\pi \: {h}^{3} }{12}). \frac{dh}{dt} } \\ \\ \implies\mathcal{ \frac{\pi}{h} {h}^{2} \frac{dh}{dt} }$

Now rate of change of volume , i.e.,

$\implies \mathcal{ \frac{dV}{dt} = 5 {m}^{3} \: and \: h = 4m } \\ \implies \mathcal{5 = \frac{\pi}{4}(4)^{2}. \frac{dh}{dt} } \\ \implies \mathcal{ \frac{dh}{dt} = \frac{5}{4\pi} = \frac{35}{88}m/h(\pi = \frac{22}{7}) }$

Thus , the rate of change of water level is 35/88 m/h.

$\Large \fbox\mathcal{Ans = \frac{35}{88} m/h }$