A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $ an ^{

Question

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $	an ^{-1}\left(\frac{1}{2}ight)$. Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in $\mathrm{m} / \mathrm{min}$.), at which the level of water is rising at the instant when the depth of water in the tank is $10 \mathrm{~m}$; is\

Professor PC Thomas Classes · Answer

(a) : Let $r$ be the radius of the cone when level of water is 10 m .\ Given, $\frac{d V}{d t}=5 \mathrm{~m}^{3} / \mathrm{min}$ and $	an ^{-1} \frac{1}{2}=	heta$\ $\Rightarrow 	an 	heta=\frac{1}{2}=\frac{r}{h} \quad 	herefore r=\frac{h}{2}$\ Now, $V=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi \frac{h^{3}}{4}$\ $	herefore \quad \frac{d V}{d t}=\frac{\pi}{12} \cdot 3 h^{2}\left(\frac{d h}{d t}ight)$\ $\Rightarrow \quad 5=\frac{\pi}{4}(100) \frac{d h}{d t}$\ $\Rightarrow \quad \frac{d h}{d t}=\frac{1}{5 \pi} \mathrm{~m} / \mathrm{min}$\