A hemispherical tank of radius 2 metres is initially full of water and has an outlet of $12 \mathrm{

Question

A hemispherical tank of radius 2 metres is initially full of water and has an outlet of $12 \mathrm{~cm}^{2}$ cross-sectional area at the bottom. The outlet is opened at same instant. The flow through the outlet is according to the law $v(t)=$ $0 \cdot 6 \sqrt{2 g h (t)}$, where $v(t)$ and $h(t)$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time $t$, and $g$ is the acceleration due to gravity. If the time it takes to empty the tank is $\frac{m \pi}{n \sqrt{g}}\left(10^{5}ight)$ units, then m+n is\. (Hint : For a differential equation by relating the decrease of water level to the outflow).

Professor PC Thomas Classes · Answer

Let the height of the water level above the base at time $t$ be $h$ and let if fall $d h$ in time $d t$. Let the decrease in volume be $d v$.\ If $r$ is the radius of the surface of the water at time $t$ $$ r^{2}=R^{2}-(R-h)^{2}=2 R h-h^{2} $$ and $d v=-\pi r^{2} d h$\ $\Rightarrow \frac{d v}{d t}=-\pi r^{2} \frac{d h}{d t}$\ we have given velocity of water\ flow is $\frac{d v}{d t}=\frac{3}{5} \sqrt{2 g h} A$\ From (1) and (2), we get\ $-\pi r^{2} \frac{d h}{d t}=\frac{3}{5} \sqrt{2 g h} A$\ $\Rightarrow \frac{d h}{d t}=-\frac{3 \sqrt{2 g h} A}{5 \pi r^{2}}=\frac{-3 \sqrt{2 g h} A}{5 \pi\left(2 R h-h^{2}ight)}$\ $\Rightarrow\left(2 R \sqrt{h}-h^{\frac{3}{2}}ight) d h=\frac{-3 A}{5 \pi} \sqrt{2 g} d t$\ Integrating both sides, we get\ $\frac{4 R}{3} h^{\frac{3}{2}}-\frac{2}{5} h^{\frac{5}{2}}=\frac{-3 A}{5 \pi} t \sqrt{2 g}+c$\ when $h=R, t=0$\ $	herefore \frac{4 R}{3} R^{\frac{3}{2}}-\frac{2}{5} R^{\frac{5}{2}}=\frac{-3 A}{5 \pi} \sqrt{2 g}(0)+c$\ $\Rightarrow c=\left(\frac{4}{3}-\frac{2}{5}ight) R^{\frac{5}{2}}=\frac{14}{15} R^{\frac{5}{2}}$\ We have to find $t$, so that $h=0$. Let $t=T$ when $h=0$ then\ $0=\frac{-3 A}{5 \pi} \sqrt{2 g} T+\frac{14}{15} R^{\frac{5}{2}} \Rightarrow T=\frac{14}{15} R^{\frac{5}{2}} \frac{5 \pi}{3 A \sqrt{2 g}}$\ But $A=12$ and $R=200$\ $	herefore T=\frac{14}{15}(200)^{\frac{5}{2}} \frac{5 \pi}{3 	imes 12 \sqrt{2 g}}$\ $=\frac{14}{15}(2)^{2} \sqrt{2} \frac{10^{5} 	imes 5 \pi}{3 	imes 12 \sqrt{2 g}}=\frac{14 \pi}{27 \sqrt{g}}\left(10^{5}ight)$ units\