Question

Water is drained from a vertical cylinderical tank by opening a valve at the base of the tank.It is known that the rate at which the water level drops is proportional to the square root of water depth $y$ ,where the constant of proportionality $k>0$ depends on the acceleration due to gravity and the geometry of the hole.If $t$ is measured in minutes and $k=\frac{1}{15}$, then the time($t$) to drain the tank if the water is 4 m deep to start with is

Answer 1

The rate of drop of water level is $\frac{dy}{dt}=-k\sqrt{y}\text{Or,}\underset{4}{\overset{0}{\int }}\frac{dy}{\sqrt{y}}=-k\underset{0}{\overset{t}{\int }}dt2\sqrt{y}{|}_4^0=-kt=-\frac{t}{15}0-4=-\frac{t}{15}\Rightarrow t=60\text{min}$