Question

Water is drained from a vertical cylindrical 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 $k =1 / 15$ then the time to drain the tank if the water is 4 meter deep to start with is -

• A. $30 \min$
• B. $45 \min$
• C. $60 \min$
• D. $80 \min$

Answer 1

Answer: (C)

Given $\frac{ dy }{ dt }=- k \sqrt{ y } \Rightarrow \int \frac{1}{\sqrt{ y }} dy =- k \int dt$
$2 \sqrt{ y }=- kt + c$
Now at $t=0, y=4$ so $c=4$.
$\therefore 2 \sqrt{ y }=\frac{- t }{15}+4 \left(\right.$ as $\left.k =\frac{1}{15}\right)$
When $y=0, t=60 \min$.