Question

Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. The tank initially contains 350 liters and 20 liters leak out during the first day. When will the tank be half empty? How much water will there be after 4 days?

Answer 1

Answer:

Therefore the tank will be half empty after 10.10 days.

Therefore there will be 2735.53 liters water in the tank after 4 days.

Step-by-step explanation:

Given that,the square root of the volume of remaining water in the tank is proportional to the rate of water leakage.

Let V be volume of water at any instant time t.

$\therefore \frac{dV}{dt} \propto \sqrt V$

$\Rightarrow \frac{dV}{dt}= k \sqrt V$

where k is constant of proportionality.

$\Rightarrow \frac{dV}{\sqrt V}= k \ dt$

Integrating both sides

$\Rightarrow \int \frac{dV}{\sqrt V}= \int k \ dt$

$\Rightarrow 2\sqrt V=kt+C$     [ C is integrating constant]

At t=0, the volume of water is 350 liters

$2\sqrt{350}=k.0+C$

$\Rightarrow C=2\sqrt{350}$

The equation becomes

$\Rightarrow 2\sqrt V=kt+2\sqrt{350}$

Again at t=1, the volume of water is(350-20)liters=330 liters

$2\sqrt {330}=k.1+2\sqrt{350}$

$\Rightarrow k=2\sqrt {330}-2\sqrt{350}$

The equation becomes

$2\sqrt V=2(\sqrt{330}-\sqrt{350} ) t+2\sqrt{350}$

$\Rightarrow \sqrt V=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}$

Now when the tank is half empty,then V= (350÷2) liters = 175 liters

$\sqrt {175}=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}$

$\Rightarrow (\sqrt{330}-\sqrt{350} ) t=\sqrt{175}-\sqrt{350}$

$\Rightarrow t=\frac{\sqrt{175}-\sqrt{350}}{ (\sqrt{330}-\sqrt{350} )}$

$\Rightarrow t=10.10$ days

Therefore the tank will be half empty after 10.10 days.

After 4 days,

$\sqrt V=(\sqrt{330}-\sqrt{350} ) (4)+\sqrt{350}$

$\Rightarrow \sqrt V= 16.54$

$\Rightarrow V= 373.53$ liters

Therefore there will be 2735.53 liters water in the tank after 4 days.