ISI2016-DCG-70

Question

Water pours into a rectangular tank of $20\:metres$ depth which was initially half-filled. The rate at which the height of the water rises is inversely proportional to the height of the water at that instant. If the height of the water after an hour is observed to be $12\:metres$, how much time, in hours, will be required to fill up the tank?

• A. $\frac{75}{11}$
• B. $\frac{125}{11}$
• C. $\frac{25}{3}$
• D. $5$

Answer 1

Let the rate of change of height of the tank be denoted by $\frac{dH}{dt}$

Then according to the question,   $\frac{dH}{dt} =\frac{K}{H}$  -----$\left ( 1 \right )$

Givent at   $t=0$,  $H=10$   and at $t=1$,  $H=12$

So, integrating by limits mentioned above, we get    $\int_{10}^{12}HdH=K\int_{0}^{1}dt$

$\Rightarrow$   $K=\frac{144-100}{2}=22$

To find the complete time to fill up the tank, we will use again equation $\left ( 1 \right )$

Now,    $\int_{10}^{20}HdH=22\times \int_{0}^{t}dt$

$\therefore$     $t=\frac{\left ( 400-100 \right )}{44}$    $\Rightarrow$     $t=\frac{75}{11}$

Option A is correct