Question

A reservior of water is being emptied in such a manner that the rate at which the water is flowing out at any instant is proportional to the quantity of in water in it at that instant. If one third of water flows out in 5 minutes, find the fraction of quantity of water that will be left in reservoir after 10 minutes.

Answer 1

Given that,

Rate of water flowing out $\propto$ Quantity of water present in reservoir at that instant (Q)

$\Rightarrow$ $\frac{-dQ}{dt}\propto Q$

$\Rightarrow \frac{-dQ}{dt}=KQ$ ; $K$ is proportionality constant.

$\Rightarrow \frac{dQ}{Q}=-Kdt$ $-\left(i\right)$

On integrating both sides of $\left(i\right)$:

$\Rightarrow$$\int \frac{dQ}{Q}=-\int Kdt$

${\log }_{e}Q=-Kt+C$ ; is the constant of integration

At, $t=0$ let $Q=Q_0$ (initial quantity of water present)

$\Rightarrow C={\log }_e{Q}_0$

$\therefore$ ${\log }_{e}Q=-Kt+{\log }_e{Q}_0$

$\Rightarrow -{\log }_{e}Q+{\log }_e{Q}_0=Kt$

$\Rightarrow K=\frac{1}{t}{\log }_e\left(\frac{Q_0}{Q}\right)$ $-\left(ii\right)$

As given in the question,

In $t=5min$, $(1/3{)}^{rd}$ of water flows out.

So, water remaining after $t=5min$=$1-\frac{1}{3}=\left({\frac{2}{3}}^{rd}\right)$

Putting the values in $e{q}^n\left(ii\right)$

$K=\frac{1}{5}{\log }_e\left(\frac{Q_0}{\frac{2}{3}{Q}_0}\right)$

$\Rightarrow K=\frac{1}{5}{\log }_e\left(\frac{3}{2}\right)$ $-\left(iii\right)$

& After, $t=10min$, $Q=?$

so, $K=\frac{1}{10}{\log }_e\left(\frac{Q_0}{Q}\right)$ $-\left(iv\right)$

$\left(iii\right)$ & $\left(iv\right)$ $\Rightarrow$ $\frac{1}{10}{\log }_e\left(\frac{Q_0}{Q}\right)=\frac{1}{5}{\log }_e\left(\frac{3}{2}\right)$

$\Rightarrow$ ${\log }_e\left(\frac{Q_0}{Q}\right)=2{\log }_e\left(\frac{3}{2}\right)$

$\Rightarrow {\log }_e\left(\frac{Q-0}{Q}\right)={\log }_e{\left(\frac{3}{2}\right)}^2$

Taking antilog on both sides:-

$\Rightarrow$ $\frac{Q_0}{Q}=\frac{9}{4}\Rightarrow \frac{Q}{Q_0}=\frac{4}{9}$

So after $10$ min ${\left(\frac{4}{9}\right)}^{th}$ of the water will be left.