$A$ and $B$ are two separate reservoirs of water. Capacity of reservoir $A$ is double the capacity o

Question

$A$ and $B$ are two separate reservoirs of water. Capacity of reservoir $A$ is double the capacity of reservoir $B$. Both the reservoirs are filled completely with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at the time. One hour after the water is released, the quantity of water in reservoirs $A$ is $1 \frac{1}{2}$ times the quantity of water in reservoir $B$. After how many hours do both the reservoirs have the same quantity of water?

Professor PC Thomas Classes · Answer

Let at any instant $t$, the volume of the reservoirs of $A$ be $V_{A}$ and that of $B$ be $V_{B}$.\ The initial capacity of $A$ be $V_{A}^{\circ}$ and $B$ be $V_{B}^{\circ}$\ $	herefore \frac{V_{A}^{\circ}}{V_{B}^{\circ}}=2$\ Also, $\frac{d V_{A}}{d t}=-K_{A} V_{A}$ and $\frac{d V_{B}}{d t}=-K_{B} V_{B}$\ $\Rightarrow \frac{d V_{A}}{V_{A}}=-K_{A} d t$ and $\frac{d V_{B}}{V_{B}}=-K_{B} d t$\ $\Rightarrow \quad \log V_{A}=-K_{A} t+C$\ and $\log V_{B}=-K_{B} t+C$\ At $t=0 V_{A}=V_{A}^{\circ}, V=V_{B}^{\circ}$\ $	herefore \log \frac{V_{A}}{V_{A}^{\circ}}=-K_{A} t$ and $\log \frac{V_{B}}{V_{B}^{\circ}}=-K_{B} t$\ Given that at $t=1$\ $\frac{V_{A}}{V_{B}}=\frac{3}{2}, \frac{V_{A}^{\circ}}{V_{B}^{\circ}}=2 \Rightarrow \frac{V_{A}}{V_{B}} \cdot \frac{V_{B}^{\circ}}{V_{A}^{\circ}}=\frac{3}{4}$\ Putting the value at $t=1$, we get\ $\log \frac{V_{A}}{V_{A}^{\circ}}=-K_{A} ; \log \frac{V_{B}}{V_{B}^{\circ}}=-K_{B}$\ $\Rightarrow \log \frac{V_{A}}{V_{A}^{\circ}}-\log \frac{V_{B}}{V_{B}^{\circ}}=K_{B}-K_{A}$\ $\Rightarrow \log \left(\frac{V_{A} / V_{A}^{\circ}}{V_{B} / V_{B}^{\circ}}ight)=K_{B}-K_{A}$\ $\Rightarrow \log \left(\frac{3}{4}ight)=K_{B}-K_{A}$\ Now, at $t=t_{1}$ say, $V_{A}=V_{B}$\ $\Rightarrow \log \frac{V_{A}}{V_{A}^{\circ}}=-K_{A} t_{1}$\ $\log \frac{V_{B}}{V_{B}^{\circ}}=-K_{B} t_{1}$\ $\Rightarrow \log \left(\frac{V_{A} / V_{A}^{\circ}}{V_{B} / V_{B}^{\circ}}ight)=\left(K_{B}-K_{A}ight) t_{1}$\ $=\left[\log \left(\frac{3}{4}ight)ight] t_{1}$\ $\Rightarrow \log \left(\frac{1}{2}ight)=\left(\log \frac{3}{4}ight) 	imes t_{1}$\ $\Rightarrow t_{1}=\frac{\log (1 / 2)}{\log (3 / 4)} \quad 	herefore t_{1}=\frac{\log 2}{\log (4 / 3)}$\