Question

A & B are two separate reservoirs of water. The 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 that time. One hour after the water is released, the quantity of water in reservoir A is $1.5$ times the quantity of water in reservoir B. After how many hours do both the reservoirs have the same quantity of water?

(Enter n if the answer is $T=lo{g}_{4/3}n$)

Answer 1

Volume of A = $\left(V_A\right)$ = $2\times$ Volume of B$\left(V_B\right)$

Let water present at any time in A & B be $x$ & $y$ respectively.

$\frac{-dm}{dt}=K_{1}x$ ; $\frac{-dy}{dt}=K_{2}y$

$x=C_1{e}^{-k_{1}t}$ ; $y=C_2{e}^{-k_{2}t}$

At $t$ =0. ; $x=2y$

$\Rightarrow C_1=2C_2$........(1)

At $t$ = 1. ; $x=1.5y$

$\Rightarrow C_1{e}^{-k_1}=3/2C_2{e}^{-k_2}$

$\Rightarrow 2C_2{e}^{-k_1}=3/2C_2{e}^{k_2}$

$\Rightarrow e^{-k_1}=3/4e^{-k_2}$

$\Rightarrow e^{k_1-k_2}=4/3$

$\Rightarrow k_1-k_2=ln\left(4/3\right)$

When $x=y$

$C_1{e}^{k_{1}t}=C_2{e}^{k_{2}t}$

$\Rightarrow 2C_2{e}^{-k_{1}t}=C_2{e}^{-k_{2}t}$

$\Rightarrow 2e^{-k_{1}t}=e^{-k_{2}t}$

$\Rightarrow e^{k_1-k_{2}t}=2$

$\Rightarrow t=\frac{ln2}{k_1-k_2}$

$\Rightarrow T=\frac{ln2}{ln\left(2/3\right)}=lo{g}_{4/3}2$