Question

$A$ and $B$ are two separate reservoir 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 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 resevoirs have the same quantity of water ?

• A. ${\log }_{\left(\frac{3}{4}\right)}\left(\frac{1}{2}\right)$
• B. ${\log }_{\left(1/2\right)}\left(\frac{1}{2}\right)$
• C. ${\log }_{3/4}\left(2\right)$
• D. $None$

Answer 1

Answer: (A)

${\log }_{\left(\frac{3}{4}\right)}\left(\frac{1}{2}\right)$

$\frac{dV}{dt}\propto V$ for each reservoir

$\frac{dV}{dt}\propto -V_A\Rightarrow \frac{d{V}_A}{dt}=-K_1{V}_A$ $(K_1$ is the proportional constant$).$

$\Rightarrow {\int }_{V_A}^{{V^{\prime }}_A}\frac{d{V}_A}{V_A}=K_1{\int }_0^{t}dt\Rightarrow \log \frac{{V^{\prime }}_A}{V_A}=K_{1}t\Rightarrow {V^{\prime }}_A=V_A.e^{-K_{1}t}$ ...(1)

Similary for $B,$ ${V^{\prime }}_B=V_B.e^{-K_{2}t}$ ...(2)

On dividing eqution (1) by (2), we get

$\frac{{V^{\prime }}_A}{{V^{\prime }}_B}=\frac{V_A}{V_B}.e^{-(K_1-K_2)t}$

It is given that at $t=0,V_A=2V_B$ and at $t=\frac{3}{2},{V^{\prime }}_A=\frac{3}{2}{V^{\prime }}_B$

Thus, $\frac{3}{2}=2.e^{-(K_1-K_2)t}\Rightarrow e^{-(K_1-K_2)}=\frac{3}{4}$ ...(3)

Now, let at $t=t_0$ both the reservoirs have some quantity of water.

Then, ${V^{\prime }}_A={V^{\prime }}_B$

From equation (3),

$2e^{-(K_1-K_2)}=1\Rightarrow 2.{\left(\frac{3}{4}\right)}^{t_0}=1\Rightarrow t_0={\log }_{3/4}\left(\frac{1}{2}\right)={\log }_{4/3}2$