SATHEE: Differential Equations 3 Question 14

Differential Equations 3 Question 14

14. $A$ 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 reservoir $A$ 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?

(1997, 7M)

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Answer:

Correct Answer: 14. $A \to p; B \to s; C \to q; D \to r$

Solution:

$\frac{d V}{d t} \propto V$ for each reservoir.

$\frac{d V}{d x} \propto - V_{A} \Rightarrow \frac{d V_{A}}{d t} = - K_{1} V_{A}$

$\begin{aligned} \Rightarrow \int_{V_{A}}^{V^{'}} \frac{d V_{A}}{V_{A}} = - K_{1} \int_{0}^{t} d t \\ \Rightarrow \log \frac{V_{A}^{'}}{V_{A}} = - K_{1} t \Rightarrow V_{A}^{'} = V_{A} \cdot e^{- K_{1} t} \end{aligned}$

Similarly for $B, V_{B}^{'} = V_{B} \cdot e^{- K_{2} t}$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{V_{A}^{'}}{V_{B}^{'}} = \frac{V_{A}}{V_{B}} \cdot e^{- (K_{1} - K_{2}) t}$

It is given that at $t = 0, V_{A} = 2 V_{B}$ and at

$t = \frac{3}{2}, V_{A}^{'} = \frac{3}{2} V_{B}^{'}$

Thus, $\frac{3}{2} = 2 \cdot e^{- (K_{1} - K_{2}) t} \Rightarrow e^{- (K_{1} - K_{2})} = \frac{3}{4}$

Now, let at $t = t_{0}$ both the reservoirs have some quantity of water. Then,

$V_{A}^{'} = V_{B}^{'}$

From Eq. (iii), $2 e^{- (K - K_{2})} = 1$

$\Rightarrow 2 \cdot {\frac{3}{4}}^{t_{0}} = 1$

Differential Equations 3 Question 14

Answer:

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Differential Equations 3 Question 14

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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