SATHEE: Differential Equations 3 Question 11

Differential Equations 3 Question 11

11. A hemispherical tank of radius $2 m$ is initially full of water and has an outlet of $12 c m^{2}$ cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $v (t) = 0.6 \sqrt{2 g h (t)}$ , where $v (t)$ and $h (t)$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time $t$ and $g$ is the acceleration due to gravity. Find the time it takes to empty the tank.

(2001, 10M)

Hint Form a differential equation by relating the decreases of water level to the outflow.

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Answer:

Correct Answer: 11. $\frac{14 π \times 10^{5}}{27 \sqrt{g}}$ unit

Solution:

Let $O$ be the centre of hemispherical tank. Let at any instant $t$ , water level be $B A B_{1}$ and at $t + d t$ , water level is $B A^{'} B_{1}$ . Let $∠ O_{1} O B_{1} = θ$ .

$\Rightarrow A B_{1} = r \cos θ$ and $O A = r \sin θ$ decrease in the water volume in time $d t = π A B_{1}^{2} \cdot d (O A)$

$[π r^{2}$ is surface area of water level and $d (O A)$ is depth of water level]

$\begin{aligned} = π r^{2} \cdot \cos^{2} θ \cdot r \cos θ d θ \\ = π r^{3} \cdot \cos^{3} θ d θ \end{aligned}$

Also, $h (t) = O_{2} A = r - r \sin θ = r (1 - \sin θ)$

Now, outflow rate $Q = A \cdot v (t) = A \cdot 0.6 \sqrt{2 g r (1 - \sin θ)}$

Where, $A$ is the area of the outlet.

Thus, volume flowing out in time $d t$ .

$\Rightarrow Q d t = A \cdot (0.6) \cdot \sqrt{2 g r} \cdot \sqrt{1 - \sin θ} d t$

We have, $π r^{3} \cos^{3} θ d θ = A (0.6) \cdot \sqrt{2 g r} \cdot \sqrt{1 - \sin θ} d t$

$\Rightarrow \frac{π r^{3}}{A (0.6) \sqrt{2 g r}} \cdot \frac{\cos^{3} θ}{\sqrt{(1 - \sin θ)}} d θ = d t$

Let the time taken to empty the tank be $T$ .

Then, $T = \int_{0}^{π / 2} \frac{π r^{3}}{A (0.6) \cdot \sqrt{2 g r}} \cdot \frac{\cos^{3} θ}{\sqrt{1 - \sin θ}} d θ$

$= \frac{- π r^{3}}{A (0.6) \sqrt{2 g r}} \int_{0}^{π / 2} \frac{1 - \sin^{2} θ (- \cos θ)}{\sqrt{1 - \sin θ}} d θ$

Let $t_{1} = \sqrt{1 - \sin θ}$

$\Rightarrow d t_{1} = \frac{- \cos θ}{\sqrt{1 - \sin θ}} d θ$

$∴ T = \frac{- 2 π r^{3}}{A (0.6) \sqrt{2 g r}} \int_{1}^{0} [1 - {(1 - t_{1}^{2})}^{2}] d t_{1}$

$\begin{aligned} \Rightarrow T = \frac{- 2 π r^{3}}{A (0.6) \sqrt{2 g r}} \int_{1}^{0} [1 - (1 + t_{1}^{4} - 2 t_{1}^{2})] d t_{1} \\ \Rightarrow T = \frac{- 2 π r^{3}}{A (0.6) \sqrt{2 g r}} \int_{1}^{0} [1 - 1 - t_{1}^{4} + 2 t_{1}^{2}] d t_{1} \\ \Rightarrow T = \frac{2 π r^{3}}{A (0.6) \sqrt{2 g r}} \int_{1}^{0} (t_{1}^{4} - 2 t_{1}^{2}) d t_{1} \\ \Rightarrow T = \frac{2 π r^{3}}{A (0.6) \sqrt{2 g r}} \frac{t_{1}^{5}}{5} - \frac{2 t_{1}^{3}}{3} \\ \Rightarrow T = \frac{2 π \cdot r^{5 / 2}}{A \frac{6}{10} \sqrt{2 g r}} \cdot 0 - \frac{1}{5} - 0 + \frac{2}{3} \\ \Rightarrow T = \frac{2 π \cdot 2^{5 / 2} {(10^{2})}^{5 / 2}}{12 \cdot \frac{3}{5} \cdot \sqrt{2} \cdot \sqrt{g}} \frac{2}{3} - \frac{1}{5} \\ = \frac{2 π \times 10^{5} \cdot 4 \cdot 5}{(12 \times 3) \sqrt{g}} \frac{10 - 3}{15} \\ = \frac{2 π \times 10^{5} \times 7}{3 \cdot 3 \cdot \sqrt{g} \cdot 3} = \frac{14 π \times 10^{5}}{27 \sqrt{g}} unit \end{aligned}$

Differential Equations 3 Question 11

Answer:

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Differential Equations 3 Question 11

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Medical Preparation

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