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There are three drain pipes $Q1$, $Q2$ and $Q3$, all of equal capacity, fitted to a tank of height $6$ meters. The tank is in the shape of a cuboid. $Q1$ is fitted at the bottom of the tank, while $Q3$ is fitted at a height of $4 m$ above $Q1$, and $Q2$ is fitted in between $Q3$ and $Q1$. $Q1$ alone can empty the full tank in $P$ minutes and if all the three pipes are in operation, the full tank can be emptied in $\frac{2P}{3}$ minutes. What is the height (above $Q1$) at which $Q2$ is fitted to the tank?

Let the length be $l$ and breadth be $b$ for the tank. Let $Q2$ be fitted at a height of $h$ above $Q1$.

Efficiency of $Q1$ = $\frac{6lb}{P}$ = $\frac{\text{Total volume of the cuboid tank}}{\text{Time taken by $Q1$ to drain the tank completely}}$

When $Q3$ will operate alone it will only drain top $2m$ of the tank.
When $Q2$ will operate alone it will only drain top $(6-h)m$ of the tank.

As per the question :-
$\frac{2P}{3}(\text{Efficiency of $Q1$ + Efficiency of $Q2$ + Efficiency of $Q3$}) = 6lb = \text{Total volume of the cuboid tank drained}$
but how can I find the efficiency of $Q2$ and $Q3$ to use in the above equation.

Please help!!!

Thanks in advance!

Ganit
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1 Answers1

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Contrary to one's understanding of fluid dynamics. the wording of the question suggests that all three pipes have the same constant rate of outflow, let's say $r$ metres per minute, regardless of the volume of water above the pipe.

On this assumption, then, let the height of $Q_2$ above $Q_1$ be $x$.

From the data we have $$\frac6r=P\implies \frac1r=\frac P6$$

The time to empty the top $2$ metres above $Q_3$ with all three pipes operational is $$t_1=\frac{2}{3r}$$ The time to empty the next $4-x$ metres with two pipes operational is $$t_2=\frac{4-x}{2r}$$ The time to empty the remaining $x$ metres with one pipe operational is $$t_3=\frac xr$$

But $$t_1+t_2+t_3=\frac{2P}{3}$$ $$\implies \frac23\cdot\frac P6+\frac{4-x}{2}\cdot\frac P6+\frac{Px}{6}=\frac{2P}{3}$$

From which we get $$x=\frac83$$

David Quinn
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