There is a uniform container of height $6m$ full of water. It has three identical drainage pipes $A,B,C$ attached to it. Each of them can independently empty the container (the part above it) in $T$ minutes. Pipe $A$ is fixed at the base where as pipe $B$ is fixed at the height of $5m$. Pipe $C$ is fixed somewhere between them. If together they can empty the container in $2T/3$ minutes, find position of pipe $C$. (Ignore all variations due to pressure variance at different heights).
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How can a pipe located at height 5m above the base empty the container? – zoli Nov 19 '17 at 12:42
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@zoli It can only empty the volume above it. Being able to empty the container in T minutes indicates rate of drainage of the pipe. – Ajax Nov 19 '17 at 12:44
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The rstes of emptying are $\frac6T$ for $A$, $\frac1T$ for $B$ and $\frac xT$ for $C$, $x$ being the height of $C$ above the base.
The pipes work together first, then only two of them work together, then, finally only one of them works as shown below:
When more then one pipes work together then the rates add up.
If we open all the three pipes then they will work together until the water level descends to pipe $B$. Let $t_{A,B,C}$ denote this time. Then
$$t_{A,B,C}\left(\frac1T+\frac6T+\frac xT\right)=1$$ because the there is $1$ meter difference between the uppere two pipes..
Let $t_{B,C}$ denote denote the time so that the water level descends to the height $x$.
Then
$$t_{A,C}\left(\frac6T+\frac xT\right)=5-x$$ because only pipe $B$ and $C$ work together as long as the water level is above $x$ and below level $5$. Finally,
$$t_A\frac6T=x$$ and
$$t_{A,B,C}+t_{A,C}+t_A=\frac23T. $$
Now, we have four linear equations and four unknowns. So, the task can be compelted.
zoli
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