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Water is flowing at the rate of 15km per hour through a pipe of diameter 14cm into a rectangular tank which is 50m long and 44m wide. Find the time in which the level of water in the tank will rise by 21cm.

Answer
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Hint-Using the formula of volume of pipe ,first find out the height of the pipe and then solve it further.

Complete step-by-step answer:
Given that tank is in the form of a cuboid where
Length=l=50m
Breadth=b=44m
Height=h=21cm=$\frac{{21}}{{100}}$ m (because the level of the water in the tank should raise by 21cm)
So, the volume of the tank=
$\begin{gathered}
  l \times b \times h \\
    = 50 \times 44 \times \frac{{21}}{{100}} \\
   = \frac{{44 \times 21}}{2} \\
   = 462c{m^3} \\
    \\
\end{gathered} $
Now , also let us try to find out the volume of the pipe,
The pipe is in the form of a cylinder
So, the volume of the cylinder is given by the equation
$V = \pi {r^2}h$
So, now lets substitute these values here
We are given with the diameter of the pipe, so let us find out the radius of the cylinder
Radius=$\frac{{diameter}}{2} = \frac{{14}}{2} = 7cm$
So, from this we get the volume of the pipe to be equal to
Volume=$\frac{{22}}{7} \times {\left( {\frac{7}{{100}}} \right)^2}h$
Volume$ = \frac{{22 \times 7}}{{10000}}h$
Now, since the water is flowing through the pipe and then reaching the tank, we can write that
The Volume of pipe=Volume of tank
So, we get$\frac{{22 \times 7}}{{10000}}h = 462c{m^3}$
So, we get h=$\frac{{22 \times 21 \times 10000}}{{22 \times 7}}$
                   h=30,000m
Now since the rate is expressed in km/hr we will convert this height h to km
So, we get height h=30km
S0,
We have 15km travels in pipe in 1 hour
1km travels in pipe in =$\frac{1}{{15}}hour$
The total height travelled was 30km
So, the time taken to travel 30 kilometres=$\frac{{30}}{{15}} = 2$ hours
So, the total time taken is 2 hours for the tank to be filled
Note: Whenever we have been given with the diameter of a quantity make sure to obtain the value of the radius from this and then apply it in the formula, do not apply the value of diameter in the formula

Last updated date: 01st Oct 2023
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