Question

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 1

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