Question

The surface of water in a swimming pool, when it is full of water, is rectangular with length and breadth 36 m and 10.5 m respectively. The depth of water increases uniformly from 1 m at one end to 1.75 m at the other end. The water in the pool is emptied by a cylindrical pipe of radius 7 cm at the rate of 5 km/h. The time (in hours) to empty water in the pool is (take $\pi =\dfrac{22}{7}$).
(a)$6\dfrac{1}{4}$
(b)$6\dfrac{1}{2}$
(c)$6\dfrac{3}{4}$
(d)$6\dfrac{4}{5}$

Answer 1

Hint: We are given the length, breadth and height of the pool so using these dimensions find the volume of the pool. We are asked to find the time in which a cylindrical pipe of radius 7 cm will evacuate the pool which will be found by dividing the volume of the pool to the result of the multiplication of cross sectional area of the pipe and the rate at which water is evacuating.

Complete step-by-step answer:
We are given the length and breadth of the pool as 36 m and 10.5 m and it is also given that the depth of water increases uniformly from 1 m at one end to 1.75 m at the other end. This information is depicting in the below figure:

In the above figure, we have shown the length by AC, breadth by AF, the depth of the water is shown by FG which is equal to 1m and HI which is equal to 1.75m.
We have also given a cylindrical pipe through which we are evacuating the pool at the rate of 5km/h and the radius of the cylinder is given as 7 cm. We are asked to find time in which we can empty the whole swimming pool.
First of all, we are going to find the volume of the pool. Volume of the pool is calculated by multiplying the area of the trapezium (FHIG) by the breadth of the pool (AF).
$\text{Volume of pool}=\text{Area of trapezium}\times \text{breadth}$
Area of trapezium is equal to half of the sum of parallel sides multiplied by the separation between the parallel sides.
\[\begin{align}
  & \text{Area of trapezium}=\dfrac{1}{2}\left( 1+1.75 \right)\left( 36 \right) \\
 & \Rightarrow \text{Area of trapezium}=49.5{{m}^{2}} \\
\end{align}\]
Breadth of the pool (AF) is equal to 10.5m
Volume of pool $=49.5\times 10.5{{m}^{3}}=519.75{{m}^{3}}$ …………. Eq. (1)
Let us assume that “t” hours are taken by the cylindrical pipe to evacuate the pool so volume of the water coming out from the cylindrical pool is equal to:
Volume of water evacuated from pipe $=\left( \text{Area of cross-section} \right)\left( \text{rate} \right)\left( t \right)$……….. Eq. (2)
The radius of the cylindrical pipe is given as 7 cm so the area of cross section of the pipe is:
$\begin{align}
  & \pi {{r}^{2}} \\
 & =\dfrac{22}{7}{{\left( 7 \right)}^{2}} \\
 & =154c{{m}^{2}} \\
\end{align}$
The area is in $c{{m}^{2}}$ we will convert it to ${{m}^{2}}$ by multiplying the above result by 0.0001.
$0.0154{{m}^{2}}$
Rate of the water coming from the pipe is 5 km/h or 5000 m/h. Now, substituting the values of area of cross section and rate in eq. (2) we get,
Volume of water evacuated from pipe =$0.0154\times 5000\times t$…….. Eq. (3)
To get the value of “t” we are going to equate eq. (1) and eq. (3).
Volume of pool = Volume of water evacuated from pipe
$519.75{{m}^{3}}$= $0.0154\times 5000\times t$
Dividing $1.54\times 5000$ on both the sides of the above equation we get,
$\begin{align}
  & \dfrac{519.75}{0.0154\times 5000}=t \\
 & \Rightarrow \dfrac{519.75}{15.4\times 5}=t \\
 & \Rightarrow 6.75hours=t \\
\end{align}$
Now, converting the decimal of the above time into fraction we get,
$\dfrac{27}{4}hours$
The answer is given in an improper fraction so we are going to convert the above proper fraction into improper fraction.
$6\dfrac{3}{4}hours$
Hence, the correct option is (c).

Note: You might get wrongly understood the language of the problem which says that the depth of water increases uniformly from 1 m at one end to 1.75 m at the other end instead of understanding the way we have described above you might have thought that the water level in swimming pool is initially at the height of 1 m and then increases to 1.75 m and then you find the volume of the pool by taking the swimming pool as a cuboid and you have taken the depth as 1.75 m so be careful while reading the question otherwise your whole question will be wrong altogether.