Question

Water in a canal $30dm$ wide $12dm$ deep is flowing with the velocity $20km/hr$. How much area will it irrigate in 30 minutes if $9cm$ of standing water is required.

Answer 1

Hint: With the dimensions of the canal, we can find the cross sectional area. Then we can find the distance of water flowing in 30 minutes by multiplying the time with the velocity of flow. Then we can find the volume by multiplying the cross sectional area with the distance travelled in 30 minutes. Now have the volume of the water and the area can be found out by dividing the volume with height of the water required.

Complete step-by-step answer:
We are given that the canal is $30dm$ wide $12dm$ deep. Now we can find the cross sectional area of the canal.
${A_{canal}} = width \times depth$
On substituting the values, we get
\[ \Rightarrow \]${A_{canal}} = 30dm \times 12dm$
On multiplication we get,
\[ \Rightarrow \]${A_{canal}} = 360d{m^2}$
Now we have the velocity of the water is $20km/hr$. Then the distance covered by the water in 30 minutes is given by,
$dis\tan ce = speed \times time$
On substituting values we get,
\[ \Rightarrow \]$dis\tan ce = 20km/hr \times 30\min $
We know that, $30\min = \dfrac{1}{2}hr$
\[ \Rightarrow \]$dis\tan ce = 20km/hr \times \dfrac{1}{2}hr$
On multiplication, we get the distance as.
\[ \Rightarrow \]$dis\tan ce = 10km$
We know that 1km is 10000 dm.
So we can write the distance as,
\[ \Rightarrow \]$dis\tan ce = 10 \times 10000dm = {10^6}dm$
Now we can find the volume of water
$Vol = {A_{canal}} \times dis\tan ce$
On substituting the values, we get
\[ \Rightarrow \]$Vol = 360d{m^2} \times {10^6}dm$
On multiplication we get,
\[ \Rightarrow \]$Vol = 360 \times {10^6}d{m^3}$
We know that $1dm = {10^{ - 1}}m$and hence, $1d{m^3} = {10^{ - 3}}{m^3}$. So, the volume will become,
\[ \Rightarrow \]$Vol = 360 \times {10^3}{m^3}$

For a land to irrigate, it must have 9cm of standing water. So the area that can be irrigated by the given volume of water can be found out by dividing the volume by the height of the water.
$ \Rightarrow Area = \dfrac{{vol}}{h}$
On substituting the values, we get
$ \Rightarrow Area = \dfrac{{360 \times {{10}^3}{m^3}}}{{9cm}}$
We know that$1cm = {10^{ - 1}}m$. Applying this relation, we get
$ \Rightarrow Area = \dfrac{{360 \times {{10}^3}{m^3}}}{{9 \times {{10}^{ - 1}}m}}$
On simplification, we get
$ \Rightarrow Area = 4 \times {10^5}{m^2}$
Thus, the area irrigated in 30 minutes is $4 \times {10^5}{m^2}$.

Note: We are assuming that the rate of flow or velocity of the water is constant and the canal is of uniform dimension. We are using 2 different units of time and 4 different units of length. We prefer to change all the units into SI units. It is important that the unit must be the same while doing some calculations. Here we assume that area is filled throughout with water with a uniform height which is not practically possible.