Question

The water in a canal, $30$ dm wide and $12$ dm deep, is flowing with a velocity of $20$ km per hour. How much area will it irrigate in $30$ minutes, if $9$ cm of standing water is desired?

Answer 1

Hint: From the given question, we have to first convert the units from dm to m. Given the width and depth of the canal, we have to calculate the water flowing in the canal irrigated with respect to the time and speed given and then find the area irrigated by the water.

Formula used: The conversion of the unit from decimetre into centimetre:
$1$ dm =$10$ cm.
The conversion of the unit from centimetre to metre:
$100$ cm =$1$ m.
The conversion of the unit from kilometre to metre:
$1$ km =$1000$ m.

Complete step-by-step solution:
By the given data, we have the width and depth or height of the canal in the units of decimetre.
First we have to convert the units of width and depth or height into the unit of metre.
Given that the width of the canal =$30$ dm =$30 \times 10 = 300$ cm = $3$ m.
Now, the depth of the canal = $12$ dm =$12 \times 10 = 120$ cm = $1.2$ m.
It is given that the water is flowing with a velocity $20$ km/hour=$20 \times 1000 = 20000$ m/hour.
The length of cuboid is equal to the distance travelled in $30$ minutes with the speed of $20000$ m\hour = $20000 \times \dfrac{{30}}{{60}}$ = $20000 \times \dfrac{1}{2}$ = $10000$ m.
Now, the volume of the canal is given by the formula, \[{\text{Volume}} = {\text{length}} \times {\text{width}} \times {\text{depth}}\]
So, the volume of water to be used for irrigation, ${\text{V}} = 10000 \times 3 \times 1.2$${{\text{m}}^3}$.
Therefore, ${\text{V}} = 36000$${{\text{m}}^{\text{3}}}$ .
The water accumulated in the field forms a cuboid of base area equal to the area of the field and height equal to $\dfrac{9}{{100}}{\text{m}}$.
Let ${\text{x}}{{\text{m}}^{\text{2}}}$ be the area irrigated in $\dfrac{1}{2}{\text{hour}}\left( { = 30\min } \right)$. Then, we obtain
${\text{x}} \times \dfrac{9}{{100}} = 36000$
Solve it for x,
 $ \Rightarrow {\text{x}} = 36000 \times \dfrac{{100}}{9}$
Simplifying we get,
$ \Rightarrow {\text{x}} = 4000 \times 100$
Multiplying the terms,
$ \Rightarrow {\text{x}} = 400000$${{\text{m}}^2}$

$\therefore $ The canal irrigates $400000$${{\text{m}}^2}$ area in $\dfrac{1}{2}{\text{hour}}\left( { = 30\min } \right)$

Note: We have remembered that the process of conversion depends on the specific situation and the intended purpose. A conversion factor is used to change the units of a measured quantity without changing its value. The unit bracket method of unit conversion consists of a fraction in which the denominator is equal to the numerator, but they are in different units. Because of the identity property of multiplication, the value of a quantity will not change as long as it is multiplied by one. Also, if the numerator and denominator of a fraction are equal to each other, then the fraction is equal to one. So as long as the numerator and denominator of the fraction are equivalent, they will not affect the value of the measured quantity.