# Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes, if 8 cm of standing water is needed for irrigation?

Last updated date: 20th Mar 2023

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Hint: Find out the total volume of water flowing in 10 minutes with the given speed and then equate it with the final volume of the irrigated field required.

Given, the width and depth of the canal are 6 m and 1.5 m respectively. And the water is flowing in the canal at the speed of 4 km/hr (4000 m/hr).

Length of the water column thus formed in 10 minutes (i.e. $\dfrac{1}{6}$ hour) $ = \dfrac{1}{6} \times 4000 = \dfrac{{2000}}{3}$m.

Volume of water flowing in $\dfrac{1}{6}$ hour $ = $Volume of cuboid of length $\dfrac{{2000}}{3}$m, width 6 m and depth 1.5 m.

Volume of cuboid=lwh

Volume of water flowing in $\dfrac{1}{6}$ hour $ = \dfrac{{2000}}{3} \times 6 \times 1.5 = 6000{m^3}$

Let $x{\text{ }}{{\text{m}}^2}$ is the area irrigated in $\dfrac{1}{6}$ hour, then we have:

$

\Rightarrow x \times \dfrac{8}{{100}} = 6000, \\

\Rightarrow x = 750 \times 100, \\

\Rightarrow x = 75000{\text{ }}{{\text{m}}^2} \\

$

Thus, the area needed is $75000{\text{ }}{{\text{m}}^2}$.

Note: Volume of water is conserved in the above scenario. The volume of water flown from the canal in the given time period is equal to the volume of water standing on the field.

Given, the width and depth of the canal are 6 m and 1.5 m respectively. And the water is flowing in the canal at the speed of 4 km/hr (4000 m/hr).

Length of the water column thus formed in 10 minutes (i.e. $\dfrac{1}{6}$ hour) $ = \dfrac{1}{6} \times 4000 = \dfrac{{2000}}{3}$m.

Volume of water flowing in $\dfrac{1}{6}$ hour $ = $Volume of cuboid of length $\dfrac{{2000}}{3}$m, width 6 m and depth 1.5 m.

Volume of cuboid=lwh

Volume of water flowing in $\dfrac{1}{6}$ hour $ = \dfrac{{2000}}{3} \times 6 \times 1.5 = 6000{m^3}$

Let $x{\text{ }}{{\text{m}}^2}$ is the area irrigated in $\dfrac{1}{6}$ hour, then we have:

$

\Rightarrow x \times \dfrac{8}{{100}} = 6000, \\

\Rightarrow x = 750 \times 100, \\

\Rightarrow x = 75000{\text{ }}{{\text{m}}^2} \\

$

Thus, the area needed is $75000{\text{ }}{{\text{m}}^2}$.

Note: Volume of water is conserved in the above scenario. The volume of water flown from the canal in the given time period is equal to the volume of water standing on the field.

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