A canal is $ 300 \mathrm{~cm} $ wide and $ 120 \mathrm{~cm} $ deep. The water in the canal is flowing with a speed of $ 20 \mathrm{~km} / \mathrm{h} $. How much area will it irrigate in 20 minutes if $ 8 \mathrm{~cm} $ of standing water is desired?

Given:

A canal is \( 300 \mathrm{~cm} \) wide and \( 120 \mathrm{~cm} \) deep. The water in the canal is flowing with a speed of \( 20 \mathrm{~km} / \mathrm{h} \).

\( 8 \mathrm{~cm} \) of standing water is desired.

To do:

We have to find the area it will irrigate in 20 minutes.

Solution:

Width of the canal $b=300 \mathrm{~cm}$

$=0.3 \mathrm{~m}$

Depth of the canal $h=120 \mathrm{~cm}$

$=1.2 \mathrm{~m}$

Speed of the water flow $=20 \mathrm{~km} / \mathrm{hr}$

Flow of water in 20 minutes $l=8\ cm$

$=0.08\ m$

Therefore,

Volume of the water flown in the canal in one hour $=$ Width of the canal $\times$ Depth of the canal $\times$ Speed of the canal

$= 3 \times 1.2 \times 20 \times 1000\ m^3$

$= 72000\ m^3$

Volume of water flown in 20 minutes $=\frac{20}{60}\times72000$

$=24000\ m^3$

Height of the water in the field $=8 \mathrm{~cm}$

$=\frac{8}{100} \mathrm{~m}$

Area of the field irrigated $=$ Volume of the water $\div$ Height of the water in the field

$=\frac{24000}{\frac{8}{100}}$

$=\frac{24000 \times 100}{8}$

$=3000 \times 100$

$=300000 \mathrm{~m}^{2}$

The area will it irrigate in 20 minutes is $300000\ m^2$. 

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Updated on: 10-Oct-2022

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