Water in a canal $30\ dm$ wide and $12\ dm$ deep, is flowing with a velocity of $100\ km$ per hour. How much area will it irrigate in $30$ minutes if $8\ cm$ of standing water is desired?

Given:

Water in a canal $30\ dm$ wide and $12\ dm$ deep, is flowing with a velocity of $100\ km$ per hour.

To do:

We have to find the area it will irrigate in $30$ minutes if $8\ cm$ of standing water is desired.

Solution:

Width of the canal $(b) = 30\ dm$

$= 3\ m$

Depth of the canal $(h) = 12\ dm$

$= 1.2\ m$

Speed of the water $= 100\ km/hr$

Length of the water flow in (30 minutes) $\frac{1}{2}\ hr =\frac{100 \times 1000}{2}$

$=50000 \mathrm{~m}$

Therefore,

Volume of water $=l b h$

$=3 \times 1.2 \times 50000 \mathrm{~m}^{3}$

The water standing in the field $=8\ cm$

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

Area of the field $=\frac{\text { Volume of water }}{\text { Depth of water }}$

$=\frac{3 \times 1.2 \times 50000}{\frac{8}{100}}$

$=\frac{3 \times 12 \times 50000 \times 100}{10 \times 8}$

$=9 \times 250000$

$=2250000 \mathrm{~m}^{2}$

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Updated on: 10-Oct-2022

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