Question

water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10km/h.
How much area will it irrigate in 30 minutes; if 8cm standing water is needed?


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Answer 1

Solution: This means that canal is cuboidal in shape. In such cases assume length to be speed of water flowing (because water covers that much length to come out of the canal)

Volume of water coming out of canal

→ lbh → 6 m × 1.5 m × 10,000 m/h

→ 90,000 m³/h

Now this much water has been flowing for 30 mins or ½ hr.

→ 90,000 m³/h × ½ h

→ 45,000 m³

So 45,000 m³ of water has been stagnant on field. Now Area × height = Volume

→ Area × 8/100 m = 45,000 m³

→ Area = 45,000 m³ × 100/8 m

→ Area = 5,62,500 m² or 0.5,625 km²

Answer 2

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Given :

• Length (L) = 10 km/h = 10000 m/s
• Breadth (B) = 6 m
• Height or Depth (H) = 1.5 m
• Time = 30 mins = ½ hr

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To Find :

• Area that will be irrigated

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Solution :

Formula for Volume is :

$\large \bigstar {\boxed{\sf{Volume \: = \: LBH}}} \\ \\ \implies {\sf{Volume \: = \: 6 \: \times \: 1.5 \: \times \: 10000}} \\ \\ \implies {\sf{Volume \: = \: 9 \: \times \: 10000}} \\ \\ \implies {\sf{Volume \: = \: 90000 \: m^3s^{-1}}} \\ \\ \small {\underline{\sf{\: \: \: \: \dag \: \:Convert \: Volume \: to \: m^3 \: from \: m^3s^{-1} \: \: \: \:}}} \\ \\ \implies {\sf{Volume \: = \: 90000 \: \times \: \dfrac{1}{2}}} \\ \\ \implies {\sf{Volume \: = \: 45000 \: m^3}}$

As we know that,

• Area = Length * Breadth
• Volume = Length * Breadth * Height
• Volume = Area * height

If Standing Water needed is 8 cm

Then, height (H) = 8 cm = 8/100 m

Use relation of Area and volume :

$\large \star {\boxed{\sf{Volume \: = \: Area \: \times \: H}}} \\ \\ \implies {\sf{45000 \: = \: Area \: \times \: \dfrac{8}{100}}} \\ \\ \implies {\sf{Area \: = \: \dfrac{45000 \: \times \: 100}{8}}} \\ \\ \implies {\sf{Area \: = \: \dfrac{4500000}{8}}} \\ \\ \implies {\sf{Area \: = \: 562500 \: m^2}}$

$\therefore$ Area is 5,62,500 m²