The field capacity of a certain soil is 20-Take dry unit weight of soil as 15kN-m3 and unit weight of water as 10kN-m3-

Depth of water applied, =γ_dγ_w× d×[Field capacity Moisture content before irrigation] ⇒ 50=1510× d×[0.20 0.10] d=50×1015×0.1= 333.33 mm≃ 333 mm nbsp; ...

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Hydrology & Irrigation

Classes of Soil Water and Different Soil Moisture Levels

The field cap...

Question

The field capacity of a certain soil is 20% and the moisture content of the soil before irrigation is 10%. What is the approximate depth upto which the soil will be wetted with an application of 50 mm of water?
[Take dry unit weight of soil as $15 k N / m^{3}$ and unit weight of water as $10 k N / m^{3}$ ]

250 mm

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333 mm

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367 mm

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433 mm

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Solution

The correct option is B 333 mm
Depth of water applied,

$= \frac{γ_{d}}{γ_{w}} \times d \times [Field capacity - Moisture content before irrigation]$

$\Rightarrow 50 = \frac{15}{10} \times d \times [0.20 - 0.10]$

$d = \frac{50 \times 10}{15 \times 0.1} = 333.33 m m ≃ 333 m m$

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The field capacity of a certain soil is 20% and the moisture content of the soil before irrigation is 10%. What is the approximate depth upto which the soil will be wetted with an application of 50 mm of water? [Take dry unit weight of soil as 15kN/m3 and unit weight of water as 10kN/m3]

The correct option is B 333 mm Depth of water applied, =γdγw×d×[Field capacity - Moisture content before irrigation] ⇒50=1510×d×[0.20−0.10] d=50×1015×0.1=333.33mm≃333mm

The correct option is B 333 mm
Depth of water applied,

$= \frac{γ_{d}}{γ_{w}} \times d \times [Field capacity - Moisture content before irrigation]$

$\Rightarrow 50 = \frac{15}{10} \times d \times [0.20 - 0.10]$

$d = \frac{50 \times 10}{15 \times 0.1} = 333.33 m m ≃ 333 m m$