The field capacity of a certain soil is $15 %$ and the moisture content of the soil before irrigation is $8 %$ . If the dry unit weight of soil is $15.3 k N / m^{3}$ , the depth upto which the soil profile will be wetted with an application of $60 m m$ of water is ________ mm.
549.58

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Solution

The correct option is A 549.58
Depth of water applied,
$= \frac{γ_{d}}{γ_{w}} \times d \times (Field capacity -Moisture content before irrigation)$

$\Rightarrow 60 = \frac{15.3}{9.81} \times d \times (0.15 - 0.08)$

$\Rightarrow d = 549.58 m m$

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The field capacity of a certain soil is 15% and the moisture content of the soil before irrigation is 8%. If the dry unit weight of soil is 15.3 kN/m3, the depth upto which the soil profile will be wetted with an application of 60 mm of water is ________ mm.549.58

The correct option is A 549.58Depth of water applied, =γdγw×d×(Field capacity -Moisture content before irrigation) ⇒60=15.39.81×d×(0.15−0.08) ⇒d=549.58mm

The field capacity of a certain soil is $15 %$ and the moisture content of the soil before irrigation is $8 %$ . If the dry unit weight of soil is $15.3 k N / m^{3}$ , the depth upto which the soil profile will be wetted with an application of $60 m m$ of water is ________ mm.
549.58

The correct option is A 549.58
Depth of water applied,
$= \frac{γ_{d}}{γ_{w}} \times d \times (Field capacity -Moisture content before irrigation)$

$\Rightarrow 60 = \frac{15.3}{9.81} \times d \times (0.15 - 0.08)$

$\Rightarrow d = 549.58 m m$