A soil has specific gravity of its solids equal to 2.65. The mass density of water is $1000 k g / m^{3}$ . Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/ $m^{3}$ , round off to 1 decimal place) would be _____ .
2094.862

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Solution

The correct option is A 2094.862
$G_{s} = 2.65$

$ρ_{w} = 1000 k g / m^{3}$

$η_{a} = 0$

w = 10% = 0.10

$γ_{d} = [\frac{(1 - η_{a}) G_{s} ρ_{w}}{1 + w G_{s}}]$

$= \frac{[(1 - 0) \times 2.65 \times 1000]}{1 + 0.1 \times 2.65}$

$= 2094.862 k g / m^{3}$

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A soil has specific gravity of its solids equal to 2.65. The mass density of water is 1000 kg/m3. Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m3, round off to 1 decimal place) would be _____ .2094.862

The correct option is A 2094.862 Gs=2.65 ρw=1000 kg/m3 ηa=0 w = 10% = 0.10 γd=[(1−ηa)Gsρw1+wGs] =[(1−0)×2.65×1000]1+0.1×2.65 = 2094.862 kg/m3

A soil has specific gravity of its solids equal to 2.65. The mass density of water is $1000 k g / m^{3}$ . Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/ $m^{3}$ , round off to 1 decimal place) would be _____ .
2094.862

The correct option is A 2094.862
$G_{s} = 2.65$

$ρ_{w} = 1000 k g / m^{3}$

$η_{a} = 0$

w = 10% = 0.10

$γ_{d} = [\frac{(1 - η_{a}) G_{s} ρ_{w}}{1 + w G_{s}}]$

$= \frac{[(1 - 0) \times 2.65 \times 1000]}{1 + 0.1 \times 2.65}$

$= 2094.862 k g / m^{3}$