In its natural condition a soil sample has a mass of 1.980 kg and a volume of $0.001 m^{3}$ . After being completely dried in an oven; the mass of the sample is 1.800 kg. Specific gravity is 2.7. Unit weight of water is $10 k N / m^{3}$ . The degree of saturation of soil is

0.65

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0.70

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0.54

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0.61

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Solution

The correct option is C 0.54
$γ_{t} = \frac{1.980 \times 9.81 \times 10^{- 3}}{0.001} k N / m^{3}$

$w = \frac{1.980 - 1.800}{1.800} = 0.1$

We know that,

$γ_{t} = \frac{G (1 + w)}{1 + e} \cdot γ_{w}$

$\Rightarrow \frac{1.980 \times 9.81 \times 10^{- 3}}{0.001} = \frac{2.7 (1 + 0.1)}{1 + \frac{w G}{S}} \cdot 9.81$
$[∵ e = \frac{w G}{S}]$

$\Rightarrow S = 0.54$

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In its natural condition a soil sample has a mass of 1.980 kg and a volume of 0.001 m3. After being completely dried in an oven; the mass of the sample is 1.800 kg. Specific gravity is 2.7. Unit weight of water is 10 kN/m3. The degree of saturation of soil is

The correct option is C 0.54 γt=1.980×9.81×10−30.001kN/m3 w=1.980−1.8001.800=0.1 We know that, γt=G(1+w)1+e⋅γw ⇒1.980×9.81×10−30.001=2.7(1+0.1)1+wGS⋅9.81 [∵e=wGS] ⇒S=0.54

In its natural condition a soil sample has a mass of 1.980 kg and a volume of $0.001 m^{3}$ . After being completely dried in an oven; the mass of the sample is 1.800 kg. Specific gravity is 2.7. Unit weight of water is $10 k N / m^{3}$ . The degree of saturation of soil is