What is the dry unit weight of a clay soil when the void ratio of a sample there of is 0.50, the degree of saturation is 70% and the specific gravity of soil grains is 2.7? Take the value of $γ_{w}$ to be $9.81 k N / m^{3}$ .

$13.651 k N / m^{2}$

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19.951 k N / m^{2}

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23.651 k N / m^{2}

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29.95 k N / m^{2}

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Solution

The correct option is B $19.951 k N / m^{2}$
(b)
$B u l k u n i t w e i g h t; γ_{b} = \frac{(G + S e) γ_{w}}{1 + e}$
$= \frac{2.7 + 0.4 \times 0.5}{1 + 0.5}$

$= 19.47 k N / m^{3}$

$∴$ Se = wG

water content
$w = \frac{0.7 \times 0.5}{2.7} = 0.12962$

$γ_{d} = \frac{γ_{b}}{1 + w} = \frac{19.947}{1 + 0.12962}$

$= 17.658 k N / m^{3}$
Note : Dry unit weight has been asked but no answer is matching. But bulk unit weight is matching.

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What is the dry unit weight of a clay soil when the void ratio of a sample there of is 0.50, the degree of saturation is 70% and the specific gravity of soil grains is 2.7? Take the value of γw to be 9.81kN/m3.

The correct option is B 19.951kN/m2(b) Bulk unit weight ; γb=(G+Se)γw1+e =2.7+0.4×0.51+0.5 =19.47kN/m3 ∴ Se = wG water content w=0.7×0.52.7=0.12962 γd=γb1+w=19.9471+0.12962 =17.658kN/m3 Note : Dry unit weight has been asked but no answer is matching. But bulk unit weight is matching.

What is the dry unit weight of a clay soil when the void ratio of a sample there of is 0.50, the degree of saturation is 70% and the specific gravity of soil grains is 2.7? Take the value of $γ_{w}$ to be $9.81 k N / m^{3}$ .