A soil has dry unit weight of $15.5 k N / m^{2}$ , specific gravity of 2.65 and degree of saturation of 72%. Considering the unit weight of water as $10 k N / m^{3}$ , the water content of the soil (in % round off to two decimal places) is
19.26

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Solution

The correct option is A 19.26
Given,

$γ_{d} = 15.5 k N / m^{3}$

G = 2.65

S = 72% = 0.72

$γ_{w} = 10 k N / m^{3}$

We know that,

$γ_{d} = \frac{G}{1 + e} γ_{w}$

$15.5 = \frac{2.65}{1 + e} \times 10$

$e = 0.709$

$A l s o, S e = w G$

$w = \frac{S e}{G} = \frac{0.72 \times 0.709}{2.65}$

$w = 0.1926$

$\Rightarrow W a t e r c o n t e n t (w) = 19.26$ %

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A soil has dry unit weight of 15.5 kN/m2, specific gravity of 2.65 and degree of saturation of 72%. Considering the unit weight of water as 10 kN/m3 , the water content of the soil (in % round off to two decimal places) is 19.26

The correct option is A 19.26Given, γd=15.5kN/m3 G = 2.65 S = 72% = 0.72 γw=10kN/m3 We know that, γd=G1+eγw 15.5=2.651+e×10 e=0.709 Also, Se=wG w=SeG=0.72×0.7092.65 w=0.1926 ⇒Water content (w)=19.26 %

A soil has dry unit weight of $15.5 k N / m^{2}$ , specific gravity of 2.65 and degree of saturation of 72%. Considering the unit weight of water as $10 k N / m^{3}$ , the water content of the soil (in % round off to two decimal places) is
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