As a part of control check during construction of an earthen dam, a core of compacted soil was cut and measured its weight and volume as 1800 gm and 1000 cc respectively. Its water content was determined and found to be 10%. If the dam gets fully saturated due to shower of rains, the change in water content without change in volume is in %. (Assume G=2.65)

13.4

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15.2

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19.8

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7.06

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Solution

The correct option is B 15.2
$W_{s o i l} = 1800 g m$

$V = 1000 c c$

$w = 10 %$

$γ = \frac{1800}{1000} = 1.8 g / c c, γ_{d} = \frac{γ}{1 + w} = \frac{1.8}{1 + 0.1} = 1.636 g / c c$

$γ_{d} = \frac{G γ_{w}}{1 + e} \Rightarrow e = \frac{G γ_{w}}{γ_{d}} - 1$

$∴ e = \frac{2.65 \times 1}{1.636} - 1 = 0.62$

$∴$ Dams gets fully saturated without change in volume

$e . S_{r} = w_{s a t} . G$

$W_{s a t} = \frac{0.62}{2.65} = 23.396 % = 23.4 %$

Change in water content $= 23.4 % - 10 % = 13.4 %$

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The correct option is B 15.2Wsoil=1800gm V=1000cc w=10% γ=18001000=1.8g/cc, γd=γ1+w=1.81+0.1=1.636g/cc γd=Gγw1+e⇒e=Gγwγd−1 ∴e=2.65×11.636−1=0.62 ∴ Dams gets fully saturated without change in volume e.Sr=wsat.G Wsat=0.622.65=23.396%=23.4% Change in water content =23.4%−10%=13.4%