The dry density of a soil is 1.5 g/cc. If the saturation water content were 50% then its saturated density and submerged density would, respectively, be

1.5 g/cc and 1.0 g/cc

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2.0 g/cc and 1.0 g/cc

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2.25 g/cc and 1.25 g/cc

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2.50 g/cc and 1.50 g/cc

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Solution

The correct option is C 2.25 g/cc and 1.25 g/cc
(c)
$γ_{s a t} = γ_{d} (1 + w) = 1.5 \times 1.5$

= 2.25 g/cc

$γ_{s u b} = γ_{s a t} - γ_{w} = 2.25 - 1.0$

= 1.25 g/cc

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The dry density of a soil is 1.5 g/cc. If the saturation water content were 50% then its saturated density and submerged density would, respectively, be

The correct option is C 2.25 g/cc and 1.25 g/cc(c) γsat=γd(1+w)=1.5×1.5 = 2.25 g/cc γsub=γsat−γw=2.25−1.0 = 1.25 g/cc

The correct option is C 2.25 g/cc and 1.25 g/cc
(c)
$γ_{s a t} = γ_{d} (1 + w) = 1.5 \times 1.5$

= 2.25 g/cc

$γ_{s u b} = γ_{s a t} - γ_{w} = 2.25 - 1.0$

= 1.25 g/cc