Question

A saturated specimen of undisturbed inorganic clay has a volume of 19.2 cm and mass 32.5g . After oven-drying at 105° C for 24 hours, the mass reduces to 20.9 g. For the soil in the natural state, find (i)W, (ii) G (iii) e, (iv) Ysal and (v) Yd
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Answer 1

Answer:

Water Content = weight of water/ weight of soil solid = (weight of wet mass - weight of dry mass)/ weight of dry mass = (417-225)/225 = 0.85. In percentage, water content = 0.8533*100 = 85.33%.

The volume of soil solid =  mass of soil solid /(specific gravity of soil solid * density of water) = 225 g /(2.70*1 g per cubic cm) = 83.33 cubic cm.

Volume of void = total volume - volume of soil solid = 276 - 83.33 = 192.67...

Explanation:

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Answer 2

Given:

A saturated specimen of undisturbed inorganic clay has a volume of 19.2 cm and a mass of 32.5g. After oven-drying at 105° C for 24 hours, the mass reduces to 20.9 g.

Find:

find (i)W, (ii) G (iii) e, (iv) Ysal and (v) Yd

​ Solution:

Water Content = weight of water/ weight of soil solid

= (weight of wet mass - weight of dry mass)/ weight of dry mass

$= (417-225)/225 = 0.85.$

In percentage, water content  $=0.8533*100 = 85.33%.$

The volume of soil solid

=  mass of soil solid /(specific gravity of soil solid * density of water) $= 225 g /(2.70*1 g per cubic cm) = 83.33 cubic cm.$

The Volume of void = total volume -  the volume of soil solid

$= 276 - 83.33 = 192.67$

Weight of can + wet soil = 150.63 gms,

Weight of can + dry soil = 131.58 gms,

Weight of can = 26.48 gms.