0
If $W_{L}$=51%, $W_{p}$=32% $W_{s}$=19% Soil specimen shrinks from volume 42.5 cc at $w_{L}$ to 26.8 cc of shrinkage limit.Determine gravity
0  upvotes
0

enter image description here

Reduction in volume R=42.5-26.8

=15.7cc

=15.7gm

% reduction in water content =51 to 1g=32% w=\frac{M_{w}}{M_{s}}=0.32

$M_{s}=\frac{M_{w}}{0.32}=\frac{15.7}{0.32}$=49.06gm

Initial $M_{w}$ in soil sample=0.51 Ms=$0.51 \times 49.06$

=25.02gm Volume of water initially

$V_{s}$=volume of solids=42.5-25.05

=17.478 cc

G$\frac{M_{s}}{V_{s}\times P_{w}} =\frac{49.06}{17.478 \times 1}$

=2.806

0  upvotes
Please log in to add an answer.

Next up

Read More Questions

If you are looking for answer to specific questions, you can search them here. We'll find the best answer for you.

Search

Study Full Subject

If you are looking for good study material, you can checkout our subjects. Hundreds of important topics are covered in them.

Know More