Question

9 A saturated specimen of clay was immersed in
mercury and displaced volume was 21.8 cc. The
weight of the sample was 32.2 gm. After oven
drying for 48 hours, weight reduced to 20.2 gm
while volume came down to 11.6 cc. The
shrinkage limit of the soil is​

Answer 1

Answer:

Explanation:

=> Here, It is given that

-> dry weight of soil, W_d = 20.2 gm

-> The initial total weight of soil in stage-I , W1 = 32.2 gm

-> the initial volume of the soil in stage-I , V1 = 21.8 cc

-> volume of soil in dry state or Volume reduction after oven drying, V_d = 11.6 cc

-> γ_w = the density of water = 1 g / cm^3

-> The shrinkage limit of the soil ω_s = ?

=> The volume of the soil at shrinkage limit is equal to the total volume of oven-dried soil. So, the shrinkage limit of the soil :

ω_s =$\frac{(W1 - W_d)}{W_d} - \frac{(V1 - V_d) γ_w }{W_d}]* 100$

=$\frac{(32.2 - 20.2)}{20.2} - \frac{(21.8 - 11.6)* 1 }{20.2}]* 100$

= 8.91 %

Thus,  the shrinkage limit of the soil is 8.91 %.