Question

The laboratory tests on a soil sample yields the following results: natural moisture content = 18% , liquid limit = 60%, plastic limit = 25%, percentage of clay sized fraction = 25%. The liquidity index and activity (as per the expression proposed by Skempton) of the soil, respectively, are

• A. -0.2 and 1.4
• B. 0.2 and 1.4
• C. -1.2 and 0.714
• D. 1.2 and 0.714

Answer 1

The correct option is A -0.2 and 1.4

Given $\Rightarrow w_L$= 60 % = 0.6

$w_p$ = 25 % = 0.25

% of clay sized fraction 25% = 0.25

$w_n=18$ % = 0.18

$I_L=\frac{w_n-w_p}{w_L-w_p}$

$=\frac{0.18-0.25}{0.6-0.25}=-0.2$


Activity =
$\frac{l_p}{\%ofclaysizedfraction}$

$=\frac{w_L-w_p}{0.25}=\frac{0.6-0.25}{0.25}=1.4$