Question
Download Solution PDFA soil has a liquid limit of 60%, shrinkage limit of 20% and plastic limit of 30%. If it has natural moisture content of 40%, the liquidity index of the soil will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Liquidity index is given by (IL)
Liquidity index = \(\rm \frac{{natural\;moisture\;content - plastic\;limit}}{{liquid\;limit - plastic\;limit}}\)
\({{\rm{I}}_{\rm{L}}} = \frac{{{{\rm{W}}_{\rm{n}}} - {\rm{\;}}{{\rm{W}}_{\rm{p}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)
Calculation:
Given,
wL = 60 %
wP = 30 %
ws = 20 %
wn = 40 %
IP = wL - wP = 60 - 30 = 30%
\({{\rm{I}}_{\rm{L}}} = \frac{{{{\rm{W}}_{\rm{n}}} - {\rm{\;}}{{\rm{W}}_{\rm{p}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)
\(I_L= \;\frac{{40 - 30}}{{60 - 30}}\; = \;0.33\)
Consistency index is given by (IC)
\({{\rm{I}}_{\rm{C}}} = \frac{{{{\rm{W}}_{\rm{L}}} - {\rm{\;}}{{\rm{W}}_{\rm{n}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)
Last updated on Jun 4, 2025
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