A 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:

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OSSC JE Civil Mains Official Paper: (Held On: 16th July 2023)
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  1. 0.42
  2. 0.33
  3. 0.5
  4. 0.38

Answer (Detailed Solution Below)

Option 2 : 0.33
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Concept:

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%

\({{\rm{I}}_{\rm{L}}} = \frac{{{{\rm{W}}_{\rm{n}}} - {\rm{\;}}{{\rm{W}}_{\rm{p}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)

\(I_L= \;\frac{{40 - 30}}{{60 - 30}}\; = \;0.33\)

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Consistency index is given by (IC)

\({{\rm{I}}_{\rm{C}}} = \frac{{{{\rm{W}}_{\rm{L}}} - {\rm{\;}}{{\rm{W}}_{\rm{n}}}}}{{{{\rm{I}}_{\rm{P}}}}}\)

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