Question

A soil sample has following properties:

Natural water content = 30 %
Plasticity index = 40 %
Liquidity index = 50 %

The estimated plastic limit (in %) of the soil is (rounded off to one decimal place).

Hint:

Use \(LI = (w-PL)/PI\) and rearrange to find \(PL\) directly from the given water content, plasticity index and liquidity index.

Answer 1

Correct Answer: 10

This is a straightforward application of the liquidity index formula, but instead of jumping straight to $PL$, let's think about what liquidity index physically means and rearrange the formula that way.

The liquidity index measures, as a fraction of the plasticity range, how far the soil's current water content $w$ has moved above the plastic limit:

\[ LI = \frac{w - PL}{LL - PL} = \frac{w - PL}{PI} \]

A liquidity index of $LI = 0.50$ means the natural water content sits exactly halfway between $PL$ and $LL$. So the gap between $w$ and $PL$ equals half the plasticity index:

\[ w - PL = LI \times PI = 0.50 \times 40 = 20\% \]

We are told $w = 30\%$, so:

\[ PL = w - 20 = 30 - 20 = 10\% \]

Let's summarize:

• $LI \times PI$ directly gives the gap between natural water content and plastic limit, a quicker route than solving the fraction equation term by term.
• Here that gap works out to 20%, so the plastic limit sits 20 percentage points below the natural water content of 30%.
• Check: $LL = PL + PI = 10 + 40 = 50\%$, and $w = 30\%$ indeed lies at the midpoint of $[10\%, 50\%]$, consistent with $LI = 0.5$.

So the estimated plastic limit is $PL = 10.0\%$.