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

Step 1: Recall the definition of liquidity index.
The liquidity index (\(LI\)) tells us where the natural water content of a soil sits between its plastic limit (\(PL\)) and liquid limit (\(LL\)). It is defined as
\[ LI = \frac{w - PL}{PI} \]
where \(w\) is the natural water content and \(PI = LL - PL\) is the plasticity index.

Step 2: Substitute the given values.
Here \(w = 30\%\), \(PI = 40\%\), and \(LI = 50\% = 0.50\) (as a fraction).
\[ 0.50 = \frac{30 - PL}{40} \]

Step 3: Solve for the plastic limit.
\[ 30 - PL = 0.50 \times 40 = 20 \]
\[ PL = 30 - 20 = 10 \]

Step 4: Check the result makes sense.
With \(PL = 10\%\), \(LL = PL + PI = 10 + 40 = 50\%\). The natural water content \(w = 30\%\) lies between \(PL = 10\%\) and \(LL = 50\%\), and indeed \(LI = (30-10)/40 = 0.5\), confirming the value is consistent.

Final Answer:
\[ \boxed{PL = 10.0\%} \]