Liquid limit flow curve — in a Casagrande liquid limit test, water content was 70% at 10 blows and 20% at 100 blows. Using the standard log N versus w linear relation, the liquid limit (at 25 blows) is approximately:

Difficulty: Medium

Correct Answer: 50%

Explanation:


Introduction / Context:
The Casagrande liquid limit determination relies on an empirical linear relation between water content w and the logarithm (base 10) of the number of blows N required to close the groove by 12.7 mm. Given two reliable (N, w) data points, one can interpolate or extrapolate to find w at N = 25, the reported liquid limit LL.


Given Data / Assumptions:

  • w1 = 70% at N1 = 10 blows.
  • w2 = 20% at N2 = 100 blows.
  • Assume w = a - b * log10(N) is valid over this range.


Concept / Approach:

Because log10(10) = 1 and log10(100) = 2, the change in w over one log cycle is w2 - w1 = 20 - 70 = -50%. The slope b is therefore 50% per log decade. Compute w at N = 25, where log10(25) ≈ 1.39794. Linear interpolation on the log scale provides the answer without plotting.


Step-by-Step Solution:

Write w = a - b * log10(N).From points: 70 = a - b1 and 20 = a - b2, so b = 50 and a = 120.At N = 25: log10(25) ≈ 1.39794.Compute w = 120 - 50 * 1.39794 ≈ 120 - 69.897 ≈ 50.1%.


Verification / Alternative check:

Graphically, plotting the two points on w versus log10(N) and reading at N = 25 would yield essentially 50%, matching the calculation. Minor rounding differences do not change the selected option.


Why Other Options Are Wrong:

35% and 65% do not satisfy the linear log relation between the given points. “None of these” is false because 50% matches the computed LL. 42% would correspond to a different slope or data pair.


Common Pitfalls:

Interpolating linearly on N instead of on log10(N); rounding log10(25) too coarsely; mixing percent units between decimal and percent forms.


Final Answer:

50%

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