The soil has a liquid limit of 50% and plastic limit of 25%, respectively. If the volumetric shrinkages at the liquid limit and plastic limit are 50% and 30%, respectively, determine the dry volume of soil, if the volume at liquid limit is 1 ml.
Try it first: pick an option, then read the explanation.
Detailed Explanation
Volumetric shrinkage relates a soil's volume at some wetter state to its final dry volume once all further water loss stops changing the volume (i.e., once the shrinkage limit has been passed). It is defined as the percentage reduction in volume relative to the dry volume, not relative to the original (wetter) volume.
Given: volumetric shrinkage at the liquid limit, Vs = 50%; volume at the liquid limit, VL = 1 ml.
Vs = (VL − Vd)/Vd ⇒ 0.5 = (1 − Vd)/Vd ⇒ 0.5 Vd = 1 − Vd ⇒ 1.5 Vd = 1
Why "0.667 ml" Is Correct
Solving 1.5 Vd = 1 gives Vd = 1/1.5 = 0.6666 ≈ 0.667 ml. Note that the plastic-limit volumetric shrinkage (30%) given in the question isn't actually needed for this particular calculation — the dry volume can be found directly from the liquid-limit data alone.
Key Point
Volumetric shrinkage is always expressed relative to the final dry volume (the denominator is Vd), not the starting volume — a common source of sign/setup errors.
Key Concepts for Students
- Volumetric shrinkage, Vs = (V − Vd)/Vd, can be evaluated at any reference state (liquid limit, plastic limit, etc.) as long as the volume at that state and the dry volume are known or solvable.
- This question also tests whether you can identify which given values are actually needed — the plastic-limit shrinkage percentage here is extra information, not required for this specific calculation.
- The dry volume, once reached at or below the shrinkage limit, does not decrease further no matter how much more the sample is dried.
- This relationship is the same volumetric-shrinkage definition used throughout the shrinkage-limit test to compute the shrinkage ratio and specific gravity of solids.