Difficulty: Medium
Correct Answer: 18.2%
Explanation:
Introduction / Context:
Determining water content from pycnometer readings requires careful volume–mass bookkeeping. When a wet soil is used in the pycnometer and the specific gravity of solids (Gs) is known, we can split the wet mass into dry soil and soil water by using displacement concepts and the known capacity of the pycnometer.
Given Data / Assumptions:
Concept / Approach:
When the wet soil is placed in the pycnometer and then topped up with water, the added water occupies the remaining volume. The difference between the water-only fill and the added water gives the volume of the wet soil (solids plus soil water). Using Gs, we relate dry mass to volume of solids. Two equations in unknowns Wd and Ww then yield the moisture content w = Ww / Wd.
Step-by-Step Solution:
Verification / Alternative check:
Check volumes: V_s = 220 / 2.75 = 80 cm^3; V_w (soil) = 40 cm^3; total 80 + 40 = 120 cm^3, consistent with the capacity difference method. Hence the calculation is self-consistent.
Why Other Options Are Wrong:
24.2% and 53.8% result from incorrect separation of wet mass or misuse of Gs. 'None of these' is incorrect because a consistent solution gives 18.2%.
Common Pitfalls:
Forgetting that water density equals 1 g/cm^3 (so grams convert to cm^3); mixing up added water with total water; using formulas meant for dry-soil pycnometer readings directly on wet-soil data without adjusting volumes.
Final Answer:
18.2%
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