In a sieve analysis of soil, the total weight of soil taken was 500 g. The mass of soil retained over 4.75 mm sieve was 100 g, mass retained over 2 mm sieve was 150 g, and the mass retained over 425-micron sieve was 200 g. The effective size of the soil will be:
- A75 microns
- B2 mm
- C425 micronsCorrect
- D4.75 mm
Solution
Concept
Effective size D10 is the sieve size through which 10% of the soil passes.
Formula
% finer = 100 - (cumulative mass retained / total mass) × 100
Step-by-step solution
1. Total mass = 500 g. 2. Cumulative retained on 4.75 mm = 100 g, so % finer = 100 - 20 = 80%. 3. Cumulative retained on 2 mm = 100 + 150 = 250 g, so % finer = 100 - 50 = 50%. 4. Cumulative retained on 425 micron = 250 + 200 = 450 g, so % finer = 100 - 90 = 10%. 5. The size at which 10% passes is 425 microns, so D10 = 425 microns.
Explanation
Effective size means D10, the particle size at which 10% of the soil is finer. Here exactly 10% of the soil passes the 425 micron sieve. So the effective size is 425 microns.
Why the other options are wrong
A: 75 microns is not a sieve used in the data, and the 10% finer point falls at 425 microns. B: 2 mm has 50% finer, so it gives D50. D: 4.75 mm has 80% finer, not 10%.
Shortcut
Add the retained masses until the total reaches 90% (450 g). That sieve size is D10.
Common mistake
Using individual % retained instead of cumulative % finer, or confusing D10 with D60.
Correct answer: C) 425 microns