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## Q. 2.1

Following are the results of a sieve analysis. Make the necessary calculations and draw a particle-size distribution curve.

 U.S. sieve no. Mass of soil retained on each sieve (g) 4 0 10 40 20 60 40 89 60 140 80 122 100 210 200 56 Pan 12

## Verified Solution

The following table can now be prepared.

 U.S. sieve (1) Opening (mm) (2) Mass retained on each sieve (g) (3) Cumulative mass retained above Percent  each sieve (g) (4) Percent finer ${}^a$ (5) 4 4.75 0 0 100 10 2.00 40 0 + 40 = 40 94.5 20 0.850 60 40 + 60 = 100 86.3 40 0.425 89 100 + 89 = 189 74.1 60 0.250 140 189 + 140 = 329 54.9 80 0.180 122 329 + 122 = 451 38.1 100 0.150 210 451 + 210 = 661 9.3 200 0.075 56 661 + 56 = 717 1.7 Pan – 12 717 + 12 = 729 = ΣM 0

$\frac{{}^a \sum M-col.4}{\sum M}\times 100=\frac{729-col .4}{729}\times 100$
The particle-size distribution curve is shown in Figure 2.28.