Allied & ICAR Exam Bank • ALLIED_1213 NAT • 1 Mark

Section 5: Irrigation and Drainage Engineering

Canal Design and Silt Theories

Subtopic: Lacey's Silt Factor

Question Statement:

For an alluvium bed material having an average particle diameter $d = 0.36\text{ mm}$, Lacey's silt factor $f = 1.76 \sqrt{d}$ is:

Answer Key:

1.056

Step-by-Step Solution & Explanation

$f = 1.76 \sqrt{0.36} = 1.76 \times 0.6 = 1.056$.
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