Question

A soil has coefficient of uniformity of 6, particle size corresponding to 60% finer and 30% finer are respectively 0.85 and 0.35. Then the coefficient of curvature of soil is ..........

• A. 1.017
• B. 2.907
• C. 2.420
• D. 6.000

Hint:

Use: $C_u = \dfrac{D_{60}}{D_{10}}$ and $C_c = \dfrac{(D_{30})^2}{D_{10} \cdot D_{60}}$ to classify soil based on gradation.

Answer 1

The Correct Option is A

Given:
$C_u = 6$
$D_{60} = 0.85$
$D_{30} = 0.35$
We use the formula for coefficient of curvature:
$C_c = \dfrac{(D_{30})^2}{D_{10} \cdot D_{60}}$
We need to find $D_{10}$ first using the formula for coefficient of uniformity:
$C_u = \dfrac{D_{60}}{D_{10}}$
$⇒ D_{10} = \dfrac{D_{60}}{C_u} = \dfrac{0.85}{6} = 0.1417$
Now, calculate $C_c$:
$C_c = \dfrac{(0.35)^2}{0.1417 \cdot 0.85} = \dfrac{0.1225}{0.12045} \approx 1.017$