Question:
A soil yielded a maximum dry unit weight of \(18\;\text{kN/m}^3\) at a moisture content of \(16\%\) during a Standard Proctor Test. What is the degree of saturation of the soil if its specific gravity is \(2.7\)? (Take \(\gamma_w=10\;\text{kN/m}^3\)).
A soil yielded a maximum dry unit weight of \(18\;\text{kN/m}^3\) at a moisture content of \(16\%\) during a Standard Proctor Test. What is the degree of saturation of the soil if its specific gravity is \(2.7\)? (Take \(\gamma_w=10\;\text{kN/m}^3\)).
Show Hint
For compaction numericals,
\[
\gamma_d=\frac{G\gamma_w}{1+e}
\]
followed by
\[
w=\frac{Se}{G}.
\]
These two equations are sufficient to determine the degree of saturation.
Updated On: Jul 23, 2026
- \(98.42\%\)
- \(86.40\%\)
- \(84.32\%\)
- \(75.71\%\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The dry unit weight is related to the void ratio by
\[
\boxed{
\gamma_d
=
\frac{G\gamma_w}{1+e}
}
\]
Also,
\[
\boxed{
w=\frac{Se}{G}
}
\]
where
\[
S=\text{Degree of saturation}.
\]
Step 1: Calculate the void ratio. Given, \[ \gamma_d=18\;\text{kN/m}^3, \] \[ G=2.7, \] \[ \gamma_w=10\;\text{kN/m}^3. \] Using \[ 18 = \frac{2.7\times10}{1+e} \] \[ 1+e = \frac{27}{18} = 1.5 \] \[ e=0.5 \]
Step 2: Calculate the degree of saturation. Given, \[ w=16\%=0.16 \] Using \[ w=\frac{Se}{G} \] \[ S = \frac{wG}{e} = \frac{0.16\times2.7}{0.5} = 0.864 \] Hence, \[ S=86.4\% \] Therefore, \[ \boxed{\text{Degree of Saturation}=86.40\%.} \] Thus, the correct option is \[ \boxed{(B)\;86.40\%.} \]
Step 1: Calculate the void ratio. Given, \[ \gamma_d=18\;\text{kN/m}^3, \] \[ G=2.7, \] \[ \gamma_w=10\;\text{kN/m}^3. \] Using \[ 18 = \frac{2.7\times10}{1+e} \] \[ 1+e = \frac{27}{18} = 1.5 \] \[ e=0.5 \]
Step 2: Calculate the degree of saturation. Given, \[ w=16\%=0.16 \] Using \[ w=\frac{Se}{G} \] \[ S = \frac{wG}{e} = \frac{0.16\times2.7}{0.5} = 0.864 \] Hence, \[ S=86.4\% \] Therefore, \[ \boxed{\text{Degree of Saturation}=86.40\%.} \] Thus, the correct option is \[ \boxed{(B)\;86.40\%.} \]
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