Question:
If an infinite slope of clay at a depth \(5\) m has cohesion of \(1\) t/m\(^2\) and unit weight of \(2\) t/m\(^3\), then the stability number will be
If an infinite slope of clay at a depth \(5\) m has cohesion of \(1\) t/m\(^2\) and unit weight of \(2\) t/m\(^3\), then the stability number will be
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For slope stability problems,
\[
\boxed{
N_s=\frac{c}{\gamma H}
}
\]
Higher cohesion gives a larger stability number and hence a more stable slope.
Updated On: Jul 23, 2026
- \(0.1\)
- \(0.2\)
- \(0.3\)
- \(0.4\)
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The Correct Option is A
Solution and Explanation
Concept:
The stability number is defined as
\[
\boxed{
N_s=\frac{c}{\gamma H}
}
\]
where
\[
c=\text{Cohesion},
\]
\[
\gamma=\text{Unit weight of soil},
\]
\[
H=\text{Height (or depth) of slope}.
\]
Step 1: Write the given data. \[ c=1\text{ t/m}^2 \] \[ \gamma=2\text{ t/m}^3 \] \[ H=5\text{ m} \]
Step 2: Calculate the stability number. \[ N_s = \frac{1}{2\times5} = \frac{1}{10} = 0.1 \] Hence, \[ \boxed{N_s=0.1} \] Therefore, the correct option is \[ \boxed{(A)\;0.1.} \]
Step 1: Write the given data. \[ c=1\text{ t/m}^2 \] \[ \gamma=2\text{ t/m}^3 \] \[ H=5\text{ m} \]
Step 2: Calculate the stability number. \[ N_s = \frac{1}{2\times5} = \frac{1}{10} = 0.1 \] Hence, \[ \boxed{N_s=0.1} \] Therefore, the correct option is \[ \boxed{(A)\;0.1.} \]
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