✅ Answer: B – Resistance of slopes against failure
Focuses on equilibrium and shear strength.
✅ Answer: A – Excessive shear stress
Failure when τ > τ_max = c + σ′tanφ.
✅ Answer: B – Circular failure
Typical for cohesive soils.
✅ Answer: B – Layered or jointed rock slopes
Failure along weak planes.
✅ Answer: A – Shear strength / shear stress
Measures margin against failure.
✅ Answer: C – >1
Usually F.S. = 1.3–1.5 for long-term.
✅ Answer: B – Slope stability analysis
Divides slip surface into slices for computation.
✅ Answer: A – Fellenius method
Simplified method assuming circular failure.
✅ Answer: A – Inter-slice shear forces
Improves accuracy over Fellenius.
✅ Answer: A – ΣM_resisting / ΣM_driving
Standard approach for rotational failures.
✅ Answer: A – Minimum factor of safety
Represents probable failure surface.
✅ Answer: B – Cohesion + friction
Shear strength of soil.
✅ Answer: A – Soil weight component parallel to slope
Downslope component of gravity.
✅ Answer: B – Decreases factor of safety
Steeper slope = higher driving force.
✅ Answer: B – Increases stability
Higher shear strength = more resistance.
✅ Answer: B – Reduces stability
Raises pore pressure → lowers σ′.
✅ Answer: D – r_u = u / σ_total
Used to estimate pore pressure effects.
✅ Answer: A – N_s = c / (γH)
Non-dimensional slope stability parameter.
✅ Answer: A – Taylor’s chart
Graphical design aid for slope stability.
✅ Answer: D – F.S. = N_cr / N_s = (c_req / c)
Relates actual stability to critical.
✅ Answer: B – 1.3–1.5
For long-term stability.
✅ Answer: A – Rapid drawdown in reservoirs
Pore pressure remains high = reduced stability.
✅ Answer: B – Infinite extent, parallel planes
Used for long uniform slopes.
✅ Answer: A – F.S. = tanφ / tanβ
Stability depends purely on slope angle.