GATE AG 2025 • Q47 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Soil Mechanics

Subtopic: Index properties of soils

Question Statement:

An undisturbed soil sample: volume 100 cm³, void ratio 0.60. Oven-dry mass reduces from 185 g to 165 g (no volume change). Specific gravity 2.64. Find the degree of saturation (%). (Rounded to 2 decimal places)

Official Answer Key:

52.70 to 53.40

Step-by-Step Solution & Explanation

Dry (solid) mass = 165 g, so Vs = Ms/(Gs×ρw) = 165/2.64 = 62.5 cm³. Vv = e×Vs = 0.60×62.5 = 37.5 cm³. Mass of water = 185−165 = 20 g ⇒ Vw = 20 cm³ (ρw=1 g/cm³). Degree of saturation S = Vw/Vv×100 = 20/37.5×100 = 53.33%.

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