PEC EPE Engineering Geology Numerical Questions

PEC EPE Engineering Geology Numerical Questions

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Other PEC EPE Exam Past Paper Questions

    Calculate the volume of a sedimentary rock layer with a length of 20 meters, width of 10 meters, and thickness of 2 meters.

    Solution:

    Volume = Length × Width × Thickness

    Volume = 20 m × 10 m × 2 m = 400 cubic meters

    Answer: Volume of the sedimentary rock layer = 400 cubic meters

    A sample of soil has a porosity of 40%. Calculate the volume of void spaces in a 10 cubic meter soil sample.

    Solution:

    Volume of void spaces = Porosity × Total volume

    Volume of void spaces = 0.40 × 10 cubic meters = 4 cubic meters

    Answer: Volume of void spaces in the soil sample = 4 cubic meters

    Calculate the weight of a rock sample with a density of 2,500 kg/m³ and a volume of 5 cubic meters.

    Solution:

    Weight = Density × Volume

    Weight = 2,500 kg/m³ × 5 cubic meters = 12,500 kg

    Answer: Weight of the rock sample = 12,500 kg

    A geological survey team measures the strike of a rock layer to be 30 degrees and the dip to be 20 degrees. Calculate the true dip of the rock layer.

    Solution:

    True dip = Dip / Cos(Strike)

    True dip = 20 degrees / Cos(30 degrees) ≈ 23.09 degrees

    Answer: True dip of the rock layer ≈ 23.09 degrees

    A rock sample weighs 500 N in air and 300 N when fully submerged in water. Calculate the specific gravity of the rock.

    Solution:

    Specific Gravity = Weight in air / (Weight in air – Weight in water)

    Specific Gravity = 500 N / (500 N – 300 N) = 500 N / 200 N = 2.5

    Answer: Specific Gravity of the rock ≈ 2.5

    The water content of a soil sample is 15%. Calculate the dry unit weight of the soil if its saturated unit weight is 18 kN/m³.

    Solution:

    Dry unit weight = Saturated unit weight × (1 – Water content)

    Dry unit weight = 18 kN/m³ × (1 – 0.15) = 18 kN/m³ × 0.85 ≈ 15.3 kN/m³

    Answer: Dry unit weight of the soil ≈ 15.3 kN/m³

    Calculate the void ratio of a soil sample if its porosity is 35%.

    Solution:

    Void ratio = Porosity / (1 – Porosity)

    Void ratio = 0.35 / (1 – 0.35) = 0.35 / 0.65 ≈ 0.5385

    Answer: Void ratio of the soil sample ≈ 0.5385

    A soil sample has a total volume of 20 cubic meters and a void ratio of 0.6. Calculate the volume of void spaces in the sample.

    Solution:

    Volume of void spaces = Void ratio × Total volume

    Volume of void spaces = 0.6 × 20 cubic meters = 12 cubic meters

    Answer: Volume of void spaces in the soil sample ≈ 12 cubic meters

    A geological formation is composed of limestone (density = 2,700 kg/m³) overlain by shale (density = 2,000 kg/m³). Calculate the average density of the formation if the thickness of limestone is 15 meters and the thickness of shale is 10 meters.

    Solution:

    Average density = (Density of limestone × Thickness of limestone + Density of shale × Thickness of shale) / (Total thickness)

    Average density = (2,700 kg/m³ × 15 m + 2,000 kg/m³ × 10 m) / (15 m + 10 m) ≈ 2,333.33 kg/m³

    Answer: Average density of the geological formation ≈ 2,333.33 kg/m³

    The porosity of a soil sample is 20% and its dry unit weight is 16 kN/m³. Calculate its saturated unit weight if the water content is 12%.

    Solution:

    Saturated unit weight = Dry unit weight / (1 – Water content)

    Saturated unit weight = 16 kN/m³ / (1 – 0.12) = 16 kN/m³ / 0.88 ≈ 18.18 kN/m³

    Answer: Saturated unit weight of the soil ≈ 18.18 kN/m³

    A rock sample weighs 600 N in air and 500 N when fully submerged in water. Calculate the apparent specific gravity of the rock.

    Solution:

    Apparent Specific Gravity = Weight in air / (Weight in air – Buoyant force)

    Buoyant force = Weight in air – Weight in water

    Buoyant force = 600 N – 500 N = 100 N

    Apparent Specific Gravity = 600 N / (600 N – 100 N) = 600 N / 500 N = 1.2

    Answer: Apparent Specific Gravity of the rock = 1.2

    A soil sample has a total volume of 15 cubic meters and a void ratio of 0.7. Calculate the volume of solids in the sample.

    Solution:

    Volume of solids = Total volume / (1 + Void ratio)

    Volume of solids = 15 cubic meters / (1 + 0.7) = 15 cubic meters / 1.7 ≈ 8.82 cubic meters

    Answer: Volume of solids in the soil sample ≈ 8.82 cubic meters

    Calculate the weight of a rock sample with a density of 2,400 kg/m³ and a volume of 4 cubic meters.

    Solution:

    Weight = Density × Volume

    Weight = 2,400 kg/m³ × 4 cubic meters = 9,600 kg

    Answer: Weight of the rock sample = 9,600 kg

    A geological survey team measures the strike of a rock layer to be 45 degrees and the dip to be 30 degrees. Calculate the true dip of the rock layer.

    Solution:

    True dip = Dip / Cos(Strike)

    True dip = 30 degrees / Cos(45 degrees) ≈ 42.43 degrees

    Answer: True dip of the rock layer ≈ 42.43 degrees

    A rock sample weighs 800 N in air and 700 N when fully submerged in water. Calculate the specific gravity of the rock.

    Solution:

    Specific Gravity = Weight in air / (Weight in air – Buoyant force)

    Buoyant force = Weight in air – Weight in water

    Buoyant force = 800 N – 700 N = 100 N

    Specific Gravity = 800 N / (800 N – 100 N) = 800 N / 700 N ≈ 1.14

    Answer: Specific Gravity of the rock ≈ 1.14

    The water content of a soil sample is 20%. Calculate the dry unit weight of the soil if its saturated unit weight is 20 kN/m³.

    Solution:

    Dry unit weight = Saturated unit weight × (1 – Water content)

    Dry unit weight = 20 kN/m³ × (1 – 0.20) = 20 kN/m³ × 0.80 = 16 kN/m³

    Answer: Dry unit weight of the soil ≈ 16 kN/m³

    Calculate the void ratio of a soil sample if its porosity is 25%.

    Solution:

    Void ratio = Porosity / (1 – Porosity)

    Void ratio = 0.25 / (1 – 0.25) = 0.25 / 0.75 ≈ 0.3333

    Answer: Void ratio of the soil sample ≈ 0.3333

    A soil sample has a total volume of 25 cubic meters and a void ratio of 0.5. Calculate the volume of void spaces in the sample.

    Solution:

    Volume of void spaces = Void ratio × Total volume

    Volume of void spaces = 0.5 × 25 cubic meters = 12.5 cubic meters

    Answer: Volume of void spaces in the soil sample ≈ 12.5 cubic meters

    A geological formation is composed of sandstone (density = 2,200 kg/m³) overlain by shale (density = 2,100 kg/m³). Calculate the average density of the formation if the thickness of sandstone is 12 meters and the thickness of shale is 8 meters.

    Solution:

    Average density = (Density of sandstone × Thickness of sandstone + Density of shale × Thickness of shale) / (Total thickness)

    Average density = (2,200 kg/m³ × 12 m + 2,100 kg/m³ × 8 m) / (12 m + 8 m) ≈ 2,160 kg/m³

    Answer: Average density of the geological formation ≈ 2,160 kg/m³

    The porosity of a soil sample is 30% and its dry unit weight is 18 kN/m³. Calculate its saturated unit weight if the water content is 10%.

    Solution:

    Saturated unit weight = Dry unit weight / (1 – Water content)

    Saturated unit weight = 18 kN/m³ / (1 – 0.10) = 18 kN/m³ / 0.90 ≈ 20 kN/m³

    Answer: Saturated unit weight of the soil ≈ 20 kN/m³