Oregon Agricultural and Biological
Engineering Exam Practice Questions
And Correct Answers (Verified Answers)
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1. An agricultural and biological engineer is designing a drainage system
for a cultivated field where excessive water accumulation has reduced
crop productivity. Which soil property most directly influences the
rate at which water can move through the soil profile?
A. Soil color
B. Soil temperature only
C. Hydraulic conductivity
D. Soil organic matter color
Hydraulic conductivity describes the ability of soil to transmit water under
a hydraulic gradient and is therefore a fundamental parameter in
agricultural drainage and irrigation design.
2. A field has a soil water content of 0.30 m³/m³ and a permanent wilting
point of 0.12 m³/m³. What is the approximate volumetric water
content available to plants above the wilting point?
A. 0.12 m³/m³
B. 0.18 m³/m³
,C. 0.30 m³/m³
D. 0.42 m³/m³
Plant-available water over this simplified range is the difference between
the current water content and the permanent wilting point:
0.30−0.12=0.180.30-0.12=0.18 m³/m³.
3. A sprinkler irrigation system applies water at an average rate of 12
mm/h. If the crop root zone can accept water at a maximum
infiltration rate of 8 mm/h, what is the most likely result if the system
continues operating at 12 mm/h?
A. Increased infiltration efficiency without runoff
B. Potential runoff or surface ponding
C. Immediate reduction of soil porosity to zero
D. Complete prevention of evaporation
When application intensity exceeds the soil infiltration capacity, water can
accumulate at the surface, potentially producing ponding, runoff, and
inefficient irrigation.
4. An irrigation canal carries water at a flow rate of 0.50 m³/s. If the canal
cross-sectional flow area is 1.25 m², what is the average flow velocity?
A. 0.25 m/s
B. 0.40 m/s
C. 0.75 m/s
D. 1.25 m/s
Continuity gives Q=AVQ=AV. Therefore,
V=Q/A=0.50/1.25=0.40V=Q/A=0.50/1.25=0.40 m/s.
5. A centrifugal pump supplies irrigation water at 0.08 m³/s against a
total dynamic head of 20 m. If the pump efficiency is 70% and water
has a specific weight of 9.81 kN/m³, what is the approximate required
input power?
,A. 11.2 kW
B. 22.4 kW
C. 27.5 kW
D. 39.2 kW
Hydraulic power is Ph=γQH=(9.81)(0.08)(20)=15.70P_h=\gamma
QH=(9.81)(0.08)(20)=15.70 kW. Dividing by 0.70 gives an input power of
approximately 22.4 kW.
6. A drip irrigation system is designed to deliver water directly to the
root zone. Compared with many conventional surface irrigation
systems, the principal advantage of drip irrigation is:
A. Elimination of all maintenance requirements
B. Reduced application to non-target areas and improved control of water
delivery
C. Elimination of all soil evaporation
D. Unlimited flow without pressure regulation
Drip irrigation can deliver water close to the crop root zone, reducing
unnecessary wetting and improving control of application when the system
is properly designed and maintained.
7. An agricultural field requires a net irrigation depth of 40 mm. If the
irrigation system has an application efficiency of 80%, what gross
depth must be applied?
A. 32 mm
B. 50 mm
C. 64 mm
D. 80 mm
Gross application is calculated as net required depth divided by application
efficiency: 40/0.80=5040/0.80=50 mm.
, 8. A soil contains 0.25 m³ of water per cubic meter of soil. If the soil bulk
volume under consideration is 100 m³, what is the approximate
volume of water contained in the soil?
A. 4 m³
B. 20 m³
C. 25 m³
D. 40 m³
Volumetric water content equals water volume divided by bulk soil
volume. Therefore, water volume is 0.25(100)=250.25(100)=25 m³.
9. Soil bulk density is calculated from:
A. Soil mass divided by particle mass
B. Water volume divided by soil volume
C. Dry soil mass divided by total bulk soil volume
D. Void volume divided by particle volume
Bulk density is the oven-dry mass of soil divided by its total bulk volume,
including both solids and pore spaces.
10. A soil has a particle density of 2.65 Mg/m³ and a bulk density of
1.59 Mg/m³. What is its approximate total porosity?
A. 20%
B. 40%
C. 60%
D. 75%
Total porosity can be estimated from n=1−ρb/ρsn=1-\rho_b/\rho_s. Thus,
n=1−(1.59/2.65)=0.40n=1-(1.59/2.65)=0.40, or approximately 40%.
11. A soil sample contains a high proportion of clay. Compared with
a coarse sandy soil, the clay soil generally has:
Engineering Exam Practice Questions
And Correct Answers (Verified Answers)
Plus Rationale 2027 Q&A| Instant
Download Pdf.
1. An agricultural and biological engineer is designing a drainage system
for a cultivated field where excessive water accumulation has reduced
crop productivity. Which soil property most directly influences the
rate at which water can move through the soil profile?
A. Soil color
B. Soil temperature only
C. Hydraulic conductivity
D. Soil organic matter color
Hydraulic conductivity describes the ability of soil to transmit water under
a hydraulic gradient and is therefore a fundamental parameter in
agricultural drainage and irrigation design.
2. A field has a soil water content of 0.30 m³/m³ and a permanent wilting
point of 0.12 m³/m³. What is the approximate volumetric water
content available to plants above the wilting point?
A. 0.12 m³/m³
B. 0.18 m³/m³
,C. 0.30 m³/m³
D. 0.42 m³/m³
Plant-available water over this simplified range is the difference between
the current water content and the permanent wilting point:
0.30−0.12=0.180.30-0.12=0.18 m³/m³.
3. A sprinkler irrigation system applies water at an average rate of 12
mm/h. If the crop root zone can accept water at a maximum
infiltration rate of 8 mm/h, what is the most likely result if the system
continues operating at 12 mm/h?
A. Increased infiltration efficiency without runoff
B. Potential runoff or surface ponding
C. Immediate reduction of soil porosity to zero
D. Complete prevention of evaporation
When application intensity exceeds the soil infiltration capacity, water can
accumulate at the surface, potentially producing ponding, runoff, and
inefficient irrigation.
4. An irrigation canal carries water at a flow rate of 0.50 m³/s. If the canal
cross-sectional flow area is 1.25 m², what is the average flow velocity?
A. 0.25 m/s
B. 0.40 m/s
C. 0.75 m/s
D. 1.25 m/s
Continuity gives Q=AVQ=AV. Therefore,
V=Q/A=0.50/1.25=0.40V=Q/A=0.50/1.25=0.40 m/s.
5. A centrifugal pump supplies irrigation water at 0.08 m³/s against a
total dynamic head of 20 m. If the pump efficiency is 70% and water
has a specific weight of 9.81 kN/m³, what is the approximate required
input power?
,A. 11.2 kW
B. 22.4 kW
C. 27.5 kW
D. 39.2 kW
Hydraulic power is Ph=γQH=(9.81)(0.08)(20)=15.70P_h=\gamma
QH=(9.81)(0.08)(20)=15.70 kW. Dividing by 0.70 gives an input power of
approximately 22.4 kW.
6. A drip irrigation system is designed to deliver water directly to the
root zone. Compared with many conventional surface irrigation
systems, the principal advantage of drip irrigation is:
A. Elimination of all maintenance requirements
B. Reduced application to non-target areas and improved control of water
delivery
C. Elimination of all soil evaporation
D. Unlimited flow without pressure regulation
Drip irrigation can deliver water close to the crop root zone, reducing
unnecessary wetting and improving control of application when the system
is properly designed and maintained.
7. An agricultural field requires a net irrigation depth of 40 mm. If the
irrigation system has an application efficiency of 80%, what gross
depth must be applied?
A. 32 mm
B. 50 mm
C. 64 mm
D. 80 mm
Gross application is calculated as net required depth divided by application
efficiency: 40/0.80=5040/0.80=50 mm.
, 8. A soil contains 0.25 m³ of water per cubic meter of soil. If the soil bulk
volume under consideration is 100 m³, what is the approximate
volume of water contained in the soil?
A. 4 m³
B. 20 m³
C. 25 m³
D. 40 m³
Volumetric water content equals water volume divided by bulk soil
volume. Therefore, water volume is 0.25(100)=250.25(100)=25 m³.
9. Soil bulk density is calculated from:
A. Soil mass divided by particle mass
B. Water volume divided by soil volume
C. Dry soil mass divided by total bulk soil volume
D. Void volume divided by particle volume
Bulk density is the oven-dry mass of soil divided by its total bulk volume,
including both solids and pore spaces.
10. A soil has a particle density of 2.65 Mg/m³ and a bulk density of
1.59 Mg/m³. What is its approximate total porosity?
A. 20%
B. 40%
C. 60%
D. 75%
Total porosity can be estimated from n=1−ρb/ρsn=1-\rho_b/\rho_s. Thus,
n=1−(1.59/2.65)=0.40n=1-(1.59/2.65)=0.40, or approximately 40%.
11. A soil sample contains a high proportion of clay. Compared with
a coarse sandy soil, the clay soil generally has:
