Environmental Engineering (Test 3)

Gate Exam : Ce Civil Engineering

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Environmental Engineering

Environmental Engineering
| Environmental Engineering |
Q.1
Match the given water properties in Group-I to the given titrants shown in Group-II

A. P − 1,Q − 2,R − 3,S − 4
B. P − 2,Q − 1,R − 4,S − 3
C. P − 3,Q − 4,R − 1,S − 2
D. P − 4,Q − 3,R − 2,S − 1
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.2
A 1 hour rainfall of 10 cm has return period of 50 year. The probability that 1 hour of rainfall 10 cm or more will occur in each of two successive years is
A. 0.04
B. 0.02
C. 0.2
D. 0.0004
Answer : Option D
Explaination / Solution:

Return period of rainfall, T = 50 years
Probability of occurrence once in 50 years, p = 1/50 = 0.02
Probability of occurrence in each of 2 successive years  = p= (0.02)2 = 0.0004

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Q.3
An isohyet is a line joining points of
A. Equal temperature
B. Equal humidity
C. Equal rainfall depth
D. Equal evaporation
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.4
Identical surcharges are placed at ground surface at sites X and Y , with soil conditions shown alongside and water table at ground surface. The silty clay layers at X and Y are identical. The thin sand layer at Y is continuous and freedraining with a very large discharge capacity. If primary consolidation at X is estimated to complete in 36 months, what would be the corresponding time for completion of primary consolidation at Y ?

A. 2.25 months
B. 4.5 months
C. 9 months
D. 36 months
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.5
In compacted cylindrical bituminous mix, VMA = 15%(void mineral aggregate); Vv = 4.5% (air void) The magnitude of VFB (void filled bituminous) is 
A. 24
B. 30
C. 54
D. 70
Answer : Option D
Explaination / Solution:

VFB = VB/Vv = ((15-4.5)/15)×100% = 70%
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Q.6
Total suspended particulate matter (TSP) concentration in ambient air is to be measured using a high volume sampler. The filter used for this purpose had an initial dry weight of 9.787g. The filter was mounted in the sampler and the initial air flow rate through the filter was set at 1.5min/m3. Sampling continued for 24 hours. The airflow after 24 hours was measured to be 1.4 min/m3. The dry weight of the filter paper after 24 hour sampling was 10.283g. Assuming a linear decline in the air flow rate during sampling, what is the 24 hour average TSP concentration in the ambient air?
A. 59.2 μg/m3
B. 118.6 μg/m3
C. 237.5 μg/m3
D. 574.4 μg/m3
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.7
Chlorine gas (8 mg/L as Cl2) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2 mg/L (as Cl2) and 7.5, respectively, what is the concentration of residual OCl- ions in the water? Assume that the chlorine gas added to the water is completely converted to HOCl and OCl-. Atomic Weight of Cl: 35.5

A. 1.408 × 10-5 moles/L
B. 2.817 × 10-5 moles/L
C. 5.634 × 10-5 moles/L
D. 1.127 × 10-5 moles/L
Answer : Option B
Explaination / Solution:
No Explaination.


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Q.8
The ordinates of a 2-h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 m3/s. Use trapezoidal rule for numerical integration, if required.
A storm of 6.6 cm occurs uniformly over the catchment in 3 hours. If Φ - index is equal to 2 mm/h and base flow is 5 m3/s, what is the peak flow due to the storm?
A. 41.0 m3/s
B. 43.4 m3/s
C. 53.0 m3/s
D. 56.2 m3/s
Answer : Option C
Explaination / Solution:
No Explaination.


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Q.9
Elevation and temperature data for places are tabulated below

Based on this data, lapse rate can be referred as 
A. Super-adiabatic
B. Sub-adiabatic
C. Neutral
D. Inversion
Answer : Option A
Explaination / Solution:

Ambient lapse rate = ((21.25 - 15.70)/(444 - 4))×1000 = 12.6 oC/km[>9.8 oC/km]
When the ambient lapse rate exceeds the adiabatic lapse rate, the ambient lapse rate is said to be super adiabatic.
 

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Q.10
Anaerobically treated effluent has MPN of total coliform as 106 / mL. After chlorination, the MPN value declines to 102 / mL. The percent removal (%R) and log removal (logR) of total coliform MPN is
A. %R = 99.90; logR = 4
B. %R = 99.90; logR = 2
C. %R = 99.99; logR = 4
D. %R = 99.99; logR = 2
Answer : Option D
Explaination / Solution:
No Explaination.


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CE Civil Engineering