This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Flood Frequency Studies – Set 2”.
1. From past flood data of a catchment in India, it was found that a peak flood of 165 m3/s had a return period of 30 years. Find the probability of flood equal to or greater than 165 m3/s occurring zero times in the next 25 years?
a) 0.29
b) 0.43
c) 0.57
d) 0.96
View Answer
Explanation: Given return period, T= 30 years
Probability of occurrence, p = \(\frac{1}{T} = \frac{1}{30}\)
Probability of non-occurrence, q = 1-p = \(1-\frac{1}{30}=\frac{29}{30}\)
؞ Probability of not occurring in next 25 years = qn=(\(\frac{29}{30})^{25}\)=0.4285≅0.43
2. From past flood data of an area, it was found that a peak flood of 80 m3/s had a return period of 10 years. Find the probability of flood equal to or greater than 80 m3/s occurring at least once in the next 8 years?
a) 0.38
b) 0.43
c) 0.57
d) 0.9
View Answer
Explanation: Given return period, T= 10 years
Probability of occurrence, p = 1/T = 1/10
Probability of non-occurrence, q=1-p=1-\(\frac{1}{10}=\frac{9}{10}\)
؞ Probability of occurring at least once in next 8 years =1-qn = \(1-(\frac{9}{10})^8\)=0.5695≅0.57
3. What is the probability of a 15 year flood occurring once in the next 15 years?
a) 0.38
b) 0.65
c) 0.78
d) 0.93
View Answer
Explanation: Given return period, T= 15 years
Probability of occurrence, p = 1/T = 1/15
Probability of non-occurrence, q=1-p=1-\(\frac{1}{15}=\frac{14}{15}\)
∴Probability of occurence of flood once in 15 years=\(_1^{15}C.p^1.q^{15-1}=15.(\frac{1}{15})^1.(\frac{14}{15})^{14}\)
=0.3806≅0.38
4. The probability of a flood greater than or equal to a 20 year flood occurring thrice in the next three years is k x 10-3. What is the value of ‘k’?
a) 0.0125
b) 0.125
c) 1.25
d) 12.5
View Answer
Explanation: Given return period, T= 20 years
Probability of occurrence, p = 1/T = 1/20
Probability of non-occurrence, q=1-p=1-\(\frac{1}{20}=\frac{19}{20}\)
∴Probability of occurence of flood three times in 3 years=\(_3^3C.p^3.q^{3-3}=1.(\frac{1}{20})^3.(\frac{19}{20})^0\)
=0.000125=0.125*10-3
5. A culvert is design for a service life of 5 years for a flood peak of return period 120 years. What is the hydrologic risk of this design?
a) 4%
b) 19%
c) 81%
d) 96%
View Answer
Explanation: Given return period, T= 120 years
Probability of occurrence, p = \(\frac{1}{T} = \frac{1}{120}\)
Probability of non-occurrence, q=1-p=1-\(\frac{1}{120}=\frac{119}{120}\)
∴Hydrologic risk for a 5 year design life=1-q5=1-\((\frac{119}{120})^5\)=0.0409≅4%
6. A spillway is to be designed for a peak flood for the next 18 years. The acceptable risk in the design is 12%. What should be the return period of the flood to be used in the design?
a) 120 years
b) 130 years
c) 140 years
d) 150 years
View Answer
Explanation: Given, risk = 12%
⇒0.12=1-qn=1-q18⇒q18=0.88 ⇒q=\(\sqrt[18]{0.88}\)=0.9929233
Now, p=1-q=1-0.9929233=0.0070767
∴Retrun period, T=\(\frac{1}{p}=\frac{1}{0.0070767}\)=141.3≅140 years
7. The flood peak data (in m3/s) of a catchment outlet for 7 years is – 80, 105, 92, 77, 85, 75, 90. What is the return period (in months) of a flood of magnitude greater than or equal to 80 m3/s? Use Weibull’s formula.
a) 16.8
b) 18.7
c) 19.2
d) 24
View Answer
Explanation: Arranging the given data in decreasing order – 105, 92, 90, 85, 80, 77, 75.
Total number of data, N = 7
Rank number of 80 m3/s flood, m = 5
So, probability of occurrence of flood ≥ 80 m3/s, p=\(\frac{m}{N+1}=\frac{5}{7+1}=\frac{5}{8}\)
∴Return period=\(\frac{1}{p}=\frac{1}{\frac{5}{8}}=\frac{8}{5}\)=1.6 years=19.2 months
8. The flood peak data (in m3/s) of a catchment outlet for 11 years is – 45, 51, 37, 66, 58, 40, 57, 62, 37, 48, 31. What is the flood value (in m3/s) which has a return period of 2 years? Use Weibull’s formula.
a) 37
b) 48
c) 51
d) 62
View Answer
Explanation: Arranging the given data in decreasing order – 66, 62, 58, 57, 51, 48, 45, 40, 37, 37, 31.
Total number of data, N = 11
Return period, T = 2 years
From Weibull’s formula, \(T=\frac{N+1}{m} ⇒m=\frac{N+1}{T}=\frac{11+1}{2}=\frac{12}{2}\)=6
The rank of 6 corresponds to a flood of 48 m3/s.
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