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 m^{3}/s had a return period of 30 years. Find the probability of flood equal to or greater than 165 m^{3}/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 = q

^{n}=(\(\frac{29}{30})^{25}\)=0.4285≅0.43

2. From past flood data of an area, it was found that a peak flood of 80 m^{3}/s had a return period of 10 years. Find the probability of flood equal to or greater than 80 m^{3}/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-q

^{n}= \(1-(\frac{9}{10})^8\)=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-q

^{5}=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-q

^{n}=1-q

^{18}⇒q

^{18}=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 m^{3}/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 m^{3}/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 m

^{3}/s flood, m = 5

So, probability of occurrence of flood ≥ 80 m

^{3}/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 m^{3}/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 m^{3}/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 m

^{3}/s.

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