Question 1 |

A flood control structure having an expected life of n years is designed by
considering a flood of return period T years. When T=n , and n\rightarrow \infty , the
structure's hydrologic risk of failure in percentage is .
(round off to one decimal place)

25.2 | |

68.4 | |

78.2 | |

63.5 |

Question 1 Explanation:

Risk of failure =1-q^n=1-(1-p)^n=1-\left ( 1-\frac{1}{T} \right )^n

For T=n\rightarrow \infty

Risk of failure =1-\frac{1}{e}=0.632

% risk of failure = 0.632 x 100=63.2%

For T=n\rightarrow \infty

Risk of failure =1-\frac{1}{e}=0.632

% risk of failure = 0.632 x 100=63.2%

Question 2 |

Superpassage is a canal cross-drainage structure in which

natural stream water flows with free surface below a canal | |

natural stream water flows under pressure below a canal | |

canal water flows with free surface below a natural stream | |

canal water flows under pressure below a natural stream |

Question 2 Explanation:

Cross-section of a super passage

Question 3 |

Muskingum method is used in

hydrologic reservoir routing | |

hydrologic channel routing | |

hydraulic channel routing | |

hydraulic reservoir routing |

Question 3 Explanation:

Muskingum method is used in hydrological channel routing

Question 4 |

In a homogeneous unconfined aquifer of area 3.00 km^2, the water table was at an elevation
of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm^2), the water
table rose to 103.20 m. After this recharge, ground water pumping took place and the
water table dropped down to 101.020 m. The volume of ground water pumped after the
natural recharge, expressed (in Mm^2 and round off to two decimal places), is ______.

0.5 | |

1 | |

1.5 | |

2.5 |

Question 4 Explanation:

\begin{aligned} V_R&=0.9Mm^3\\ V&=3 \times (103.2-102)\\ y_s \; \text{or}\; y_R&=\frac{VR}{V}=\frac{0.9}{3.6}\\ \text{Now,}\;\; y_s&=\frac{V_D}{V}\\ V_D&=\frac{0.9}{3.6}[3 \times (103.2-101.2)]\\ V_D&=1.5Mm^3 \end{aligned}

Question 5 |

The probability that the annual maximum flood discharge will exceed 25000 m^3/s, at least once in next 5 years is found to be 0.25. The return period of this flood event (in years, round off to 1 decimal place) is ____

12.2 | |

16.4 | |

17.9 | |

20.6 |

Question 5 Explanation:

\begin{aligned} \text{Risk} &=1-q^n \\ 0.25&=1-q^5 \\ q&=0.944087 \\ 1-p&=0.944087 \\ 1-\frac{1}{T}&=0.944087 \\ T&\simeq 17.9 \; \text{years} \end{aligned}

Question 6 |

A culvert is designed for a flood frequency of 100 years and a useful life of 20 years. The risk involved in the design of the culvert (in percentage, up to two decimal places) is______

9.45 | |

18.21 | |

27.47 | |

36.36 |

Question 6 Explanation:

Risk = The probabality of a flood to occur at least once in n-successive years.

\begin{aligned} \therefore \;\; \text{Risk }&=1-q^{n} \\ &=1-\left ( 1-P \right )^{n} \\ &=1-\left ( 1-\frac{1}{T} \right )^{n} \\ &=1-\left ( 1-\frac{1}{100} \right )^{20} \\ &=1-\left ( 0.99 \right )^{20} \\ &=0.18209=18.209\% \end{aligned}

\begin{aligned} \therefore \;\; \text{Risk }&=1-q^{n} \\ &=1-\left ( 1-P \right )^{n} \\ &=1-\left ( 1-\frac{1}{T} \right )^{n} \\ &=1-\left ( 1-\frac{1}{100} \right )^{20} \\ &=1-\left ( 0.99 \right )^{20} \\ &=0.18209=18.209\% \end{aligned}

Question 7 |

For routing of flood in a given channel using the Muskingum method, two of the routing coefficients are estimated as C_{0}=-0.25 and C_{1}= 0.55. The value of the third coefficient C_{2} would be ______

0.2 | |

0.5 | |

0.7 | |

0.9 |

Question 7 Explanation:

In Muskingum flood routing method

\begin{aligned}C_{0}+C_{1}+C_{2}&=1 \\ \Rightarrow \;\; C_{2}&=1-\left ( -0.25 \right )-0.55=0.7 \end{aligned}

\begin{aligned}C_{0}+C_{1}+C_{2}&=1 \\ \Rightarrow \;\; C_{2}&=1-\left ( -0.25 \right )-0.55=0.7 \end{aligned}

Question 8 |

The type of flood routing (Group I) and the equation(s) used for the purpose (Group II) are given below.

The correct match is

The correct match is

P - 1; Q - 1, 2 & 3 | |

P - 1; Q - 1 & 2 | |

P - 1 & 2; Q - 1 | |

P - 1 & 2; Q - 1 & 2 |

Question 9 |

The Muskingum Model of routing a flood through a stream reach is expressed as O_{2}=K_{0}I_{2}+K_{1}I_{1}+K_{2}O_{1}, where K_{0}, K_{1} and K_{2} are the routing coefficients for the concerned reach, I_{1} and I_{2} are the inflows to reach, and O_{1} and O_{2} are the outflows from the reach corresponding to time step 1 and 2 respectively. The sum of K_{0}, K_{1} and K_{2} of the model is

-1 | |

-0.5 | |

0.5 | |

1 |

Question 10 |

In reservoirs with an uncontrolled spillway, the peak of the plotted outflow hydrograph

lies outside the plotted inflow hydrograph | |

lies on the recession limb of the plotted inflow hydrograph | |

lies on the peak of the inflow hydrograph | |

is higher than the peak of the plotted inflow hydrograph |

Question 10 Explanation:

There are 10 questions to complete.