Question 1 |

A tube-well of 20 cm diameter fully penetrates a horizontal, homogeneous and isotropic confined aquifer of infinite horizontal extent. The aquifer is of 30 m uniform thickness. A steady pumping at the rate of 40 litres/s from the well for a long time results in a steady drawdown of 4 m at the well face. The subsurface flow to the well due to pumping is steady, horizontal and Darcian and the radius of influence of the well is 245 m. The hydraulic conductivity of the aquifer (in m/day,round off to integer) is ______________

45 | |

82 | |

36 | |

12 |

Question 1 Explanation:

\begin{aligned} 40 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s} &=\frac{2 \pi k \times 30 \mathrm{~m} \times 4 \mathrm{~m}}{\ln \left(\frac{245}{0.1}\right)} \\ k &=4.14 \times 10^{-4} \mathrm{~m} / \mathrm{s} \\ \text{or, } \qquad \qquad \qquad \qquad k &=35.77 \mathrm{~m} / \mathrm{d} \\ &\approx 36 \mathrm{~m} / \mathrm{d} (\text{ round off to nearest integer}) \end{aligned}

Question 2 |

Two reservoirs are connected through a homogeneous and isotropic aquifer having hydraulic conductivity (K) of 25 m/day and effective porosity (\eta) of 0.3 as shown in the figure (not to scale). Ground water is flowing in the aquifer at the steady state.

If water in Reservoir 1 is contaminated then the time (in days, round off to one decimal place) taken by the contaminated water to reach to Reservoir 2 will be _____________

If water in Reservoir 1 is contaminated then the time (in days, round off to one decimal place) taken by the contaminated water to reach to Reservoir 2 will be _____________

2000 | |

2400 | |

2800 | |

3200 |

Question 2 Explanation:

\begin{aligned} k&=25 \mathrm{~m} / \mathrm{d} \\ n&=0.3 \\ i&=\frac{20 \mathrm{~m}}{2000 \mathrm{~m}}=0.01 \\ V&=k i \\ V_{s}&=\frac{V}{n} \\ t&=\frac{l}{V_{s}} \\ t&=2400 \mathrm{days} \end{aligned}

Question 3 |

A catchment may be idealised as a rectangle. There are three rain gauges located inside the catchment at arbitrary locations. The average precipitation over the catchment is estimated by two methods:

(i) Arithmetic mean (P_A), and

(ii) Thiessen polygon (P_T).

Which one of the followingstatements is correct?

(i) Arithmetic mean (P_A), and

(ii) Thiessen polygon (P_T).

Which one of the followingstatements is correct?

P_A is always smaller than P_T | |

P_A is always greater than P_T | |

P_A is always equal to P_T | |

There is no definite relationship between P_A \; and \; P_T |

Question 3 Explanation:

The result from Thiessen polygon method is more accurate than arithmetic mean method.
But there is no any close relationship between values obtained by Thiessen polygon method and Arithmetic mean method.

Therefore, there is no any relation between P_A and P_T.

Therefore, there is no any relation between P_A and P_T.

Question 4 |

Rainfall depth over a watershed is monitored through six number of well distributed rain gauges. Gauged data are given below

The Thiessen mean value (in mm, up to one decimal place) of the rainfall is ______

The Thiessen mean value (in mm, up to one decimal place) of the rainfall is ______

479.1 | |

258.8 | |

147.3 | |

369.4 |

Question 4 Explanation:

Thiessen mean value of rainfall;

P_{avg}=\frac{\sum_{i=1}^{6}P_{i}A_{i}}{\sum_{i=1}^{6}A_{i}}

P_{avg}=\frac{470\times 95+465\times 100+435\times 98+525\times 80+480\times 85+510\times 92}{95+100+98+80+85+92} =479.09 mm

P_{avg}=\frac{\sum_{i=1}^{6}P_{i}A_{i}}{\sum_{i=1}^{6}A_{i}}

P_{avg}=\frac{470\times 95+465\times 100+435\times 98+525\times 80+480\times 85+510\times 92}{95+100+98+80+85+92} =479.09 mm

Question 5 |

A catchment is idealized as a 25 km x 25 km square. It has five rain gauges, one at each corner and one at the center, as shown in the figure.

During a month, the precipitation at these gauges is measured as G_{1} = 300 mm, G_{2} = 285 mm, G_{3} = 272mm, G_{4} = 290mm and G_{5} = 288mm. The average precipitation (in mm, up to one decimal place) over the catchment during this month by using the Thiessen polygon method is ______

During a month, the precipitation at these gauges is measured as G_{1} = 300 mm, G_{2} = 285 mm, G_{3} = 272mm, G_{4} = 290mm and G_{5} = 288mm. The average precipitation (in mm, up to one decimal place) over the catchment during this month by using the Thiessen polygon method is ______

287.4 | |

143.1 | |

78.1 | |

156.2 |

Question 5 Explanation:

Let sides of square as,

a=25 km

Now area of the polygon is calculated as,

\begin{aligned} A_{1}&=A_{2}=A_{3}=A_{4} \\ &=\frac{1}{2}\times \frac{a}{2}\times \frac{a}{2} \\ &=\frac{a^{2}}{8}=\frac{25^{2}}{8}=78.125 \\ A_{5}&=\text{Area of square of sides } \frac{\sqrt{2}a}{2} \\ &=\frac{a}{\sqrt{2}} =\frac{a^{2}}{2}=\frac{25^{2}}{2}=312.5\, km^{2} \\ \therefore & \text{Mean precipitation,}\\ \bar{P}&=\frac{G_{1}A_{1}+G_{2}A_{2}+G_{3}A_{3}+G_{4}A_{4}+G_{5}A_{5}}{A_{1}+A_{2}+A_{3}+A_{4}+A_{5}} \\ \therefore \;\;\bar{P}&=\frac{\left ( 300+255+272+290 \right )\times 78.125+288\times 312.5}{\left ( 4\times 78.125 \right )+312.5} \\ &= 287.375 mm\end{aligned}

Question 6 |

In a catchment, there are four rain-gauge stations, P, Q, R, and S. Normal annual precipitation values at these stations are 780 mm, 850 mm, 920 mm, and 980 mm respectively. In the year 2013, stations Q, R, and S, were operative but P was not. Using the normal ratio method, the precipitation at station P for the year 2013 has been established as 860 mm. If the observed precipitation at station Q and R for the year 2013 were 930 mm and 1010 mm, respectively; what was the observed precipitation (in mm) at station S for that year?

1010.63 | |

980.78 | |

1093.43 | |

1040.25 |

Question 6 Explanation:

Let P_{p}= Precipitation in 2013 at station P

N_{p}= Noramal annual precipitation at station P

\begin{aligned} \frac{P_{p}}{N_{p}}&=\frac{1}{3}\left [ \frac{P_{Q}}{N_{q}}+\frac{P_{R}}{N_{R}}+\frac{P_{S}}{N_{S}} \right ]\\ \frac{860}{780}&=\frac{1}{3}\left [ \frac{930}{850}+\frac{1010}{920}+\frac{P_{S}}{980} \right ] \\ \Rightarrow \;\; P_{s}&=1093.43 mm \end{aligned}

N_{p}= Noramal annual precipitation at station P

\begin{aligned} \frac{P_{p}}{N_{p}}&=\frac{1}{3}\left [ \frac{P_{Q}}{N_{q}}+\frac{P_{R}}{N_{R}}+\frac{P_{S}}{N_{S}} \right ]\\ \frac{860}{780}&=\frac{1}{3}\left [ \frac{930}{850}+\frac{1010}{920}+\frac{P_{S}}{980} \right ] \\ \Rightarrow \;\; P_{s}&=1093.43 mm \end{aligned}

Question 7 |

An isohyet is a line joining points of

equal temperature | |

equal humidity | |

equal rainfall depth | |

equal evaporation |

Question 8 |

The intensity of rainfall and time interval of a typical storm are :

The maximum intensity of rainfall for 20 minutes duration of the storm is

The maximum intensity of rainfall for 20 minutes duration of the storm is

1.5 mm/minute | |

1.85 mm/minute | |

2.2 mm/minute | |

3.7 mm/minute |

Question 8 Explanation:

Maximum intensity =\frac{2.2\times 10+1.5\times 10}{10+10}=1.85 mm/minute

Question 9 |

Match the following:

P-1 Q-3 R-2 S-4 | |

P-3 Q-4 R-1 S-2 | |

P-1 Q-2 R-4 S-3 | |

P3 Q-4 R-2 S-1 |

There are 9 questions to complete.