Question 1 |
The hyetograph in the figure corresponds to a rainfall event of 3 cm.
If the rainfall event has produced a direct runoff of 1.6 cm, the \phi-index of the event (in mm/hour,round off to one decimal place) would be
If the rainfall event has produced a direct runoff of 1.6 cm, the \phi-index of the event (in mm/hour,round off to one decimal place) would be
1.2 | |
4.2 | |
3.4 | |
2.8 |
Question 1 Explanation:
\begin{aligned} \text { Total rainfall } &=3 \mathrm{~cm} \\ \text { Total runoff } &=1.6 \mathrm{~cm} \\ \therefore \qquad \qquad \qquad\text { Total infiltration } &=3-1.6=1.4 \mathrm{~cm}\\ \therefore \qquad \qquad \qquad \qquad\qquad\text { W-index }&=\frac{\text { Total infiltration }}{\text { Total duration of storm }}\\ &=\frac{1.4}{(210 / 60)} \mathrm{cm} / \mathrm{hr} \\ &=0.4 \mathrm{~cm} / \mathrm{hr}=4 \mathrm{~mm} / \mathrm{hr}\\ \text { As } \phi \text { -index }>\text { W-index }\qquad \qquad \end{aligned}
Hence storm of intensities 4 mm/hr and 3 mm/hr will not produce rainfall exam.
\begin{aligned} \phi \text { -index } &=\frac{\text { Total infiltration in which rainfall excess occur }}{\text { Time period in which rainfall excess occur }} \\ &=\frac{\text { Total infiltration }-\text { Infiltration in which no rainfall excess occur }}{T_{\text {excess }}} \\ &=\frac{14 \mathrm{~mm}-\left(4 \times \frac{30}{60}+3 \times \frac{30}{60}\right) \mathrm{mm}}{\left(\frac{150}{60}\right) \mathrm{hr}} \\ &=4.2 \mathrm{~mm} / \mathrm{hr} \end{aligned}
Question 2 |
A 12-hour unit hydrograph (of 1 cm excess rainfall) of a catchment is of a triangular shape with a base width of 144 hour and a peak discharge of 23 \mathrm{~m}^{3} / \mathrm{s}. The area of the catchment (in \mathrm{~km}^{2},round off to the nearest integer) is _______
412 | |
632 | |
596 | |
128 |
Question 2 Explanation:
Area of hydrograph = Total direct runoff volume
\begin{aligned} \Rightarrow \frac{1}{2} \times 23 \mathrm{~m}^{3} / \mathrm{sec} \times 144 \times &3600 \mathrm{sec}=\text { Area of catchment } \times \text { Runoff depth }\\ \Rightarrow \frac{1}{2} \times 23 \times 144 \times 3600 \mathrm{~m}^{3} &=A \times \frac{1}{100} \mathrm{~m} \\ A &=596.16 \times 10^{6} \mathrm{~m}^{2} \\ \therefore \quad \text{ Area of catchment }&=596.16 \mathrm{~km}^{2} \end{aligned}
Question 3 |
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 3 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 4 |
The value of abscissa (x) and ordinate (y) of a curve are as follows:
\begin{array}{|c|c|} \hline x & y \\ \hline 2.0 & 5.00 \\ \hline 2.5 & 7.25 \\ \hline 3.0 & 10.00 \\ \hline 3.5 & 13.25 \\ \hline 4.0 & 17.00 \\ \hline \end{array}
By Simpson's 1 / 3^{\mathrm{rd}} rule, the area under the curve (round off to two decimal places) is ______________
\begin{array}{|c|c|} \hline x & y \\ \hline 2.0 & 5.00 \\ \hline 2.5 & 7.25 \\ \hline 3.0 & 10.00 \\ \hline 3.5 & 13.25 \\ \hline 4.0 & 17.00 \\ \hline \end{array}
By Simpson's 1 / 3^{\mathrm{rd}} rule, the area under the curve (round off to two decimal places) is ______________
20.67 | |
54.62 | |
38.45 | |
66.22 |
Question 4 Explanation:
d = 0.5 unit
\begin{aligned} A &=\frac{d}{3}\left[\left(y_{1}+y_{5}\right)+4\left(y_{2}+y_{4}\right)+2 y_{3}\right] \\ &=\frac{0.5}{3}[(5+17)+4(7.25+13.25)+2 \times 10] \\ &=20.67 \text { unit }^{2} \end{aligned}
\begin{aligned} A &=\frac{d}{3}\left[\left(y_{1}+y_{5}\right)+4\left(y_{2}+y_{4}\right)+2 y_{3}\right] \\ &=\frac{0.5}{3}[(5+17)+4(7.25+13.25)+2 \times 10] \\ &=20.67 \text { unit }^{2} \end{aligned}
Question 5 |
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 5 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 6 |
A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The
peak flow of 60 m^3/s occurs at 20 hours from the start of the storm. The area of catchment
is 300 km^2. The rainfall excess of the storm (in cm), is
2 | |
3.24 | |
5.4 | |
6.48 |
Question 6 Explanation:
\left [ \frac{\frac{1}{2} \times 60 m^3/s \times 90 \times 3600s}{300 \times 10^6m^2} \times 100 \right ]cm= Rainfall excess
Rainfall excess = 3.24 cm
Question 7 |
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 7 Explanation:
Cross-section of a super passage
Question 8 |
Muskingum method is used in
hydrologic reservoir routing | |
hydrologic channel routing | |
hydraulic channel routing | |
hydraulic reservoir routing |
Question 8 Explanation:
Muskingum method is used in hydrological channel routing
Question 9 |
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 9 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 10 |
The probability that a 50 year flood may NOT occur at all during 25 years life of a project
(round off to two decimal places), is _______.
0.6 | |
0.25 | |
0.78 | |
0.44 |
Question 10 Explanation:
\begin{aligned} P&=\frac{1}{T}=\frac{1}{50}=0.02 \\ q&=1-P=0.98 \end{aligned}
\therefore \; Probability of non-occurance of an event is given by
\begin{aligned} \text{Assurance} &= q^n \\ &= (0.98)^{25}\\ &= 0.603 \end{aligned}
\therefore \; Probability of non-occurance of an event is given by
\begin{aligned} \text{Assurance} &= q^n \\ &= (0.98)^{25}\\ &= 0.603 \end{aligned}
There are 10 questions to complete.