This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Aquifer Properties – Set 2”.

1. What is the correct dimension of intrinsic permeability?

a) [L]

b) [L^{2}]

c) [LT^{-1}]

d) [L^{2}T^{-1}]

View Answer

Explanation: Intrinsic permeability is directly proportional to the mean particle size of a porous medium and this has dimension of [L

^{2}]. It is generally represented by units like cm

^{2}or m

^{2}.

2. A Darcy is a common unit for which of the following?

a) Seepage velocity

b) Hydraulic gradient

c) Permeability

d) Intrinsic permeability

View Answer

Explanation: The common unit for intrinsic permeability alongside cm

^{2}and m

^{2}is Darcy. Since the values of intrinsic permeability obtained with m

^{2}are very small, it is converted to Darcy to represent the quantity is a more workable range. 1 Darcy=9.87*10

^{-13}m

^{2}.

3. What is the definition of transmissibility?

a) Discharge through an aquifer one unit width and unit thickness

b) Discharge through an aquifer of unit width under unit hydraulic gradient

c) Discharge through an aquifer of unit thickness under unit hydraulic gradient

d) Discharge through an aquifer of unit width and unit thickness under unit hydraulic gradient

View Answer

Explanation: The discharge through an aquifer of unit width and dome thickness (B) under a unit hydraulic gradient is called as transmissibility (T). It is equal to the permeability (K) for unit thickness of aquifer. It is given as T = KB, and has units of m

^{2}/s or m

^{3}/day/m.

4. In an unconfined aquifer of area 1.6 km^{2}, water table elevation is 98.6 m. After a natural recharge of 1.25 Mm^{3} the water table rose to 101.3 m. What is the specific retention of the aquifer?

a) 12%

b) 17%

c) 23%

d) 29%

View Answer

Explanation: Change in volume of aquifer, V=(101.3-98.6)*1.6*10

^{6}=4.32*10

^{6}m

^{3}

Volume of water retained, V

_{r}=1.25*10

^{6}m

^{3}

∴Specific yield,y

_{r}=\(\frac{V_r}{V}=\frac{1.25*10^6}{4.32*10^6}\)=0.2894≅0.29 or 29%

5. The water table of an aquifer, of specific yield 0.27 and area 7 km^{2}, dropped by 2.5 m after water pumping for few hours. How many million litres of water was pumped out from the aquifer?

a) 3850

b) 4725

c) 6481

d) 7560

View Answer

Explanation: Given y

_{s}= 0.27, A = 7 x 10

^{6}m

^{2}

Volume of water yielded, V

_{y}= specific yield x change in aquifer volume

∴V

_{y}=y

_{s}*V=0.27*(2.5*7*10

^{6})=4.725*10

^{6}m

^{3}

=4.725*10

^{9}litres

=4725 million litres

6. A 6 hour localized rainfall of uniform intensity 3.2 mm/hr occurred over an unconfined aquifer of 12 ha area. What will be the rise in the elevation of water table of the aquifer, if its specific retention is 12%?

a) 16 cm

b) 19 cm

c) 23 cm

d) 26 cm

View Answer

Explanation: Given yr = 0.12, A = 12 x 10

^{4}m

^{2}

Volume of water added, V

_{r}=3.2*10

^{-3}m/hr*6 hr*12*10

^{4}m

^{2}=2304 m

^{3}

Now, y

_{r}=\(\frac{Volume \, of \, water \, added}{Chnage \, in \, aquifer \, volume}\)

So change in aquifer volume, V=\(\frac{V_r}{y_r} =\frac{2304}{0.12}\)=19200 m

^{3}

∴Change in elevation of water table=\(\frac{19200}{12*10^4}\)=0.16 m=16 cm

7. Three equally spaced observation wells A, B and C are inserted in an unconfined aquifer, in that order. The head difference between the first and last well, which are 1.5 km apart, is 8 m. The permeability and porosity of the aquifer is 12 m/day and 30%, respectively. How many days will a tracer injected at well A take to be observed at well B?

a) 875

b) 1750

c) 3500

d) 7000

View Answer

Explanation: Given n = 0.3, K = 12 m/day, H = 8 m, L = 1.5 km = 1500 m

Hydraulic gradient, i=\(\frac{H}{L}=\frac{8}{1500}\)=5.33*10

^{-3}

Discharge velocity, V=Ki=12 m/day *5.33*10

^{-3}=0.06396 m/day

Seepage velocity or velocity of tracer, V

_{s}=\(\frac{V}{n}=\frac{0.06396}{0.3}\)=0.2132 m/day

∴Time taken from A to B=\(\frac{Distance}{Velocity}=\frac{\frac{L}{2}}{V_s}=\frac{1500}{2}*\frac{1}{0.2132}\)

=3517.8≅3500 days

8. Data regarding the water table position of an aquifer of 10 km^{2} area during a week is given below.

Day | Mon | Tue | Wed | Thurs | Fri | Sat | Sun | Mon |

Depth of W.T. from ground (in m) | 25 | 22.5 | 23 | 22 | 20.5 | 23.5 | 25.5 | 25 |

If the porosity and specific yield of the aquifer is 35% and 20% respectively, which of the following is incorrect? Assume the measurements are taken at the beginning of the day.

a) The aquifer yielded least amount of water on Tuesday

b) The aquifer yielded 2.75 Mm^{3} more water than it gained during the week.

c) The aquifer gained 1.5 Mm^{3} of water on Wednesday

d) The aquifer yielded a total of 9.5 Mm^{3} of water during Friday and Saturday

View Answer

Explanation: Given n = 0.35, y

_{s}= 0.2, y

_{r}= 0.35 – 0.2 = 0.15, A = 10 x 10

^{6}m

^{3}

Monday – change from 25 to 22.5 m = W.T. rises by 2.5 m

⇒Water added=y

_{r}*2.5*A=0.15*2.5*10*10

^{6}=3.75 Mm

^{3}

Tuesday – change from 22.5 to 23 m = W.T. drops by 0.5 m

⇒Water lost=y

_{s}*0.5*A=0.2*0.5*10*10

^{6}=1 Mm

^{3}

Wednesday – change from 23 to 22 m = W.T. rises by 1 m

⇒Water added=y

_{r}*1*A=0.15*1*10*10

^{6}=1.5 Mm

^{3}

Thursday – change from 22 to 20.5 m = W.T. rises by 1.5 m

⇒Water added=y

_{r}*1.5*A=0.15*1.5*10*10

^{6}=2.25 Mm

^{3}

Friday – change from 20.5 to 23.5 m = W.T. drops by 3 m

⇒Water lost=y

_{s}*3*A=0.2*3*10*10

^{6}=6 Mm

^{3}

Saturday – change from 23.5 to 25.5 m = W.T. drops by 2 m

⇒Water lost=y

_{s}*2*A=0.2*2*10*10

^{6}=4 Mm

^{3}

Sunday – change from 25.5 to 25 m = W.T. rises by 0.5 m

⇒Water added=y

_{r}*0.5*A=0.15*0.5*10*10

^{6}=0.75 Mm

^{3}

Total amount of water gained/retained during the week = 3.75 + 1.5 + 2.25 + 0.75 = 8.25 Mm

^{3}

Total amount of water lost/yielded during the week = 1 + 6 + 4 = 11 Mm

^{3}

So, the aquifer yielded (11 – 8.25 =) 2.75 Mm

^{3}water more than it gained during the week.

9. After a natural recharge of 2 Mm^{3} in an aquifer of area 5 km^{2}, the water table rose by 2.2 m. After pumping, it was observed that the water table dropped by 2.6 m. What is the volume of water (in Mm^{3}) that was pumped after the natural recharge?

a) 1.22

b) 1.69

c) 2

d) 2.37

View Answer

Explanation: Given A = 5 x 10

^{6}m

^{2}

Specific retention, y=\(\frac{2*10^6}{2.2*5*10^6}\)=0.182

Since the water table dropped by 2.6 m after pumping the volume of water pumped is,

V=y*2.6*A=0.182*2.6*5*10

^{6}=2.366*10

^{6}m

^{3}≅2.37 Mm

^{3}

10. Two observation wells 50 m apart are inserted in an aquifer of porosity 40%. The water level difference in the wells is 65 cm. The time of travel of a tracer between the wells is 8 hours. What is the intrinsic permeability of the aquifer? Take kinematic viscosity of water as 0.01 cm^{2}/s.

a) 4307

b) 5545

c) 7172

d) 9323

View Answer

Explanation: Given n = 0.4, L = 50 m, H = 65 cm = 0.65 m, t = 8 hours, v = 0.01 cm/s

Seepage velocity, V

_{s}=\(\frac{L}{t}=\frac{50 m}{8 hr}=\frac{50*10^2 cm}{8*3600 sec}\)=0.1736 cm/sec

Discharge velocity, V=n*V

_{s}=0.4*0.1736=0.06944 cm/sec

Hydraulic gradient, i=\(\frac{H}{L}=\frac{0.65}{50}\)=0.013

Permeability, K=\(\frac{V}{i}=\frac{0.06944}{0.013}\)=5.34 cm/sec

∴Intrinsic permeability,K

_{0}=\(\frac{Kv}{g}=\frac{5.34*0.01}{981}\)=5.44*10

^{-5}cm

^{2}

Since 1 darcy = 9.87 x 10

^{-9}cm

^{2}, K

_{0}=\(\frac{5.44*10^{-5} cm^2}{9.87*10^{-9} cm^2/darcy}\)≅5545 darcys

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