250+ TOP MCQs on Project Survey – Water Supply Scheme and Answers

Surveying Multiple Choice Questions on “Project Survey – Water Supply Scheme”.

1. Which of the following can be identified as the objective of water supply scheme?
a) Chlorination of water
b) Treat water
c) Safe water supply
d) Ionization of water
Answer: c
Clarification: In general, the objective of water supply scheme includes safe water supply, sufficient quantity of water, supply of water to a convenient point with reasonable cost and encouraging personal and house hold cleanliness of the users.

2. Which of the following indicates the component of a water supply scheme?
a) Impure water
b) Chlorination of water
c) Sub surface water
d) Intake of the water
Answer: d
Clarification: The protected water supply scheme consists of four components. Those include the source of water from where it is being produced, intake of the water, treatment of the water and finally the distribution of the treated water.

3. Surface water can act as a source of water in water supply scheme.
a) True
b) False
Answer: a
Clarification: The source of water is classified as surface water and sub surface water. Generally in the water supply scheme surface water is having more priority than the sub surface water. The different sources of surface water include river, streams, lakes, canals etc.

4. While considering the design period, which must be given more priority?
a) Area of land
b) Population
c) Usage of water
d) Arrangement of pipes
Answer: b
Clarification: Design period is considered based on the population present in a particular area. While considering design period, population forecast methods has to be used for the determination of the upcoming population in that area. By doing this the design period can be estimated for a particular water tank construction.

5. The design period of storage reservoir can be given as _________
a) 50 yr
b) 20 yr
c) 30 yr
d) 10 yr
Answer: a
Clarification: Every water storage structure is having certain design periods based on the population present in that particular area. A storage reservoir is generally having a design period of 50 years.

6. Which of the following can be designated as an intake structure?
a) Culvert
b) River
c) Dam
d) Reservoir
Answer: d
Clarification: Intakes are the structures which can collect water from the surface sources and are used for the treatment plant. Reservoir intake, Lake Intake and canal intakes are some of the examples of an intake structure.

7. Which type of water is generally used in the treatment of water?
a) Chlorinated water
b) Treated water
c) Raw water
d) Sulphated water
Answer: c
Clarification: Water treatment plant generally uses raw water which is obtained from different sources containing various impurities. It is not recommended to be used directly without treatment and hence it is supplied to the treatment plant for treatment.

8. Which of the following does not act as a major factor that effects per capita demand?
a) Human activity
b) Industrial activities
c) Usage of water
d) Placement of pipe
Answer: b
Clarification: The factors effecting per capita demand include cost of water, climatic condition, pressure in the distributed system, industrial activities, commercial activities and economical status of the consumers.

9. The amount of water required for 1 percent per day is determined as ___________
a) Daily demand
b) Monthly demand
c) Annual demand
d) Per capita demand
Answer: d
Clarification: Per capita demand can be defined as the amount of water required for 1 percent per day. It includes commercial, industrial, domestic, public uses and also in case of fire demand.

10. Which of the following can act as a type of variation in water demand?
a) Monthly variation
b) Annual variation
c) Crop variation
d) 10 year variation
Answer: a
Clarification: Variation in water demand is generally due to seasonal variation, monthly variation, daily and hourly variations. The demand for water in these variations is generally high and consumes more amount of water than daily consumption.

250+ TOP MCQs on Classification of Surveying and Answers

Surveying Multiple Choice Questions on “Classification of Surveying”.

1. Which of the following is made in connection with the construction of streets, water supply systems, sewers?
a) Traverse surveying
b) Hydrographic surveying
c) Cadastral surveying
d) City surveying
Answer: d
Clarification: City surveying is made in connection with the construction of streets, water supply systems and sewers. A survey which deals with bodies of water for the purpose of navigation, water supply, harbor works or for the determination of mean sea level is hydrographic surveying.

2. Which of the following is a classification based on the instrument used?
a) Topographic surveying
b) Hydrographic surveying
c) Cadastral surveying
d) Traverse surveying
Answer: d
Clarification: Topographic surveying, Hydrographic surveying, Cadastral surveying classification is based on the nature of field survey. Traverse surveying, chain surveying is classified based on the type of instrument used.

3. Determining points of strategic importance are called _______
a) Topographic surveying
b) City surveying
c) Military surveying
d) Traverse surveying
Answer: c
Clarification: Determining points of strategic importance is military surveying. City surveying is made in connection with the construction of streets, water supply systems and sewers.

4. For exploring mineral wealth which type of surveying is used?
a) Topographic surveying
b) Engineering surveying
c) Military surveying
d) Mine surveying
Answer: d
Clarification: For exploring mineral wealth mine surveying is used. Determining points of strategic importance is military surveying.

5. Determining quantities or afford sufficient data for the designing of works such as roads and reservoirs is called _______
a) Topographic surveying
b) Engineering surveying
c) City surveying
d) Cadastral surveying
Answer: b
Clarification: Determining quantities or afford sufficient data for the designing of works such as roads and reservoirs is engineering surveying. City surveying is made in connection with the construction of streets, water supply systems and sewers.

6. What consists of a horizontal and vertical location of certain points by linear and angular measurements and is made to determine the natural features of a country such as rivers, streams?
a) Topographic surveying
b) Engineering surveying
c) City surveying
d) Cadastral surveying
Answer: a
Clarification: Topographic surveying consists of the horizontal and vertical location of certain points by linear and angular measurements and is made to determine the natural features of a country such as rivers, streams etc. City surveying is made in connection with the construction of streets, water supply systems and sewers.

7. Which of the following is a classification based on the nature of the field survey?
a) Topographic surveying
b) Mine surveying
c) Military surveying
d) Chain surveying
Answer: a
Clarification: Chain surveying classification is based on instruments used. For exploring mineral wealth mine surveying is used.

8. Which of the following is not a classification based on instruments used or methods employed?
a) Chain surveying
b) Topographic surveying
c) Traverse surveying
d) Aerial surveying
Answer: b
Clarification: Topographic surveying classification is based upon the nature of the field survey. Chain surveying is classified based on the instrument used. Aerial surveying is classified based on the type of method employed.

9. A survey which deals with bodies of water for the purpose of navigation, water supply, harbor works or for the determination of mean sea level is ________
a) Topographic surveying
b) Hydrographic surveying
c) Cadastral surveying
d) City surveying
Answer: b
Clarification: Survey which deals with bodies of water for the purpose of navigation, water supply, harbor works or for the determination of mean sea level is hydrographic surveying.

250+ TOP MCQs on Compass – Bearings and Angles and Answers

Surveying Multiple Choice Questions on “Compass – Bearings and Angles”.

1. The direction of a survey line can either be established with relation to _______
a) each other
b) main station
c) arrows
d) tie station
Answer: a
Clarification: The direction of a survey line can either be established in relation to each other or with relation to any meridian.

2. What is the direction of line relative to a given meridian?
a) Bearing of a line
b) Length of a line
c) Slope of a line
d) Reciprocal of slope of a line
Answer: a
Clarification: Bearing of a line is the direction of line relative to a given meridian. A meridian is any direction such as true meridian, magnetic meridian, arbitrary meridian.

3. Which line passes through true north and true south?
a) True Meridian
b) Magnetic Meridian
c) Arbitrary Meridian
d) Dip
Answer: a
Clarification: True Meridian through a point is the line in which a plane passing that point and the north and the south poles, intersects with the surface of the earth. It, thus, passes through the true north and south.

4. Which meridian direction can be established with the help of a magnetic compass?
a) True Meridian
b) Magnetic Meridian
c) Arbitrary Meridian
d) All meridians
Answer: b
Clarification: Magnetic meridian through a point in the direction shown by a freely floating and balanced magnetic needle free from all other attractive forces. The direction of magnetic meridian can be established with the help of a magnetic compass.

5. Which meridians are used to determine the relative positions of the lines in a small area?
a) True Meridian
b) Magnetic Meridian
c) Arbitrary Meridian
d) All meridians
Answer: c
Clarification: Arbitrary Meridian is any convenient direction towards a permanent and prominent mark or signal, such as a church spire etc. Such meridians are used to determine the relative positions of the lines in a small area.

6. What is the horizontal angle which it makes with the true meridian through one of the extremities of the line?
a) True bearing
b) Magnetic bearing
c) Arbitrary bearing
d) Dip
Answer: a
Clarification: True bearing is the horizontal angle which it makes with the true meridian through one of the extremities of the line. True Meridian through a point is the line in which a plane passing that point and the north and the south poles, intersects with the surface of the earth.

7. What is the horizontal angle which it makes with the magnetic meridian through one of the extremities of the line?
a) True bearing
b) Magnetic bearing
c) Arbitrary bearing
d) Dip
Answer: b
Clarification: Magnetic bearing is the horizontal angle which it makes with the magnetic meridian through one of the extremities of the line. The direction of magnetic meridian can be established with the help of a magnetic compass.

8. What is the horizontal angle which it makes with the magnetic meridian through one of the extremities of the line?
a) True bearing
b) Magnetic bearing
c) Arbitrary bearing
d) Dip
Answer: c
Clarification: Arbitrary bearing is the horizontal angle which it makes with the arbitrary meridian through one of the extremities of the line. Arbitrary meridians are used to determine the relative positions of the lines in a small area.

9. Convert 22°30′ whole circle bearings to quadrant bearings?
a) 180 – 22°30
b) 22°30
c) 360 – 22°30
d) 270 – 22°30
Answer: b
Clarification: Reduced bearing = Whole circle bearing if R.B is less than 90°. Therefore here R.B = W.C.B = N 22°30’E.

10. Convert 122°30′ whole circle bearings to quadrant bearings?
a) 180 – 122°30
b) 122°30
c) 360 – 122°30
d) 270 – 122°30
Answer: a
Clarification: Reduced bearing = 180 – Whole circle bearing if R.B is lies between 90° and 180°. Therefore here R.B = W.C.B = N 22°30’E.

250+ TOP MCQs on Reciprocal Levelling and Answers

Surveying Multiple Choice Questions on “Reciprocal Levelling”.

1. Which of the following type of levelling is necessary across a river ravine or any obstacle requiring a long site between two points?
a) Barometric levelling
b) Trigonometric levelling
c) Reciprocal levelling
d) Spirit levelling
Answer: c
Clarification: When it is necessary to carry a levelling across a river ravine or any obstacle requiring a long site between two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal, special method that is reciprocal levelling must be used. It is used to obtain better accuracy.

2. Which of the following type of levelling is used when two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal?
a) Barometric levelling
b) Trigonometric levelling
c) Reciprocal levelling
d) Spirit levelling
Answer: c
Clarification: When it is necessary to carry a levelling across a river ravine or any obstacle requiring a long site between two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal, special method that is reciprocal levelling must be used. It is used to obtain better accuracy.

3. Which of the following error cannot be eliminated in reciprocal levelling?
a) error in instrument adjustment
b) combined effect of earth’s curvature and refraction of atmosphere
c) variations in average refraction
d) variation in temperature
Answer: d
Clarification: Reciprocal levelling must be used to obtain better accuracy. It is also used to eliminate error in instrument adjustment, combined effect of earth’s curvature and refraction of atmosphere, variations in average refraction.

4. When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When the instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L. of Q?
a) 125.555
b) 125.565
c)125.575
d)125.585
Answer: d
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 m. Therefore true elevation of Q is 126.386 – 0.801 = 125.585 m.

5. When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?
a) 0.057 m
b) 0.069 m
c) 0.058 m
d) 0.048 m
Answer: b
Clarification: Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 mts.

6. When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the error due to collimation?
a) 0.058 mts
b) 0.052 mts
c) 0.054 mts
d) 0.068 mts
Answer: c
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 – 0.069 = 0.054 m.

7. When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L of Q?
a) 125.585
b) 126.187
c) 127.187
d) 128.197
Answer: c
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Therefore true elevation of Q is 126.386 + 0.801 = 127.187 m.

8. When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect correction for curvature and refraction?
a) 0.057 m
b) 0.069 m
c) 0.058 m
d) 0.048 m
Answer: b
Clarification: Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 m.

9. When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find error due to collimation?
a) 0.178 mts
b) 0.192 mts
c) 0.194 mts
d) 0.188 mts
Answer: b
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 + 0.069 = 0.192 m.

10. When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606 and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the angular error in collimation adjustment of the instrument?
a) 39″
b) 49″
c) 59″
d) 69″
Answer: a
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 + 0.069 = 0.192 m. If ∆ is the inclination of line of the site then tan∆ = 0.192/1010 = 0.000190. Therefore ∆ =39″.

250+ TOP MCQs on Theodolite Traversing – Reduction of Readings and Answers

Surveying Assessment Questions and Answers on “Theodolite Traversing – Reduction of Readings”.

1. In which of the following transverse method angles are measured by theodolite?
a) By fast needle
b) By direct observation of angles
c) By locating details with transit and tape
d) By free needle
Answer: b
Clarification: In trans versing by direct observation of angles, angles between the lines are directly measured by a theodolite. The method is therefore accurate in comparison to the previous three methods.

2. In transversing by direct observation of angles, magnetic bearing of any one line can also be measured if required.
a) True
b) False
Answer: a
Clarification: The magnetic bearings of any one line can be measured and magnetic bearing of other lines can be calculated. The angles measured at different stations may be either included angle and deflection angle.

3. Which of the following comes under transversing by included angles?
a) Transversing by fast needle
b) Transversing by free needle
c) Transversing by direct observation of angles
d) Transversing by chain and compass
Answer: c
Clarification: Transversing by included angles and transversing by deflection angles comes under transversing by direct observation of angles.

4. __________ at a station is either of the two angles by the two survey lines meeting there.
a) Included angle
b) Deflection angle
c) Transverse angle
d) Deviated angle
Answer: a
Clarification: An included angle at a station is either of the two angles formed by the two survey lines meeting there. The method consists simply in measuring each angle directly from a backsight on the preceding station.

5. Included angles can be measured _________
a) Clockwise
b) Counter clockwise
c) Clockwise and counterclockwise
d) Clockwise or counterclockwise
Answer: d
Clarification: Included angles can be measured either clockwise or counterclockwise. But it is better to measure all angles clockwise.

6. All angles are preferred to measure clockwise because of the graduations of theodolite circle increase in this direction.
a) True
b) False
Answer: a
Clarification: It is better to measure included angles clockwise. It is because of graduations of theodolite circle increase in this direction.

7. A deflection angle is an angle in which a survey line makes with prolongation of back sight.
a) True
b) False
Answer: b
Clarification: A deflection angle is an angle in which a survey line makes with the prolongation of the preceding line.

8. Transversing by deflection angles is more suitable for surveys of roads railways, pipe lines etc.
a) True
b) False
Answer: a
Clarification: Transversing by deflection angles is more suitable for surveys of roads railways, pipe lines etc, where the survey lines make small deflection angles.

9. Deflection angle may vary from __________ to ___________
a) 0° to 90°
b) 90° to 180°
c) 0° to 180°
d) 0° to 270°
Answer: a
Clarification: A deflection angle is an angle in which a survey line makes with the prolongation of the preceding line. It may vary from 0° to 180°.

10. In following figure deflection angle at Q is teta L.
a) True
b) False
Answer: b
Clarification: The deflection angle at Q is alpha R and that at R is teta L.

To practice all areas of Surveying Assessment Questions,

250+ TOP MCQs on Area Calculation – Area by Planimeter and Answers

Surveying Multiple Choice Questions on “Area Calculation – Area by Planimeter”.

1. Planimeter is an instrument which is used for__________
a) Locating co-ordinates
b) Transferring point from paper to ground
c) Measuring area of plan
d) Sighting parallel and perpendicular points to station

Answer: c
Clarification: By the use of planimeter, the area of the land can be measured which is used for processing. Plumb bob is used for transferring point from paper to ground and alidade for sighting parallel and perpendicular points to station.

2. The formula for finding area by the use of planimeter is______________
a) Δ = M(F-I±10N+C)
b) Δ = M(F+I±10N+C)
c) Δ = M(F-I±10N-C)
d) Δ = M(F-I±10N±C)

Answer: a
Clarification: The area of the obtained figure can be calculated by, Δ = M(F-I±10N+C)
Where, F = Final reading, I = Initial reading, N = number of times the zero mark of the dial passes the fixed index mark, M = A multiplying constant, C = Constant of the instrument which when multiplied by M.

3. Which of the following methods will give the best output for area?
a) Area by double mean distances
b) Area by triangles
c) Area by co-ordinates
d) Area by planimeter

Answer: b
Clarification: The area calculated by forming triangles will be able to give the best output because it involves formation of frame work.

4. Multiplier constant(M) is also known as___________
a) Planimeter constant
b) Tacheometric constant
c) Meridian constant
d) Simpson’s constant

Answer: a
Clarification: From the formula of area by planimeter, Δ = M (F-I±10N+C) the variable M represents multiplier constant which is required for further calculations.

5. Multiplier constant is equal to__________
a) A+nꞌ
b) A*nꞌ
c) A/nꞌ
d) nꞌ/A

Answer: c
Clarification: The value of multiplier constant is given as M= A/nꞌ in which,
A = known area, nꞌ = change in wheel readings.

6. Calculate the area if I = 8.257, M = 150 sq.cm, F = 4.143, C = 31.155.
a) 6255.15
b) 2565.15
c) 2655.15
d) 2556.15

Answer: d
Clarification: The area can given by, A=M (F- I ± 10N + C)
On substituting, I = 8.257, M = 150 sq.cm, F = 4.143, C = 31.155 and N = -1
A = 150 (4.143- 8.257- 10 + 31.155) = 2556.15sq.cm.

7. Which of the following mathematical operations can be used for area computation?
a) Euler’s equation
b) Simpson’s one-third rule
c) Quadratic equation
d) Simultaneous differential equation

Answer: b
Clarification: Simpson’s one-third rule assumes that the short length of the boundary between the ordinates is parabolic arc and this method is more useful when the boundary line departs considerably from the straight line.

8. Simpson’s rule is capable of producing more accurate results.
a) True
b) False

Answer: a
Clarification: The results obtained by the use of Simpson’s rule are more accurate in all cases. The result obtained is smaller than those obtained by using the trapezoidal rule.

9. Find the value of the multiplier constant if the length of the arm is given as 45.78 m and diameter as 2.54 m.
a) 643.86 sq. m
b) 436.86 sq. m
c) 346.68 sq. m
d) 364.86 sq. m

Answer: d
Clarification: The value of multiplier constant can be given as M = L * circumference
Circumference = π * D = π *2.54 = 7.97 m. On substitution, we get
M = 45.78 * 7.97 = 364.86 sq. m.

10. If length of the arm = 23.31 m, distance between wheel and pivot = 2 m, D = 3m. Find the value of constant C.
a) 4.63 sq. m
b) 6.64 sq. m
c) 6.46 sq. m
d) 4.95 sq. m

Answer: c
Clarification: The value of the constant C can be given as, (C = frac{π (L^2-2aL+R^2)}{M}). The value of M can be given as M = L* π*D = 219.691 sq. m. On substitution, we get
C = π (23.312-2*2*23.31+1.52) / 219.691 = 6.46 sq. m.

11. If the area of the traverse is 645.32 sq. m and the change in the wheel readings can be given as 10, find the value of multiplier constant.
a) 64.532 sq. m
b) 6453.2 sq. m
c) 6.4532 sq. m
d) 0.65432 sq. m

Answer: a
Clarification: If the values of area and the change in wheel readings are given then the multiplier constant can be given as M = A/nꞌ, where A is the area and nꞌ is the change in wheel readings. On substitution, we get
M = 645.32 / 10 = 64.532 sq. m.

12. If the area of traverse is drawn to a scale 1ꞌꞌ = 23 ft, find the change in area if the original area is 497.76 sq. in.
a) 6.04 sq. m
b) 6.04 m
c) 6.04 sq. in
d) 6.04 acres

Answer: d
Clarification: Here, the scale is given as 1ꞌꞌ = 23 ft. So, 1 sq. in = 23*23 sq. ft
And the area of the field can be given as (23*23*497.76) / 43560 = 6.044 acres.

13. Find the value of I if the area of the field is given as 234.315 sq. m, M = 22.15 sq. m, F = 3.256, N = 1, C = 26.43.
a) 21.907 sq. m
b) 29.107 sq. m
c) 29.701 sq. m
d) 23.071 sq. m

Answer: b
Clarification: The area of the field can be given as A=M (F- I ± 10N + C). On substitution of the given values we get,
234.315 = 22.15 (3.256-I+10+26.43)
I = 29.107 sq. m.