Category: Surveying Objective Questions

Surveying Multiple Choice Questions on “Contour Gradient”.

1. _______ is a line lying throughout on the surface of the ground and preserving a constant inclination to the horizontal.
a) Contour gradient
b) Contour interval
c) Contour slope
d) Contour inclination
Answer: a
Clarification: Contour gradient is a line lying throughout on the surface of the ground and preserving a constant inclination to the horizontal. If the inclination of such a line is given, its direction from a point may be easily located either on the map or on the ground.

2. Which of the following cannot be used to locate the contour gradient in the field?
a) Clinometer
b) Theodolite
c) Level
d) Chain
Answer: d
Clarification: The method of locating the contour gradient on the map can be done in the filed by using clinometer, theodolite and level.

3. If a level is used to locate the contour gradient, it is not necessary to set the level on the contour gradient.
a) True
b) False
Answer: a
Clarification: Contour gradient is a line lying throughout on the surface of the ground and preserving a constant inclination to the horizontal. If a level is used to locate the contour gradient, it is not necessary to set the level on the contour gradient.

4. If the inclination of contour gradient is given, its direction from a point may be easily located either on the map or on the ground.
a) True
b) False
Answer: a
Clarification: Contour gradient is a line lying throughout on the surface of the ground and preserving a constant inclination to the horizontal. If the inclination of such a line is given, its direction from a point may be easily located either on the map or on the ground.

5. To locate the contour gradient, the level is set at a commanding position and reading on the staff at the second point is taken.
a) True
b) False
Answer: b
Clarification: To locate the contour gradient, the level is set at a commanding position and reading on the staff at the first point is taken.

6. From a single instrument station, several points at a given gradient can be located.
a) True
b) False
Answer: b
Clarification: To locate the contour gradient, the level is set at a commanding position and reading on the staff at the first point is taken. From a single instrument station, several points at a given gradient can be located.

7. In a direct method, the contour to be plotted is actually traced on the ground.
a) True
b) False
Answer: a
Clarification: In a direct method, the contour to be plotted is actually traced on the ground. Only those points are surveyed which happen to be plotted.

8. In indirect method, each contour is located by determining the positions of a series of points through which the contour passes.
a) True
b) False
Answer: a
Clarification: In indirect method, each contour is located by determining the positions of series of points through which the contour passes. The operation is also sometimes called tracing out contours.

9. The indirect method, guide points need not necessarily be on the contours.
a) True
b) False
Answer: b
Clarification: In indirect method, some suitable guide points are selected and surveyed. The guide points need not necessarily be on the contours.

10. The indirect method serves as a basis for the interpolation of contours.
a) True
b) False
Answer: a
Clarification: In indirect method, some suitable guide points are selected and surveyed. These guide points, having been plotted, serve as a basis for the interpolation of contours.

Surveying Multiple Choice Questions on “Plane Table – Three Point Problem”.

1. In three point problem, orientation and resection are done simultaneously.
a) True
b) False
Answer: a
Clarification: In the process of three point problem, the intersection of the three resectors in a point gives the location of the instrument station. Thus, the orientation and resection are accomplished in the some operation.

2. Which of the following serves as a solution for three point problem?
a) Traverse method
b) Axis method
c) Mechanical method
d) Transit method
Answer: c
Clarification: The solution for three point problem can be solved by a mechanical method, graphical method and Lehman’s method. Each one can be employed based on the problem raised due to orientation.

3. Mechanical method is also known as________
a) Graphical method
b) Axis method
c) Trial and error method
d) Tracing paper method
Answer: d
Clarification: Due to the usage of tracing paper involved in this process, mechanical method is also called as tracing paper method.

4. In graphical method, why Bessel’s method is chosen the best?
a) Due to accuracy in result
b) Due to quick output
c) Due to ease in handling
d) Due to economical issues
Answer: a
Clarification: Among the several graphical methods available, Bessel’s method is chosen the best due to its accuracy in obtaining the values which might help in producing accurate output.

5. Among the three point and two point problems, the solution from two point problem will serve better output?
a) True
b) False
Answer: b
Clarification: Three point problems are capable of producing precise values and accurate output than the solution from two point problem. Because speed is the only factor concerned in case of two point problem solutions.

6. Which process involves more labour work?
a) Mechanical method
b) Graphical method
c) Two point problem
d) Three point problem
Answer: c
Clarification: When compared to the three point problem, the solution obtained from two point problem involves more labour work as speed is the first priority.

7. Which among the following doesn’t serve as a solution for proper orientation of the plane table?
a) Resection by compass
b) Resection by three point problem
c) Resection by two point problem
d) Resection by graphical triangulation
Answer: d
Clarification: Graphical triangulation is a method usually involves locating a station point and marking the remaining points, followed by changing the instrument to another point and computing the remaining points from which the required distance between the points can be found out. It doesn’t come under the category of resection.

8. Which of the following methods of three point problem is a tedious one?
a) Lehman’s method
b) Graphical method
c) Axis method
d) Mechanical method
Answer: a
Clarification: Lehman’s process, also known as trial and error process, involves a lot of trial and error, which makes it a tedious process.

9. Which shapes are generally formed in Lehman’s method?
a) Quadrilaterals
b) Triangles
c) Polygons
d) Circle
Answer: b
Clarification: A triangle is the most commonly used shape in case of surveying as it covers an entire area in a framework which makes calculations easier.

10. Usage of Lehman’s method involves following certain rules.
a) True
b) False
Answer: a
Clarification: Lehman’s method is accompanied by certain rules which must be followed for obtaining an accurate solution for solving three point problem.

Surveying online quiz on “Tacheometric Surveying – Subtense Method Principle”.

1. In the subtense bar method, the horizontal angle subtended by two targets fixed on a horizontal bar at a known distance apart is measured at instrument station by theodolite.
a) True
b) False
Answer: a
Clarification: In the subtense bar method, the horizontal angle subtended by two targets fixed on a horizontal bar at a known distance apart is measured at instrument station by theodolite. The two targets are at a distance s apart, and each at s/2 from the centre, i.e. vertical axis.

2. The two targets are at a distance s apart, and each at s/2 from the centre, i.e. vertical axis. The horizontal angle α is measured carefully by means of a theodolite. Then what is the value of D in the subtense bar method?
a) s/2α
b) s/α
c) s/4α
d) 2s/α
Answer: b
Clarification: From the geometry, D=1/2 x s x cot α/2, where, s = the distance between the targets of subtense bar, and α = apex angle subtended by targets at O. As α is small tan α/2 = α/2, D = s/α.

3. The following readings were taken with a tachometer on to a vertical staff. Horizontal Distance Stadia Readings 46.20 m 0.780; 1.010; 1.240 51.20 m 1.860; 2.165; 2.470. Calculate the tacheometric constants.
a) 100, 0.20 m
b) 200, 0.10 m
c) 100, 0.10 m
d) 200, 0.20 m
Answer: a
Clarification: D = Ks + C, where the constant K is equal to (f /i). It is called multiplying constant of the tachometer and is generally kept as 100. The constant C is equal to (f + c). It is called an additive constant.

4. Stadia readings were taken with a theodolite on a vertical staff with the telescope inclined at an angle of depression of 3^o30′. The staff readings were 2.990, 2.055 and 1.120. The reduced level of the staff station is 100.000m, and the height of the instrument is 1.40 m. What is the reduced level of the ground at the instrument? Take constants as 100 and zero.
a) 102.050 m
b) 122.050 m
c) 112.050 m
d) 132.050 m
Answer: c
Clarification: Use D = Ks + C, where the constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called an additive constant.

5. A tacheometer is setup at an intermediate point on a traverse course PQ and the following observations are made on a staff held vertical. Staff Station Vertical Angle Staff Intercept Axial Hair Readings P + 9^o30′ 2.250 2.105 Q + 6^o00′ 2.055 1.975 The constants are 100 and 0. Compute the length PQ and the reduced level of Q. RL of P = 350.50 m.
a) 428.13 m; 335.47 m
b) 402.13 m; 335.47 m
c) 422.13 m; 305.47 m
d) 422.13 m; 335.47 m
Answer: d
Clarification: Use D = Ks + C, where the constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called an additive constant.

6. Following observations were taken with a tacheometer fitted with an anallactic lens having value of constant as 100.

Inst. Station	Staff station	Reduced bearing	Vertical angle	Staff reading
O	P	N37W	412	0.910, 1.510, 2.110
O	Q	N23E	542	1.855, 2.705, 3.555

Calculate the horizontal distance between P and Q.
a) 149.96 m
b) 140.26 m
c) 141.92 m
d) 143.56 m
Answer: a
Clarification: Use D = Ks + C, where the constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called an additive constant. Use formula a2 = b2 + c2 – 2 b c cos A. From the geometry, D=1/2 x s x cot α/2, where, s = the distance between the targets of subtense bar, and α = apex angle subtended by targets at O. As α is small tan α/2 = α/2, D = s/α.

7. The horizontal angle subtended at the theodolite station by a subtense bar with vanes 3 m apart is 0^o 10 ′40′′. Calculate the horizontal distance between the theodolite and the subtense bar?
a) 960.00 m
b) 966.87 m
c) 966.78 m
d) 906.87 m
Answer: b
Clarification: Use from the geometry, D=1/2 x s x cot α/2, where, s = the distance between the targets of subtense bar, and α = apex angle subtended by targets at O. As α is small tan α/2 = α/2, D = s/α.

8. The vertical angles to vanes fixed at 1 m and 3 m above the foot of the staff held vertically at a station P were – 1o 45′ and + 2o 30′, respectively. Find the horizontal distance and the reduced RL of P if the RL of the instrument axis is 110.00 m?
a) 26.95 m, 100.177
b) 20.95 m, 108.177
c) 26.95 m, 108.177
d) 26.95 m, 108.000
Answer: c
Clarification: Use from the geometry, D=1/2 x s x cot α/2, where, s = the distance between the targets of subtense bar, and α = apex angle subtended by targets at O. As α is small tan α/2 = α/2, D = s/α.

9. The following notes refer to a traverse run by a tacheometer fitted with an anallactic lens, with constant 100 and staff held vertical. Line, Bearing, Vertical Angle, Staff Intercept -PQ, 30^o 24′, + 5^o 06′,1.875; QR, 300^o 48′, + 3^o 48′,1.445; RS, 226^o 12′, − 2^o 36′, 1.725 respectively. Find the length and bearing of SP.
a) 191.930 m, 126o 47 ′47′′
b) 190.930 m, 125o 47 ′47′′
c) 193.930 m, 124o 47 ′47′′
d) 192.930 m, 120o 47 ′47′′
Answer: a
Clarification: Use D = Ks + C, where the constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called an additive constant. From the geometry, D=1/2 x s x cot α/2, where s = the distance between the targets of subtense bar, and α = apex angle subtended by targets at O. As α is small tan α/2 = α/2, D = s/α.

10. Distance and elevation formulae for fixed hair method assuming the line of sight as horizontal and considering an external focusing type telescope is D = Ks + C. where C is _______
a) f/i
b) i/f
c) f + c
d) f – c
Answer: c
Clarification: The constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called additive constant whose value ranges from 30 cm to 50 cm for external focusing telescopes and 10 cm to 20 cm for internal focusing telescopes.

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