Category: Surveying Objective Questions

250+ TOP MCQs on Chain Surveying – Chain Triangulation and Answers

Surveying Multiple Choice Questions on “Chain Surveying – Chain Triangulation”.

1. What is the lateral distance of an object or ground feature measured from a survey line?
a) Offset
b) Perpendicular distance
c) Side distance
d) Perpendicular offset
Answer: a
Clarification: An offset is the lateral distance of an object or ground feature measured from a survey line. When the angle of the offset is 90°, it is called perpendicular offset.

2. An offset is laid out 6° from its true direction on the field. Find the resulting displacement of the plotted point on the paper in a direction parallel to the chain line? (Given the length of offset is 10 m and scale is 5 m to 1 cm)
a) 0.209 cm
b) 0.260 cm
c) 0.0109 cm
d) 0.910 cm
Answer: a
Clarification: Distance parallel to the chain = lsin∆/ s = 10 sin 6°/ 5 = 0.209 cm.

3. An offset is laid out 6° from its true direction on the field. Find the resulting displacement of the plotted point on the paper in a direction perpendicular to the chain line? (Given length of offset is 10 m and scale is 5 m to 1 cm)
a) 0.209 cm
b) 0.260 cm
c) 0.0109 cm
d) 0.910 cm
Answer: c
Clarification: Distance perpendicular to the chain = l(1 – cos∆)/s = 10 (1 – cos 6°)/5. By simplifying we get 0.0109 cm.

4. An offset is laid out 1° 30′ from its true direction on the field. Find the degree of accuracy with which the offset should be measured so that the maximum displacement of the point on the paper from both sources may be equal?
a) 1 in 38
b) 1 in 39
c) 1 in 36
d) 1 in 37
Answer: b
Clarification: Displacement due to angular error is lsin∆. Displacement due to linear error l/r. Equating both r = 38.20. Hence, the offset should be measured with an accuracy of 1 in 39.

5. An offset is measured with an accuracy of 1 in 40. If the scale of plotting is 1 cm = 20 m, find the limiting length of the offset so that the displacement of the point on the paper from both sources of error may not exceed 0.25 mm?
a) 14.10 m
b) 14.14 m
c) 14.40 m
d) 14.44 m
Answer: b
Clarification: The total displacement of the paper = √2 l /rs. From this we can simply for l = (0.025/√2)*40*20 = 14.14 m.

6. What triangles are generally preferred to get good results in plotting?
a) Isosceles
b) Obtuse angled
c) Equilateral
d) Acute angle
Answer: c
Clarification: To get good results in plotting, the frame work should consist of triangles which are as nearly equilateral as possible.

7. The line must run through level ground as possible is the condition to be fulfilled by survey lines or survey stations.
a) True
b) False
Answer: a
Clarification: The survey stations should be so selected that a good system of lines is obtained fulfilling the following conditions such as survey stations must be mutually visible, the lines must run through level ground as possible etc.

8. The main lines should form well-conditioned triangles is the condition to be fulfilled by survey lines or survey stations.
a) True
b) False
Answer: a
Clarification: The survey stations should be so selected that a good system of lines is obtained fulfilling the following conditions such as the lines must run through the level ground as possible, the main lines should form well-conditioned triangles etc.

9. Survey lines must be as few as possible is the condition to be fulfilled by survey lines or survey stations.
a) True
b) False
Answer: a
Clarification: The survey stations should be so selected that a good system of lines is obtained fulfilling the following conditions such as the lines must run through the level ground as possible, survey lines must be as few as possible so that the frame work can be plotted conveniently, the main lines should form well-conditioned triangles etc.

10. As far as possible main survey lines should pass through the obstacles to get know how many obstacles present.
a) True
b) False
Answer: b
Clarification: The survey stations should be so selected that a good system of lines is obtained fulfilling the following conditions such as the lines must run through the level ground as possible, as far as possible, the main survey lines should not pass through obstacles.

250+ TOP MCQs on Levelling – Balancing Backsights and Foresights and Answers

Surveying Questions and Answers for Campus interviews on “Levelling – Balancing Backsights and Foresights”.

1. By balancing back sight and fore sight error due to curvature can be eliminated.
a) True
b) False
Answer: a
Clarification: When the difference in elevation between any two points is determined from a single set up back sighting on one point and fore sighting on the other, error due to curvature can be eliminated. Error due to refraction also be eliminated.

2. By balancing back sight and fore sight error due to non parallelism of the line of collimation can be eliminated.
a) True
b) False
Answer: a
Clarification: When the difference in elevation between any two points is determined from a single set up back sighting on one point and fore sighting on the other, error due to non parallelism of the line of collimation can be eliminated. Error due to refraction also is eliminated.

3. By which of the following, the difference in elevation between two points can be calculated by taking a difference between the two readings and no correction for the inclination of the line of sight is necessary?
a) Levelling
b) Centering
c) Contouring
d) Balancing
Answer: d
Clarification: If the back sight and foresight distances are balanced, the difference in elevation between two points can be directly calculated by taking a difference of two readings and no correction for the inclination of the line of sight is necessary.

4. If the observed back sight and fore sight are x1 and x2. The correction back sight on A will be equal to x1-y1, where y1= D1 tan i°. The correct fore sight on B will be equal to x2-y2 where, y2 = D2 tan i°. Then what is the correction difference in level between A and B.
a) x1 – x2
b) x2 – x1
c) x1 – x2 + (D2 tan i° – D1 tan i°)
d) x2 – x1 + (D1tan i° + D2 tan i°)
Answer: c
Clarification: The correct difference in level between A and B is (x1 – x2) – (x2 – y2) = (x1 – x2) + (y2 – y1) = x1 – x2 + (D2 tan i° – D1 tan i°).

5. If the observed back sight and fore sight are x1 and x2. The correction back sight on A will be equal to x1-y1, where y1= D1 tan i°. The correct fore sight on B will be equal to x2-y2 where, y2 = D2 tan i°. Then what is the correction difference in level between A and B, if D1 = D2?
a) x1 – x2
b) x2 + x1
c) x1 – x2 + (D2 tan i° – D1 tan i°)
d) x2 – x1 + (D1tan i° + D2 tan i°)
Answer: a
Clarification: The correct difference in level between A and B is (x1 – x2) – (x2 – y2) = (x1 – x2) + (y2 – y1) = x1 – x2 + (D2 tan i° – D1 tan i°) = x1 -‘x2.

6. If the observed back sight and fore sight are 20 m and 18 m. The correction back sight on A will be equal to 16 m, The correct fore sight on B will be equal to 14 m where then what is the correction difference in level between A and B?
a) 4 m
b) 3 m
c) 2 m
d) 6 m
Answer: c
Clarification: The correct difference in level between A and B is (20 – 4) – (18- 4) = 20 – 18 = 2 m.

7. If the staff reading at point A = ha and at a point B = hb. The correct staff reading should have been Ha and Hb, then the correction difference in elevation between A and B is given by _____
a) ha – hb
b) ha + hb
c) Ha – Hb
d) Ha + Hb
Answer: c
Clarification: The correction difference in elevation between any two points is given by taking a difference of correct staff reading. Therefore, here Ha – Hb.

8. If the staff reading at point A = ha and at a point B = hb. The correct staff reading should have been Ha and Hb, where Ha = ha – ha’ and Hb = hb – hb’ then the correction difference in elevation between A and B is given by ____________
a) ha – hb – ha’ + hb’
b) ha + hb + ha’ + hb’
c) Ha – Hb + ha’ – hb’
d) Ha + Hb
Answer: a
Clarification: The correction difference in elevation between any two points is given by taking a difference of correct staff reading. Therefore, here Ha – Hb.
Ha – Hb = (ha – ha’) – (hb – hb’) = ha – hb – ha’ + hb’.

9. If the back sight and fore sight distances are balanced, the elevation between two points is equal to the difference between the rod readings taken to the two points and correction for curvature and refraction is necessary.
a) True
b) False
Answer: b
Clarification: If the back sight and fore sight distances are balanced, the elevation between two points is equal to the difference between the rod readings taken to the two points. No correction for curvature and refraction is necessary.

10. Turning point is also called _____
a) intermediate point
b) level point
c) change point
d) end point
Answer: c
Clarification: Turning point is a point on which both minus sight and plus sight are taken on a line of direct levels. It is also called a change point.

To practice all areas of Surveying for Campus Interviews,

250+ TOP MCQs on Theodolite Traversing – Methods and Answers

Surveying Multiple Choice Questions on “Theodolite Traversing – Methods”.

1. In order to measure the magnetic bearing of a line, the theodolite should be provided with ______
a) extra telescope
b) spirit level
c) compass
d) tabular or trough compass
Answer: d
Clarification: In order to measure the magnetic bearing of a line, the theodolite should be provided with either a tabular compass or trough compass.

2. Direct angles are angles measured clockwise from the preceding line to the following line.
a) True
b) False
Answer: a
Clarification: Direct angles are angles measured clockwise from the preceding line to the following line. They are also known as angles to the right or azimuths from the back line.

3. In measuring direct angles, azimuths from the back line may vary from 0° to 180°.
a) True
b) False
Answer: b
Clarification: Direct angles are angles measured clockwise from the preceding line to the following line. They are also known as angles to the right or azimuths from the back line. They may vary from 0° to 360 °.

4. A deflection angle is an angle in which a survey line makes with the prolongation of the preceeding line.
a) True
b) False
Answer: a
Clarification: A deflection angle is an angle in which a survey line makes with the prolongation of the preceding line. It is designated as right (R) or left (L) according to its measure to the clockwise or to anti clockwise from the prolongation of the previous line.

5. How methods are there for prolongation of a straight line?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: There are three methods of prolonging a straight line. Those are named as first method, second method and third method.

6. To run a straight line between two points, when both ends are inter visible. We establish intermediate points through ________
a) line of sight
b) balancing
c) using random line
d) back sight
Answer: a
Clarification: Set instrument at A and take sight on B. Establish intermediate points C, D, E etc. In the line of sight. It possible only in case both ends are inter visible.

7. To locate the point of intersection of two straight lines, we stretch a thread or string between two stakes, where line of sight cuts the string is over required point of intersection.
a) True
b) False
Answer: a
Clarification: Let it be required to locate the point of intersection P of the two lines AB and CD. Set the instrument at A, sight B and set two stakes a and b a short distance apart on either side of the estimated position of point P. Set the instrument at C and sight D. Stretch a thread or string between ab and locate P, where the line of sight cuts the string.

8. Theodolite can be used to lay off a horizontal angle.
a) True
b) False
Answer: a
Clarification: Theodolite can be used to lay off a horizontal angle. It also used to locate the point of intersection, to run a straight line between two points, to prolong straight line etc.

9. The method of repetition is used when it is required to lay off an angle with greater precision than that possible by single observations.
a) True
b) False
Answer: a
Clarification: The method of repetition is used when it is required to lay off an angle with greater precision than that possible by single observations.

10. To measure a vertical angle, the instrument should be levelled with reference to the line of sight.
a) True
b) False
Answer: b
Clarification: To measure a vertical angle, the instrument should be levelled with reference to the altitude bubble. When the altitude bubble is on the index frame procedure starts.

250+ TOP MCQs on Area Calculation – Area by Double Meridian Distances and Answers

Surveying Puzzles on “Area Calculation – Area by Double Meridian Distances”.

1. Double meridian distance of a line is equal to __________
a) Sum of parallel distances
b) Sum of perpendicular distances
c) Sum of total areas
d) Sum of meridian distances
Answer: d
Clarification: The double meridian distance of a line is equal to the sum of the meridian distances of the two extremities, which is useful for the determination of the required area of the plot.

2. Which of the following area calculation methods is mostly used?
a) Area by double meridian
b) Area by co-ordinates
c) Area by planimeter
d) Area by Simpson’s rule
Answer: a
Clarification: Area by double mean distances involves more methods of obtaining the area, which actually increases the accuracy of the output. Each method is having its own importance which can be sorted out based on the inputs available.

3. The double parallel distance can be given as __________
a) Sum of vertical distances
b) Sum of perpendicular distances
c) Sum of parallel distances
d) Area of parallel distances
Answer: c
Clarification: The double parallel distance for a line can be given as sum of the parallel distances of its ends. The principles of finding area by D.M. D. method and D.P.D. method are identical.

4. Which of the following indicates the purpose of D.P.D?
a) Checking area computed by D.P.D
b) Checking area computed by D.M.D
c) Checking area computed by perpendiculars
d) Checking area computed by parallels
Answer: b
Clarification: The D.P.D method is employed for checking the area computed by D.M.D method. It is an independent area and has its own set of rules but some principles are identical.

5. Which of the following describes the double meridian distance?
a) Sum of latitudes
b) Sum of horizontal distances
c) Sum of parallel distances
d) Sum of meridian distances
Answer: d
Clarification: Double meridian distance of a line is equal to sum of the meridian distances of the two extremities. It is represented by the symbol M. The area calculated by double meridian distance can be having accuracy in its output.

6. Calculate the area of the figure by using the data provided. L1 = 92.69 m, L2 = 248.96 m, L3 = 146.31m, L4 = 157.43 m.
surveying-puzzles-q6
a) 29283.46 sq. m
b) 29823.46 sq. m
c) 29328.64 sq. m
d) 29238.64 sq. m
Answer: a
Clarification: From the figure, it can be observed that it consist four triangles so the total are can be given as the summation of the area of triangles.
A = 1⁄2(D1*L1) + 1⁄2(D2*L2) + 1⁄2(D3*L3) + 1⁄2(D4*L4)
= 1⁄2(112.69*92.69) + 1⁄2(242.96*24.31) + 1⁄2(146.31*170.69) + 1⁄2(157.43*109.52)
= 5222.61 + 2953.17 + 12486.82 + 8620.86 = 29283.46 sq. m.

7. Find the value of M2 if D1 = 24.86 m, D2 = 17.65 m.
a) 76.37 m
b) 67.37 m
c) 76.73 m
d) 37.76 m
Answer: b
Clarification: The value of M2 can calculated by M2 = M1 + D1 + D2. But we already know that M1 = D1. So, on substitution, we get
M2 = 24.86 + 24.86 + 17.65 = 67.37 m.

8. Find the meridian distance if m1 = 32.76, and D2 = 44.56 m.
a) 71.24 m
b) 17.24 m
c) 17.42 m
d) 71.42 m
Answer: d
Clarification: Meridian distance can be calculated by m2 = m1 + D1/2 + D2/2. On substitution we get,
m2 = 32.76 + 32.76/2 + 44.56/2
m2 = 71.42 m.

9. Find the area of a triangle if latitude distance is given as 209.96 m and meridian distance is 5.78 m.
a) 1213.86 sq. m
b) 1231.68 sq. m
c) 1213.68 sq. m
d) 1123.68 sq. m
Answer: c
Clarification: If the latitude and meridian distance are given then the area of a triangle can be calculated by the product of both i.e., Area of triangle = latitude * meridian distance = 209.96 * 5.78 = 1213.68 sq. m.

10. Find the area of the figure by double distances, if L1 = 13.99 m and L2 = 66.54 m.
surveying-puzzles-q10
a) 2247.17 sq. m
b) 2274.17 sq. m
c) 2274.71 sq. m
d) 2247.71 sq. m
Answer: a
Clarification: The formula for area by double distances can be given as
A = 1⁄2 (M1*L1+ M2*L2). We know that, M1 = D1 = 24.56 m and M2 = M1 + D1 + D2 = 24.56 + 24.56 + 13.26 = 62.38 m. On substitution, we get
A = 1⁄2(24.56*13.99 + 62.38 * 66.54) = 2247.17 sq. m.

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250+ TOP MCQs on Designation of Curve and Answers

Surveying Multiple Choice Questions on “Designation of Curve”.

1. Which of the following doesn’t represent the classification of the curve?
a) Simple
b) Compound
c) Complex
d) Reverse
Answer: c
Clarification: A curve can be expressed as a turning which is provided for a change in direction. It is classified as Simple curve, Compound curve, Reverse curve, Transition curve.

2. The formula for length of the curve can be given as ____________
a) L = R * Δ
b) L = R + Δ
c) L = R * (tan(frac{Δ}{2}))
d) L = R / Δ
Answer: a
Clarification: Length of the curve can be given as the total distance from point of curvature to point of tangent, which is given as L = R * Δ where, Δ is the deflection angle.

3. Sharpness of the curve can be determined by _________
a) Chord length
b) Radius
c) Mid-ordinate
d) Tangent
Answer: b
Clarification: The sharpness of the curve can be determined by the radius or by its degree of curvature. In India, degree of curvature method is adopted due to the circumstances.

4. Relation between radius and degree of curvature can be approximately given as __________
a) R = 5370 / D
b) R = 7530 / D
c) R = 5770 / D
d) R = 5730 / D
Answer: d
Clarification: The relation between radius and degree of curvature can be given as, R = 5730 / D. It is an approximation which can be verified by applying check if necessary.

5. The relation of radius and degree of curvature cannot be applied for small radius.
a) True
b) False
Answer: a
Clarification: We know that the relation between radius and degree of curvature is an approximate value, it cannot be applied for smaller curves and also for obtaining more accuracy in work, it is recommended to take exact value rather than approximate value.

6. The maximum curvature provided for a highway is about__________
a) 100
b) 200
c) 300
d) 500
Answer: b
Clarification: Over turning of vehicle depends upon the amount of curvature provided, which should be at a minimum rate. In general, highways are provided curvature and railway track is having curvature about 10.

7. While designing a curve, which among the following must be taken into consideration?
a) Minerals present
b) Geomorphology
c) Topography
d) Rocks present
Answer: c
Clarification: For designing a curve, topography must be given at most importance which plays a crucial role in determining its durability. Topography involves obtaining information about the folds, faults, undulations present. So that care can be taken while designing.

8. Length of the curve depends on the criteria used for defining the degree of the curve.
a) True
b) False
Answer: a
Clarification: In India, all the curves are designated based on the degree of curvature which is different from the curve designated based on radius. The criteria used will be depending upon the degree obtained by the curve, which are pre-defined.

9. Mid-ordinate is also known as __________
a) Cosine of curve
b) Sine of curve
c) Versed cosine of curve
d) Versed sine of curve
Answer: a
Clarification: The value of mid-ordinate can be given as, M = R (1 – cos (Δ/2)) in which the value (1 – cos (Δ/2)) is expressed as versed sine. Mid-ordinate is the ordinate from midpoint of long chord to midpoint of curve.

10. The formula for tangent length can be given as __________
a) T = R + tan(Δ/2)
b) T = R * tan(Δ/2)
c) T = R / tan(Δ/2)
d) T = R – tan(Δ/2)
Answer: b
Clarification: The tangent distance can be defined as the distance between point of curvature to point of intersection, which is given as T = R tan (Δ/2). Here, Δ = deflection angle which is determined by setting instrument at required points.

11. Find the value of mid-ordinate if the value of R can be given as 22.19m and the angle is given as 19˚21ꞌ.
a) 0.89 m
b) 0.98 m
c) 0.13 m
d) 0.31 m
Answer: d
Clarification: The mid-ordinate can be determined by R-R*cos (θ/2), which on substitution may obtain,
= 22.19-22.19*cos (19˚21ꞌ/2)
= 0.31 m.

12. What would be the length of the curve, if the radius of the curve is 24.69m and the angle is given as 12˚42ꞌ?
a) 9.87 m
b) 5.74 m
c) 5.47 m
d) 9.78 m
Answer: c
Clarification: The formula for finding the length of the curve can be given as l = R*(π/180)*θ. On substitution, we get
l = 24.69*(π/180)*12˚42ꞌ
l = 5.47 m.

13. Find the tangent length if the radius of the curve and its angle were given as 42.64m and 42˚12ꞌ.
a) 16.45 m
b) 16.54 m
c) 61.45 m
d) 61.54 m
Answer: a
Clarification: The value of tangent length can be found out by using the formula,
T = r*tan (θ/2). On substitution, we get
T = 42.64*tan (42˚12ꞌ/2)
T = 16.45 m.

14. What would be the value of apex distance if the angle is given as 13˚42ꞌ and the radius of the curve is given as 19.24m?
a) 0.1134 m
b) 0.831 m
c) 0.318 m
d) 0.138 m
Answer: d
Clarification: The apex distance for a simple curve can be given as
E = R*(sec (θ/2)-1). On substitution, we get
E = 19.24*(sec (13˚42ꞌ/2)-1)
E = 0.138 m.

15. If the radius of the curve is given as 14.96m and the angle is about 32˚24ꞌ, find the length of the chord.
a) 8.43 m
b) 8.34 m
c) 4.83 m
d) 3.43 m
Answer: b
Clarification: Length of the chord can be given as L = 2*r*sin (θ/2)
L = 2*14.96*sin (32˚24ꞌ/2)
L = 8.34 m.

250+ TOP MCQs on Survey Adjustments and Errors Theory – Accidental Errors Laws and Answers

Surveying Multiple Choice Questions on “Survey Adjustments and Errors Theory – Accidental Errors Laws”.

1. The laws of accidental errors follow which of the following principle?
a) Normal equation
b) Probability law
c) Laws of weight
d) Most probable value
Answer: b
Clarification: Laws of accidental errors follow the probability law, which is having a definite law for accidental error occurrence. It defines the errors and helps in expressing them in the form of equations.

2. Which of the following does not indicate the feature in laws of accidental errors?
a) Negligible errors
b) Small errors
c) Large errors
d) Positive errors
Answer: a
Clarification: The features in laws of accidental errors include the tendency of small errors to be more frequent, positive and negative errors with equal frequency and making large errors occurrence impossible.

3. Most probable value is equal to which of the following?
a) Differentiation
b) Summation
c) Arithmetic mean
d) Normal equation
Answer: c
Clarification: Most probable value is equal to the arithmetic mean, in case all the taken weights are equal and in case of unequal weights, it is equal to the weighted arithmetic mean.

4. The value of mean square error can be given as__________
a) (∑v2+n)1/2
b) (∑v2*n)1/2
c) (∑v/n)1/2
d) (∑v2/n)1/2
Answer: d
Clarification: The mean square error is the ratio which is obtained by the taking mean of all the possible errors. It is taken as. It is useful in determining the possible error occurred and helps in reducing it by distributing it equally.

5. Probability curve describes about_______________
a) Normal equation
b) Frequency of errors
c) Probability curve
d) Probability equation
Answer: b
Clarification: The probability curve, which is established from the theory of probability, describes about the features like relative frequency of the errors in the form of curve. It is the basis for many mathematical derivations.

6. Determine the probable error in a single measurement if the summation of the difference between mean and single observation is given as 8.76 in a series of 7 observations.
a) 0.98
b) 0.93
c) 9.08
d) 0.89
Answer: a
Clarification: The value of the probable error of single observation can be determined by using the formula,
Es = 0.6745*(sqrt{∑v^2/(n-1)}.) On substitution, we get
Es = 0.6745*(sqrt{8.76^2/(7-1)}.)
Es = 0.98.

7. Determine the probable error of measurements by using the different probable errors, which are given as 5.64, 2.98, 0.98 and 2.54.
a) 3.96
b) 9.63
c) 6.93
d) 9.36
Answer: c
Clarification: The probable error of measurements can be given as,
Probable error of measurement = (sqrt{E1^2+E2^2+E3^2+E4^2} ). On substitution, we get
= (sqrt{5.64^2+2.98^2+0.98^2+2.54^2} )
= 6.93.

8. What will be the mean square error, if the readings were given as 2.654, 2.987, 2,432 and 2.543.
a) 3.305
b) 0.335
c) 0.305
d) 30.35
Answer: b
Clarification: The mean square error can be given as,
M.S.E = (sqrt{∑v^2/n}). The mean of the readings can be given as,
(2.645 + 2.987 + 2.432 + 2.543) / 4 = 2.651.
The values of v are obtained by difference of each reading with the mean. So, the ∑v can be given as
∑v = (2.651-2.645) + (2.987-2.651) + (2.651-2.432) + (2.652-2.543) = 0.67
On substitution, we get
M.S.E = (sqrt{∑v^2/n})
M.S.E = (sqrt{0.67^2/4})
M.S.E = 0.335.

9. If the value of error due to the single measurement is 6.54 for 10 observations, then calculate the value of average probable error.
a) 2.086
b) 2.608
c) 0.268
d) 2.068
Answer: d
Clarification: The average probable error can be calculated by using the formula,
Em = Es / (sqrt{n}). On substitution, we get
Em = 6.54 / (sqrt{10})
Em = 2.068.

10. Find the number of observations if the mean square error and the summation of the difference between the individual and the mean series are given as 0.987 and 3.654.
a) 14
b) 12
c) 10
d) 9
Answer: a
Clarification: The mean square error can be calculated by using the formula,
M.S.E = (sqrt{∑v^2/n}). On substitution, we get
0.987 = (sqrt{3.654^2/n})
n = 14 (approximately).