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ASRB NET Surveying & Levelling: practice questions

ASRB NET — Farm Machinery & Power / Land & Water Management Engineering / Agricultural Structures & Process Engineering · India · Agricultural Scientists Recruitment Board (ICAR)

Original practice questions written to the published syllabus · not past papers
10worked here, free
30written to this exam in this subject
12syllabus units

What this subject covers

Measuring land, levels and volumes, from chain survey to total station and GNSS, and designing land grading.

In the exams: GATE AG Section 4 (surveying asked in 7 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; TNPSC Unit I; RPSC AAE Unit 1; NCEES PE surveying

Where the 30 questions fall

How our questions for ASRB NET spread across the units of Surveying & Levelling. Difficulty: 1 easy, 8 medium, 21 hard.

Land grading and levelling design7
Levelling: rise-and-fall and height-of-instrument methods3
Reciprocal levelling3
Theodolite traversing and closing error3
GPS/GNSS surveying3
Chain and tape surveying, including tape corrections2
Contouring2
Total station2
Earthwork volume computation2
Compass traversing and bearings1
Plane table surveying1
Computation of area: trapezoidal and Simpson's rules1

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Levelling: rise-and-fall and height-of-instrument methods · hard

When the bubble of a level was moved through 4 divisions, the staff reading at 80 m changed by 0.012 m. What is the sensitivity of the bubble tube per division?

  1. 7.7 seconds
  2. 123.8 seconds
  3. 30.9 seconds
  4. 7.7 minutes
Answer and working

A. 7.7 seconds

α = 206,265 × s/(nD) = 206,265 × 0.012/(4 × 80) = 7.7″ per division. Forgetting to divide by the number of divisions gives 30.9″; multiplying by n gives 123.8″; 7.7 minutes has the right number with the wrong unit.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

2 · Levelling: rise-and-fall and height-of-instrument methods · hard

A staff is read from a level. The staff reading is 3.000 m when held vertical, but the staff was actually tilted 3° from the vertical in the direction of sight. What reading was observed (to 3 decimals), and what is the lesson?

  1. 3.000 m; tilt has no effect on a horizontal sight
  2. 2.996 m; a tilted staff reads too low
  3. 3.004 m; a tilted staff always reads too high
  4. 3.157 m; the error is proportional to sin θ
Answer and working

C. 3.004 m; a tilted staff always reads too high

The horizontal line cuts the tilted staff at a graduation of 3.000/cos 3° = 3.004 m. Tilt in any direction increases the reading, which is why the staffman rocks the staff and the minimum reading is taken.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

3 · Reciprocal levelling · hard

A level set at A reads 1.250 m on a staff at a point B 2.4 km away. Correcting for curvature and refraction (0.0673d² m, d in km), what is the corrected staff reading at B?

  1. 0.798 m
  2. 0.862 m
  3. 1.088 m
  4. 1.638 m
Answer and working

B. 0.862 m

Correction = −0.0673 × 2.4² = −0.3876 m, so corrected reading = 1.250 − 0.3876 = 0.862 m. The observed reading is too large because the level surface curves away below the horizontal line of sight; adding the correction is the sign slip. 0.798 m applies curvature alone (0.0785d²); 1.088 m forgets to square d; 1.638 m adds the correction.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

4 · Contouring · hard

Reservoir areas enclosed by contours at 100, 102, 104, 106 and 108 m are 1200, 2600, 4100, 5900 and 8000 m². What is the storage between 100 m and 108 m by the prismoidal formula?

  1. 68533 m³
  2. 28400 m³
  3. 34400 m³
  4. 34267 m³
Answer and working

D. 34267 m³

Prismoidal: V = (d/3)[(A1 + A5) + 4(A2 + A4) + 2A3] = (2/3)[9200 + 4 × 8500 + 2 × 4100] = 34267 m³. The trapezoidal formula gives 34400 m³; swapping the 4 and 2 weights is a common slip; using d = 4 m double-counts. Using d = 4 m gives 68533 m³; swapping the weights gives 28400 m³.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

5 · Theodolite traversing and closing error · medium

The transit rule for balancing a traverse is preferred over Bowditch's rule when

  1. the traverse is open
  2. a compass is used for the bearings
  3. distances are measured more precisely than angles
  4. angles are measured more precisely than distances
Answer and working

D. angles are measured more precisely than distances

The transit rule distributes corrections in proportion to the size of each latitude and departure, which assumes angles are good and lengths poorer (theodolite with tape). Bowditch assumes equal effect of angular and linear errors and is the usual rule for compass traverses. Neither can be applied to an open traverse.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

6 · GPS/GNSS surveying · hard

An RTK GNSS survey gives an ellipsoidal height of 18.6 m at a point where the geoid undulation N from the geoid model is −52.3 m. What is the orthometric height (height above the geoid)?

  1. 18.6 m
  2. -70.9 m
  3. 70.9 m
  4. -33.7 m
Answer and working

C. 70.9 m

H = h − N = 18.6 − (−52.3) = 70.9 m. Where the geoid lies below the ellipsoid (negative N), orthometric heights exceed ellipsoidal heights. Adding N gives -33.7 m; using the ellipsoidal height as if it were above mean sea level is the classic mistake.

Further reading: Ghilani and Wolf, Elementary Surveying

7 · Computation of area: trapezoidal and Simpson's rules · hard

Six offsets of 2.0, 3.5, 4.2, 3.9, 3.0 and 2.4 m are taken at 10 m intervals. Using Simpson's rule for the first five ordinates and the trapezoidal rule for the last strip, what is the total area?

  1. 170.33 m²
  2. 161.33 m²
  3. 143.33 m²
  4. 168.00 m²
Answer and working

A. 170.33 m²

Simpson on O1–O5: (10/3)[2.0 + 3.0 + 4(3.5 + 3.9) + 2(4.2)] = 143.33 m². Last strip: 10(3.0 + 2.4)/2 = 27.00 m². Total = 170.33 m². Forcing Simpson onto six ordinates gives 161.33 m², which is invalid; trapezoidal throughout gives 168.00 m².

Further reading: Punmia, Jain and Jain, Surveying Vol. I

8 · Land grading and levelling design · medium

When is the plane method of land-grading design NOT the best choice?

  1. On fields so irregular that a single plane would require excessive earth movement
  2. When the field is square
  3. On nearly uniform fields
  4. When least-squares fitting can be computed quickly by software instead of by hand calculation
Answer and working

A. On fields so irregular that a single plane would require excessive earth movement

A single best-fit plane is efficient for fairly uniform fields; where the surface is strongly irregular, profile, plan-inspection or contour-adjustment methods allow varying grades and cut far less earth.

Further reading: Schwab et al., Soil and Water Conservation Engineering

9 · Land grading and levelling design · hard

Which is a well-documented agronomic drawback of deep cuts in land grading, and how is it usually managed?

  1. Cut areas expose less fertile subsoil and can yield less for some seasons; minimise cut depth and apply organic matter or fertiliser to cut areas
  2. Cut areas always yield more because of better drainage; no management is needed
  3. Deep cuts usually raise the water table everywhere
  4. Fill areas cannot be cropped for five years by law
Answer and working

A. Cut areas expose less fertile subsoil and can yield less for some seasons; minimise cut depth and apply organic matter or fertiliser to cut areas

Removing topsoil exposes subsoil lower in organic matter and nutrients, and crops on cut patches often lag until fertility is rebuilt. Designs minimise earth moved, and cut areas receive manure or fertiliser. The other options are not general truths.

Further reading: Schwab et al., Soil and Water Conservation Engineering

10 · Contouring · medium

Which statement about contour lines is FALSE?

  1. Contours cross ridge and valley lines at right angles
  2. Contours of different elevations can merge into a single line on a gentle slope
  3. Evenly spaced, straight and parallel contour lines usually indicate a uniformly sloping plane surface
  4. A closed contour with higher values inside indicates a hill
Answer and working

B. Contours of different elevations can merge into a single line on a gentle slope

Contours of different elevations merge only at a vertical cliff and cross only at an overhang; on a gentle slope they are widely spaced, never merged. The other three statements are standard characteristics.

Further reading: Punmia, Jain and Jain, Surveying Vol. I

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Surveying & LevellingThe full study manual, free ASRB NETFormat, sections and how it has changed

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Surveying & Levelling for other exams

ICAR AIEEA-PG · IndiaGATE · IndiaRajasthan PSC · IndiaTamil Nadu PSC · IndiaNCEES PE Agricultural and Biological Engineering · USA

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