JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II SURVEYING AND GEOMATICS Previous Question Paper (September - 2021 - R18)

B.Tech | II | II-I | R18 | September - 2021

Answer PaperPaper Code: 153BVMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

List out the methods of Ranging and explain with neat sketches.

15 marks

Answer

Ranging is the process of establishing intermediate points on a straight line between two terminal points when the distance between them exceeds the length of the surveying chain or tape. The two primary methods are Direct Ranging and Indirect (or Reciprocal) Ranging.

1. Direct Ranging: This method is used when the two end points are intervisible. A surveyor stands behind one end station and directs an assistant to move a ranging rod until it aligns with the line of sight between the two terminal points.

  • The surveyor stands at station A and sights station B.
  • The assistant holds a ranging rod at an intermediate point C, roughly on the line.
  • The surveyor signals the assistant to move the rod left or right until it appears to coincide with the line AB.
  • Once aligned, the assistant fixes the rod firmly in the ground.

2. Indirect Ranging: This method is used when the terminal points are not intervisible due to high ground, a hill, or a depression. It requires two intermediate points to be established through a process of mutual alignment.

  • Assume two points M and N are chosen such that both A and B are visible from them.
  • The first person at M directs the second person at N to move to N' so that M, N', and B are in a straight line.
  • The person at N' then directs the person at M to move to M' so that N', M', and A are in a straight line.
  • This process is repeated until the points M and N are perfectly aligned with A and B.

Visualizing the sketches: For direct ranging, imagine a single straight line connecting A and B with a rod C placed in the middle. For indirect ranging, imagine a zig-zag movement between two intermediate points M and N, iteratively shifting them until they fall on the line segment AB, effectively 'swinging' the line into position.

2

The following bearings were observed in a closed traverse. Find the included angles and correct the bearings of the lines.

LineFBBB
AB140° 45'318° 15'
BC216° 30'38° 00'
CD209° 15'30° 15'
DE319° 45'139° 45'
EA60° 15'240° 15'
15 marks

Answer

To correct the bearings and find the included angles, we first check for local attraction by comparing the difference between Fore Bearing (FB) and Back Bearing (BB). The theoretical difference is 180°.

LineFBBBDifference
AB140° 45'318° 15'177° 30'
BC216° 30'38° 00'178° 30'
CD209° 15'30° 15'179° 00'
DE319° 45'139° 45'180° 00'
EA60° 15'240° 15'180° 00'

Lines DE and EA have a difference of 180°, indicating stations D, E, and A are free from local attraction. We adjust the bearings starting from station A.

  • Corrected Bearings:
    • EA (FB): 60° 15', EA (BB): 240° 15' (Correct)
    • DE (FB): 319° 45', DE (BB): 139° 45' (Correct)
    • CD (FB): 209° 15', CD (BB): 29° 15' (Corrected by -1° 00' based on station D)
    • BC (FB): 215° 30', BC (BB): 35° 30' (Corrected by -1° 00' based on station C)
    • AB (FB): 140° 45', AB (BB): 320° 45' (Corrected by +2° 30' based on station B)
  • Included Angles:
    Included angle at a station = FB of next line - BB of previous line (if result is negative, add 360°).
    • Angle A = FB(AB) - BB(EA) = 140° 45' - 240° 15' + 360° = 260° 30' (Exterior) or 99° 30' (Interior)
    • Angle B = FB(BC) - BB(AB) = 215° 30' - 320° 45' + 360° = 254° 45' (Exterior) or 105° 15' (Interior)
    • Angle C = FB(CD) - BB(BC) = 209° 15' - 35° 30' = 173° 45'
    • Angle D = FB(DE) - BB(CD) = 319° 45' - 29° 15' = 290° 30' (Exterior) or 69° 30' (Interior)
    • Angle E = FB(EA) - BB(DE) = 60° 15' - 139° 45' + 360° = 280° 30' (Exterior) or 79° 30' (Interior)
3

List out the methods of contouring and explain about any one method in detail with a neat sketch.

15 marks

Answer

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4

The following offsets were taken from a survey line to a curved boundary line:

Distance (m)0510152030406080
Offset (m)2.53.84.65.26.14.75.83.92.2

Find the area between the survey line, the curved boundary line and the first, last offsets by

15 marks
a

Trapezoidal rule

7 marks

Answer

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b

Simpson’s rule

8 marks

Answer

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5

How can the height of a tower be determined when it is inaccessible?

15 marks

Answer

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6

The following data were collected while running a closed traverse ABCDA. Determine the missing data.

LineLength (m)Bearing
AB330181° 25'
BC?89° 50'
CD411355° 00'
DA827?
15 marks

Answer

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7

Explain the function of the following curves:

15 marks
a

Simple circular curve

3 marks

Answer

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b

Compound curve

3 marks

Answer

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c

Reverse curve

3 marks

Answer

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d

Transition curve

3 marks

Answer

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e

Vertical curve

3 marks

Answer

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8

Explain about the photographic mapping by using:

15 marks
a

stereo plotting instruments

8 marks

Answer

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b

mosaics.

7 marks

Answer

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This page contains the question paper of SURVEYING AND GEOMATICS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech CE II II-I, September - 2021, R18.

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