Exam Details

Subject surveying
Paper
Exam / Course pddc
Department
Organization Gujarat Technological University
Position
Exam Date January, 2019
City, State gujarat, ahmedabad


Question Paper

1
Seat No.: Enrolment
GUJARAT TECHNOLOGICAL UNIVERSITY
PD -SEMESTER 1 (NEW SYLLABUS) EXAMINATION- WINTER 2018
Subject Code:2910602 Date: 09-01-2019
Subject Name: SURVEYING
Time: 10:30 am to 01:00 pm Total Marks: 70
Instructions:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary.
3. Figures to the right indicate full marks.
Marks
Q.1

What is tacheometry? Differentiate between fixed hair method and movable hair method.
03

Explain the resection method with the help of two point problems.
04

Discuss the laws of accidental errors.
07
Q.2

Define the following terms:
Independent Quantity weight of an observation
Most Probable Value
03

List out the methods of plane table and explain any two of them.
04

What is the object of the geodetic surveying? Write uses of triangulation.
07
OR

Derive expression for horizontal and vertical distance in fixed hair method when the staff is held vertically and the measured angle is that of elevation. How will you find the R.L of staff station?
07
Q.3

Define the following terms (with Figure):
Back tangent, Apex distance, Long chord
03

Find the tacheometry constants from the observations given in table.
Station
Staff station
Vertical angle
Cross Hair reading
Distance
A
C
D


1.325,2.122
0.985,2.382
80
140
04

Two straights are AB and BC intersect at chainage the angle of deflection being 110˚. Calculate the chainage of the tangent points of a right handed circular curve of 400 m radius. (Chain was used of 20 m).
07
OR
Q.3

Define the following terms (with Figure):
Mid ordinate, Point of curve, Point of tangency
03

Give difference between following.
Latitude and Departure
Consecutive co-ordinates and Independent co-ordinates
04

Two tangents intersect at the chainage 1190 the deflection angle being Calculate all the data necessary for setting out a curve with a radius of 300 m by deflection angle method. The peg interval is 30 m.
07
2
Q.4

Derive the formula for height and distance in trigonometric levelling when two instruments are set at same level.
03

Write a short note on Plannimeter.
04

The following offsets were taken from a chain line to a irregular boundary line at a interval of 10.0 m
Offsets 0,2.50,3.50,5.00,4.60,3.20,0 m
compute the area between the chain line, the irregular boundary line and the offset by Mid-ordinate rule, Average Ordinate rule and Trapezoidal rule
07
OR
Q.4

What is trigonometric levelling? Describe their advantage and disadvantage over direct levelling.
03

What is prismoid? Derive the prismoidal formula.
04

The following perpendicular offsets were taken from a chain line to a hedge:
Distance(m)
0
6
12
18
24
30
36
Offset(m)
5.40
4.50
3.60
2.70
1.80
2.25
2.80
Calculate the area enclosed between the chain line and the offsets by Trapezoidal rule and Simpson's rule.
07
Q.5

Define following terms:
The vertical axis The horizontal axis The line of sight
03

For a closed traverse ABCDA, compute the missing data.
Line
Length
Bearing
AB
201.80
N45o30'W
BC
288.40
N60o30'E
CD
192.60
S34o45'E
DA


04

Define remote sensing and explain principle of remote sensing.
07
OR
Q.5

Write a note on Total Station.
03

Write a note on GPS.
04

A tacheometer was set up at a station C and the following readings were obtained on a staff vertically held .
Calculate the horizontal distance CD and RL of when constants of instruments are 100 and 0.15.
Inst. St.
Staff St.
Vertical angle
Hair Reading
R.L of B.M on ground
C
BM

1.500,1.800,2.450
750.50 m
C
D

0.750,1.500,2.250
07


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