I am really disappointed by your marks. I was expecting much more than what you all have got. I can't revel the marks at this point of the time, but I am really really disappointed.
Some observations,
* Many of you have not written which question you are attempting. I had to find the question number and then specify
* Many of you have tried to attempt three from each of the sections, where as the only two was allowed from Sec-A and three from Sec-B
* None of you have actually done the numerical which was directly from Bhavakatti and Basak.
* None of you actually did the correct procedure for Question number 6
* None of you could even draw the correct diagram for Question number 7
* Only few of you have correctly done the Question 8
* I tried confusing you in Question 4 (both parts) and you got confused
* First three questions were directly from my slides, only few of you could score full marks in those
* Most surprising was your understanding of Significant Figures. Only a few were able to solve Question 5 (Part-1)
Below is the answers to all the questions. Evaluate yourselves, how much you should have got in this paper.
I expect a better performance from you in the end term.
All the best.
Solution:
1. What do you mean by Surveying? (0.5)
Science and art of determining the relative position of points above, on, or beneath the surface of the earth
or establishing such points with all kinds of appropriate approaches and instruments, which aims to make the natural appearances/ features known to the people
Whether it is Science and Arts? Explain. (1)
Both.
Science
Because it uses mathematical techniques to analyze field data
Accuracy and reliability depends on understanding scientific principles underlying and affecting survey measurement
Art
Because only a surveyor who has full understanding of surveying techniques will be able to determine the most efficient methods required to obtain optimal results over a wide variety of surveying problems
List different types of Surveying based on “Type of measurement” and “Platform used”. (2)
Type: Planar and Geodetic
Platform used:
Land Survey: Measurements performed on the surface of the earth
Aerial survey: Measurements done from the aerial platform, such as aircraft, balloon
Satellite Survey: Data captured using sensors on-board satellites
What is the end product of Surveying? (0.5)
Plan/ Map
2. What is/ are the quantity measured using the following instruments, Laser Range Finder; Dumpy Level; Chain, Total Station, compass and GPS. (3)
Laser Range Finders: Distance
Dumpy Level: Difference in elevations
Chain: Distance
Total Station: Angle and Distance
Compass: Angle
GPS: Co-ordinates
Why surveying in important to Civil Engineers? (1)
Civil Engineering has been associated with planning and construction of infrastructure
Construction is based on precise measurements
Measurement is achieved using “Surveying”
Surveying is the basic requirement of “Civil Engineers”
3. List the basic methods used in surveying. Explain them in brief. (2)
Measurement of Horizontal distances, Vertical distances, Horizontal angles and Vertical angles
Explain the following terms with appropriate examples:
Representative factor
Scale
Accuracy
Precision
(2)
RF: Ratio of distance on the map to the corresponding distance on the ground
Scale represents the “number of times” the actual area is reduced
Accuracy of measurement system is the degree of closeness of measurements of a quantity to its actual (true) value
Precision of a measurement system, also called reproducibility or repeatability, is the degree to which repeated measurements under unchanged conditions show the same results
4. Solve the following,
i) A rectangular plot in plan is 20cm x 40 cm, drawn to scale of 1cm = 500 m. If the same plot is drawn on a toposheet of 1:50,000, what would be the area on the toposheet. (2)
Actual size = 20x500 by 40x500 sqm => 10000m x 20000m => 200000000 sqm => 2x108 sqm
4.ii) A 30 m tape used for measuring a line was found to be 30.01 m at the beginning and 30.03 at the end of the work. The area of the plan drawn to a scale of 1:1000 was found to be 5146 mm2. Compute the correct area of the field. (2)
Method1:
Avg length of the tape = (30.01+30.03)/2 = 30.02m
Correct length of the tape = 30.0m (l)
Measured area = 5146 sqm
Correct area = (l’/l)2 x measured area = (30.02/30.0)2 x 5146 = 5152.86 sqm
Method2:
Avg length of the tape = (30.01+30.03)/2 = 30.02m = l’
Correct length of the tape = 30.0m (l)
e1 = 1-(l’/l) = 1-(30.02/30.0) = -6.67x10-4
Correct area = (1-2e1)x computed area = (1+2x6.67x10-4)x5146 = 5152.86 sqm
Section B
5. Solve the following, (2)
i) a) 
; = 0.305
; = 0.305b) 
; = 15.2
; = 15.2c) 3.22+5.1021+103.313+7.1; = 118.7
d) What would be the number 165.1, if a precision of 0.001 is used and 0.1 is used.
165.100; 165.1
ii) State the “Principle of Surveying”. (1)
Working from whole to parts
Tie the whole of the area first
Divide the area into smaller parts
Survey the smaller parts using the reference points from whole
Work from “Whole to Part”
Choose the method which is most suitable (Operations Research!)
Make provisions of adequate checks
Record field data carefully
What is degree of accuracy and state the different order of accuracy? (1)
How much close you are to the reference value.
Represented by
1 in xxxx
ppm (parts per million)
The minimum degree of accuracy allowed for a particular survey and the range of the allowed degrees of accuracy is known as the order of accuracy.
Order of accuracies: 1st order, 2nd order, 3rd order
6. A steel tape was exactly 20 m long at 200 C when supported throughout its length under a pull of 5 kg. A line measured with this tape under a pull of 16 kg and at a mean temperature of 320C, was found to be 680 m long. Assuming the tape is supported at every 20 m; find the true length of the line. Given that (a) Cross-sectional area of the tape= 0.03 cm2, (b) Young’s Modulus=2.1x106 kg/cm2, (c) Thermal coefficient of expansion= 11 x 10-6 per 0C and (d) weight of tape = 10 g/cc. (4)
Given:
L = 20m
T0 = 200C
P0 = 5kg
Tm = 320C
Pm = 16kg
ML = 680kg
n = 1
A = 0.03 cm2
a = 11x10-6 per 0C
E = 2.1x106 cm2
Given weight = 10gm/cc
Total weight of the tape = 0.03 x 20 x 100 x 10 = 600gm = 0.6kg
Corrections per tape length:
Temp correction = 
= 11x10-6 (32-30) x 20 = 0.00264m (+)
= 11x10-6 (32-30) x 20 = 0.00264m (+)Pull correction = 
= 
= 0.00349m (+)
= = 0.00349m (+)Sag correction = 
=0.00117 (-)
=0.00117 (-)Total correction = 0.00264 + 0.00349 – 0.00117 = 0.00496m
Hence actual length of the tape = 20 + 0.00496 = 20.00496m
True length of the line = 
= 
680.169m
= 680.169m7. Answer the following,
i) Explain the term hypotenuse allowance. (1)
Used when intermediate points are to be located for taking offsets
Length of a slope is always more than the corresponding horizontal distance. Thus, it is possible to measure a larger distance along a slope such that the equivalent horizontal distance is known
ii) There is an obstacle in the form of a pond on the main chain line AB. Two points C and D were taken on the opposite sides of the pond. A line CE of length 100m was drawn on the left side of line CD and another line CF of length 80m was drawn on the right side of CD (opposite direction of CE), such that E, D and F are in a straight line. Determine the obstruction length CD, where ED=60m and DF=56m. (3)
For CEF,
CE2 = CF2 + EF2 – 2.CF.EF.Cos q
1002 = 802 + 1162 – 2x80x116.Cos q
10000 = 6400 + 13456 – 18560 Cos q
Cos q = (10000-6400-13456)/18560 => Cos q = -0.53
For CDF,
CD2 = CF2 + DF2 – 2.CF.DF.Cos q
CD2 = 802 + 562 – 2x80x56.(-0.53)
CD2 = 6400+3136+4748.8 => CD = sqrt (14284.8) => CD = 119.52
8. Determine the values of Stadia constants (additive constant and multiplying constant) from the following observations, (4)
Instrument Station | Staff reading on | Distance | Stadia Readings | |
Lower | Upper | |||
O | A | 150 | 1.255 | 2.750 |
B | 200 | 1.000 | 3.000 | |
C | 250 | 0.750 | 3.255 | |
D = 
, where S: Stadia reading difference; (f+d): Additive constant = X; (f/i): Multiplicative constant = Y
, where S: Stadia reading difference; (f+d): Additive constant = X; (f/i): Multiplicative constant = YHence D = Y.S + X
For A, 150 = Y (2.750 – 1.255) + X => 150 = 1.495 Y + X
For B, 200 = Y (3.000 – 1.000) + X => 200 = 2.000 Y + X
For C, 250 = Y (3.255 – 0.750) + X => 250 = 2.505 Y + X
Solving, Y = 99.0; X = 2
