Surveying Multiple Choice Questions - Group 3
Q.1. The staff intercept is ____
Length of staff
Folding length of staff
List count of staff
Difference between upper and lower cross hair
In diaphragm there are three cross hairs upper, middle and lower. While carry out tachometric survey distance between station and staff is find on the basis of staff intercept.
There fore correct answer is Option - 4.
There fore correct answer is Option - 4.
Q.2. Correct the following
Staff intercept is minimum when staff held normal to the line of sight.
Staff intercept is maximum when staff held normal to the line of sight.
Both of above
None of above
Staff intercept is distance between upper and lower cross hair. It becomes min. when staff held normal to the line of sight.
There fore correct ans. is Option - 3.
There fore correct ans. is Option - 3.
Q.3. The latitude and departure of a line AB are +78m and −45.1m respectively. The
whole circle bearing of the line AB is
GATE-2013
300
1500
2100
3300
Latitude of given line is +ve.
Departure of given line is -ve.
∴ line in 4th quadrant.
L cos θ = 78.
L cos θ = 45.
tan θ = -0.578.
θ = -300.
∴ WCB = 3600 - 300 = 3300
There fore correct answer is Option - 4.
Q.4. Applying correction due to local attraction, the correct bearing of line BC will be. If bearing of the given system is shown below:
| Line | FB | BB |
| AB | 126045' | 308000' |
| BC | 45015' | 227030' |
| CD | 340030' | 161045' |
| DE | 258030' | 78030' | EA | 216030' | 31045' |
GATE-2013
48015'.
50015'.
49015'.
48045'.
First find the stations which are free form local attraction.
Which are done by taking difference between FF and BB. If this difference is 1800 then the end points of that line are free from local attraction.
Here for line DE 258030' - 78030' = 180000'
There fore stations D and E are free from local attraction.
Now, start correction form line EA
FB of EA = 216030'
Correct BB of EA = 216030' - 1800 = 36030'
Observed BB of EA = 31045'
error = Observed reading - Correct reading = 31045' - 36030'
error = - 4045'
correction = + 4045'
Correct FB of AB = 126045' + 4045' = 131030'
Correct BB of AB = 131030' + 1800 = 311030'
Observed BB of AB = 3080
error = Observed reading - Correct reading = 3080 - 311030'
error = - 3030'
correction = + 3030'
Correct BB of BC = 45015' + 3030' = 48045'
There fore correct answer is Option - 4.
Which are done by taking difference between FF and BB. If this difference is 1800 then the end points of that line are free from local attraction.
Here for line DE 258030' - 78030' = 180000'
There fore stations D and E are free from local attraction.
Now, start correction form line EA
FB of EA = 216030'
Correct BB of EA = 216030' - 1800 = 36030'
Observed BB of EA = 31045'
error = Observed reading - Correct reading = 31045' - 36030'
error = - 4045'
correction = + 4045'
Correct FB of AB = 126045' + 4045' = 131030'
Correct BB of AB = 131030' + 1800 = 311030'
Observed BB of AB = 3080
error = Observed reading - Correct reading = 3080 - 311030'
error = - 3030'
correction = + 3030'
Correct BB of BC = 45015' + 3030' = 48045'
There fore correct answer is Option - 4.
Q.5. A Theodolite is placed at A and a 3 m long vertical staff is held at B. The
depression angle made at reading of 2.5m marking on staff is o 6 10'.The
horizontal distance between A and B is 2200 m. The height of instrument at A is
1.2 m and reduced level of point A is 880.88 m. Using curvature correction and
refraction correction determine the R.L. of point B (in m).
GATE-2013
602.205 m
641.205 m
642.205 m
none of the above
Let curvature correction factor, Cc = - 0.07849 x (length in km)2
Refraction Correction factor, Cr = 0.01121 x (length in km)2
Combined Correction C = Cc+Cr= -0.06728 x (length in km)2
Cc= -0.06728 x 2.22 = 0.3256 m
True staff reading = 2.5 - Cc = 2.5 - 0.3256 = 2.1743 m
Now, B'B" = (tan 6010') x 2200 = 237.701 m
HI = 880.88 + 1.2 = 882.08 m
RL of B = HI - True staff reading - B'B" = 642.205 m
There fore correct answer is Option - 3.
Q.6. Which of following error can be eliminated by reciprocal measurements in differential leveling?
1. Error due to earth curvature.
2. Error due to atmospheric - refraction.
1. Error due to earth curvature.
2. Error due to atmospheric - refraction.
GATE-2012
Both 1 and 2.
1 only.
2 only.
Neither 1 nor 2.
Curvature and Vertical Refraction Zenith angles measured for horizontal and vertical reduction of long lines require curvature and refraction corrections. The manufacturer’s specifications for most total stations indicate that the earth's curvature and atmospheric refraction are internally computed and the corrected horizontal and vertical distances displayed. This correction can be eliminated (or balanced) by taking the mean of two reciprocal measurements.
There fore correct answer is Option - 1.
There fore correct answer is Option - 1.
Q.7. Match the following related to surveying.
| Group - I | Group - II |
| P) Alidade | 1) Chain Survey |
| Q) Arrow | 2) Levelling |
| R) Bubble tube | 3) Plant table surveying |
| S) Stedia hair | 4) Theodolite |
GATE-2014
P – 3, Q – 2, R – 1, S – 4
P – 2, Q – 4, R – 3, S – 1
P – 1, Q – 2, R – 4, S – 3
P – 3, Q – 1, R – 2, S – 4
In above table Group - I consist of list of instruments used in survey work. And Group - II consist of there uses for particular survey purpose.
i.e. Alidade > used in Plant table surveying
Arrow > used in Chain Survey
Bubble tube > used in Levelling
Stedia hair > used in Theodolite
There fore correct answer is Option - 4.
i.e. Alidade > used in Plant table surveying
Arrow > used in Chain Survey
Bubble tube > used in Levelling
Stedia hair > used in Theodolite
There fore correct answer is Option - 4.
Q.8. On map long form of R. F. is____
Refractive Factor.
Refractive Fraction.
Representative Fraction.
Representative Factor.
Q.9. A levelling is carried out to established the reduced level (RL) of point R
with respect to the bench mark (BM) at P. If RL of P is + 100 m, then what is the RL (m) of R? The staff reading taken are given
below.
| Staff at | BS | IS | FS | RL |
| P | 1.655 | - | - | 100.00 |
| Q | -0.950 | - | -1.500 | - |
| R | - | - | 0.75 | ? |
GATE 2014
103.355.
103.155.
101.455.
100.355
This is carried out fly levelling. There for we carry out HI at each station.
HI at station P = 100 + 1.655 = 101.655 m
RL of station Q = 101.655 + 1.500 (-1.500 staff reading means inverted staff reading)
RL of station Q = 103.155 m
Instrument is changed from this point and shifted other place. And staff reading taken with inverted staff(i.e. -0.950 )
HI at station Q = RL of Q - 0.950 = 102.205
Staff is held at point R which is below HI.
RL of station R = HI - 0.750 = 102.205 - 0.750 = 101.455m
There fore correct answer is Option - 3.
HI at station P = 100 + 1.655 = 101.655 m
RL of station Q = 101.655 + 1.500 (-1.500 staff reading means inverted staff reading)
RL of station Q = 103.155 m
Instrument is changed from this point and shifted other place. And staff reading taken with inverted staff(i.e. -0.950 )
HI at station Q = RL of Q - 0.950 = 102.205
Staff is held at point R which is below HI.
RL of station R = HI - 0.750 = 102.205 - 0.750 = 101.455m
There fore correct answer is Option - 3.
Q.10. A line PQ is measured by 30 m chain is 1230 m. At the end chain was tested and it found that chain 12 cm too short. Find correct length of line PQ.
1225.08 m.
1230 m.
1229.88 m.
none of the above
Correct Chain length = L = 30 m.
Incorrect Chain length = L' = 30 m - 0.12 m = 29.88 m.
Incorrect PQ length = 1230.
Correct PQ length = L'/L * 1230
Correct PQ length = 1225.08 m
There fore correct answer is Option - 1.
Incorrect Chain length = L' = 30 m - 0.12 m = 29.88 m.
Incorrect PQ length = 1230.
Correct PQ length = L'/L * 1230
Correct PQ length = 1225.08 m
There fore correct answer is Option - 1.
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