Surveying Multiple Choice Questions on “Reciprocal Levelling”.
1. Which of the following type of levelling is necessary across a river ravine or any obstacle requiring a long site between two points?
a) Barometric levelling
b) Trigonometric levelling
c) Reciprocal levelling
d) Spirit levelling
Answer: c
Clarification: When it is necessary to carry a levelling across a river ravine or any obstacle requiring a long site between two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal, special method that is reciprocal levelling must be used. It is used to obtain better accuracy.
2. Which of the following type of levelling is used when two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal?
a) Barometric levelling
b) Trigonometric levelling
c) Reciprocal levelling
d) Spirit levelling
Answer: c
Clarification: When it is necessary to carry a levelling across a river ravine or any obstacle requiring a long site between two points so situated that no place for the level can be found from which the lens of foresight and backsight will be even approximately equal, special method that is reciprocal levelling must be used. It is used to obtain better accuracy.
3. Which of the following error cannot be eliminated in reciprocal levelling?
a) error in instrument adjustment
b) combined effect of earth’s curvature and refraction of atmosphere
c) variations in average refraction
d) variation in temperature
Answer: d
Clarification: Reciprocal levelling must be used to obtain better accuracy. It is also used to eliminate error in instrument adjustment, combined effect of earth’s curvature and refraction of atmosphere, variations in average refraction.
4. When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When the instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L. of Q?
a) 125.555
b) 125.565
c)125.575
d)125.585
Answer: d
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 m. Therefore true elevation of Q is 126.386 – 0.801 = 125.585 m.
5. When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?
a) 0.057 m
b) 0.069 m
c) 0.058 m
d) 0.048 m
Answer: b
Clarification: Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 mts.
6. When an instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the error due to collimation?
a) 0.058 mts
b) 0.052 mts
c) 0.054 mts
d) 0.068 mts
Answer: c
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 – 0.069 = 0.054 m.
7. When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find true R.L of Q?
a) 125.585
b) 126.187
c) 127.187
d) 128.197
Answer: c
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Therefore true elevation of Q is 126.386 + 0.801 = 127.187 m.
8. When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect correction for curvature and refraction?
a) 0.057 m
b) 0.069 m
c) 0.058 m
d) 0.048 m
Answer: b
Clarification: Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 m.
9. When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find error due to collimation?
a) 0.178 mts
b) 0.192 mts
c) 0.194 mts
d) 0.188 mts
Answer: b
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 + 0.069 = 0.192 m.
10. When an instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606 and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find the angular error in collimation adjustment of the instrument?
a) 39″
b) 49″
c) 59″
d) 69″
Answer: a
Clarification: When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 + 0.069 = 0.192 m. If ∆ is the inclination of line of the site then tan∆ = 0.192/1010 = 0.000190. Therefore ∆ =39″.