Levelling Reduction - Surveying Questions and Answers

This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Reduction in Levelling”.

1. Height of the Instrument method is a less tedious and simple process.
a) False
b) True
View Answer

Answer: b
Explanation: The above statement is true because it involves simple calculations when compared to Rise and fall method, which can be completed with minimal effort.

2. While doing construction work, which among the following is more suitable?
a) Rise and Fall method
b) Traversing
c) Height of the Instrument method(H.I)
d) Compass Surveying
View Answer

Answer: c
Explanation: Height of the Instrument method is less tedious than Rise and fall method. Moreover, this process is suitable for taking numerous readings from same instrument setting. Traversing is a process of establishing control points and Compass surveying involves in finding the bearings.

3. The formula for calculating R.L can be given as _________
a) H.I+F.S
b) H.I-F.S
c) H.I-B.S
d) H.I+B.S
View Answer

Answer: b
Explanation: By subtracting the fore sight value from height of the instrument, the value of reduced level for the next set of reading can be obtained.

4. Which of the following indicates the formula for arithmetic check?
a) ΣB.S-ΣF.S=Last R.L-First R.L
b) ΣF.S-ΣB.S=Last R.L-First R.L
c) ΣB.S+ΣF.S=First R.L-Last R.L
d) ΣF.S+ΣB.S=Last R.L-First R.L
View Answer

Answer: a
Explanation: If the difference in summation of back sight and fore sight is equal to the difference of last R.L and first R.L, then obtained set of values for finding difference in elevation are correct.

5. Rise and fall method provides check in calculations for all sights.
a) True
b) False
View Answer

Answer: a
Explanation: Since the check for intermediate sights is not available in H.I method, it may lead to errors while doing calculations.

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6. Which of the following represents a form of Bench Mark (B.M)?
a) True Benchmark
b) Assumed Benchmark
c) Datum
d) Arbitrary Benchmark
View Answer

Answer: d
Explanation: Benchmark is a point of known elevation taken as reference. It is classified as Arbitrary, G.T.S, Permanent and Temporary.

7. If the staff at the station point is not held vertically, the R.L at the observation would be ______________
a) Less than true R.L
b) Greater than true R.L
c) Equal to the true R.L
d) Two times the true R.L
View Answer

Answer: c
Explanation: Due to irregular holding of the staff, the values which are obtained by the levelling instrument might lead to decrease in R.L.

8. If the R.L of a B.M is 100m and back sight is 1.225m, find the H.I at the station?
a) 101.225m
b) -101.225m
c) 98.775m
d) -98.775m
View Answer

Answer: a
Explanation: We know that, H.I=R.L+B.S
Then, H.I= 100+1.225 H.I=101.225 m.

9. The combined correction for curvature and refraction can be given as _______
a) C = 14d²/6R
b) C = 6d²/7R
c) C = 7d²/2R
d) C = 6d²/14R
View Answer

Answer: d
Explanation: By subtracting the correction due to curvature from correction due to refraction we can get the combined correction i.e., \(C = \frac{d^2}{2R} – (\frac{1}{7}*\frac{d^2}{2R}).\)

10. Find the correction for curvature and correction for refraction, if the value of d = 2400 m?
a) 0.425, 0.604
b) 0.452, 0.064
c) 0.064, 0.452
d) 0.604, 0.425
View Answer

Answer: b
Explanation: Correction for curvature = \(\frac{d^2}{2R} = \frac{2.4^2}{2*6370}*1000 = 0.452m\) Correction for refraction \( = \frac{1}{7} * C_c = \frac{0.452}{7} = 0.064m.\)

11. Find the value of R.L, if B.M = 400 m, B.S = 1.142 m, F.S = 2.121 by using rise and fall method?
a) 400.79 m
b) 400.97 m
c) 409.79 m
d) 399.02 m
View Answer

Answer: d
Explanation: In the rise and fall method, first we must calculate the difference between B.S and F.S. We get 1.142 – 2.121 = -0.979 m, which is negative. It means we have to subtract it from the given B.M for obtaining R.L i.e., R.L = 400 – 0.979 = 399.021 m.

12. If d = 2.94 km, what would be the combined correction for curvature and refraction?
a) 1.85 km
b) 0.85 km
c) 0.58 km
d) 1.58 km
View Answer

Answer: c
Explanation: We know that combined correction for curvature can be given as \(C = \frac{6d^2}{14R}\) On substituting the value the of d in the above equation we get,
\(C = \frac{6d^2}{14R} = \frac{6*2.94^2}{14*6370} = 0.5814 km.\)

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