# Surveying Questions and Answers – Field Astronomy – Astronomical Corrections

This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Field Astronomy – Astronomical Corrections”.

1. Which of the following doesn’t belong to the set of corrections applied in astronomical corrections?
a) Correction of parallax
b) Correction of sag
c) Correction of refraction
d) Correction of semi-diameter

Explanation: Astronomical corrections are applied in the case of celestial bodies. They generally include corrections for parallax, for refraction, for semi-diameter, for the dip of horizon. All these are not applied at once but are applied based on the necessity.

2. Magnitude of refraction depends upon which of the following factors?
a) Density
b) Surface tension
c) Reflection
d) Polarisation

Explanation: Due to the curvature of the earth surface, the layers of the atmosphere can be thinner as its distance from surface increases. It may cause the deviation in the angle of ray which is equal to refraction angle. The magnitude of refraction depends on density of air, temperature, pressure of barometer, altitude.

3. Correction to the dip is always_____________
a) Zero
b) Multiplicative
c) Subtractive

Explanation: Angle of dip can be assumed as the angle between true and visible horizon. Due to the curvature of the earth, the dip angle must be subtracted from the observed angle, which makes the correction as subtractive.

4. Determine the index error for face right, if the face left and face right readings were given as 18˚24ꞌ52ꞌꞌ and 18˚23ꞌ24ꞌꞌ.
a) +24ꞌꞌ
b) +49ꞌꞌ
c) +4ꞌꞌ
d) +44ꞌꞌ

Explanation: For determining the correction for index error, calculate the mean reading = (18˚24ꞌ52ꞌꞌ + 18˚23ꞌ24ꞌꞌ) / 2 = 18˚24ꞌ8ꞌꞌ
Index error for face right can be given as 18˚24ꞌ8ꞌꞌ – 18˚23ꞌ24ꞌꞌ = +44ꞌꞌ.

5. Find the altitude correction for semi-diameter which is having index error 25˚46ꞌ21ꞌꞌ and semi-diameter 0˚26ꞌ21ꞌꞌ.
a) 26˚12ꞌ42ꞌꞌ
b) 62˚12ꞌ42ꞌꞌ
c) 26˚21ꞌ42ꞌꞌ
d) 26˚12ꞌ24ꞌꞌ

Explanation: The altitude correction for the semi-diameter can be given as the summation of index error with the semi-diameter mentioned i.e.,
25˚46ꞌ21ꞌꞌ + 0˚26ꞌ21ꞌꞌ = 26˚12ꞌ42ꞌꞌ.

6. Determine the correction for refraction if the angle of azimuth is given as 62˚21ꞌ24ꞌꞌ.
a) 10˚11ꞌ50.74ꞌꞌ
b) 0˚1ꞌ50.74ꞌꞌ
c) 0˚1ꞌ0.74ꞌꞌ
d) 50˚1ꞌ50.74ꞌꞌ

Explanation: The correction for refraction can be given as 58ꞌꞌ tan z. Where, z is the Azimuthal angle. On substitution, we get
58ꞌꞌ*tan (62˚21ꞌ24ꞌꞌ) = 0˚1ꞌ50.74ꞌꞌ.

7. Determine the correction for parallax which has to be applied to 29˚42ꞌ31ꞌꞌ for obtaining altitude of the sun, which is given as 32˚41ꞌ15ꞌꞌ.
a) 53˚51ꞌ
b) 53˚15ꞌ
c) 35˚51ꞌ
d) 5˚51ꞌ

Explanation: Correction for parallax can be given as
Correction for parallax = horizontal parallax * cos apparent latitude
Correction for parallax = 8*8cos α. On substitution, we get
Correction for parallax = 53˚51ꞌ, this has to applied for 29˚42ꞌ31ꞌꞌ in order to have the correct altitude of the sun.

8. Determine the corrected azimuth value if the azimuth is given as 54˚32ꞌ15ꞌꞌhaving b = +23ꞌꞌ with vertical angle 29˚42ꞌ31ꞌꞌ.
a) 67˚39ꞌ40ꞌꞌ
b) 67˚30ꞌ39.78ꞌꞌ
c) 67˚39ꞌ39.78ꞌꞌ
d) 76˚39ꞌ39.78ꞌꞌ

Explanation: The corrected azimuth can be determined by the summation of azimuth with correction applied.
Correction = b*tan α = 23*tan 29˚42ꞌ31ꞌꞌ = 13˚7ꞌ24.8ꞌꞌ. On summation,
Corrected azimuth = 54˚32ꞌ15ꞌꞌ + 13˚7ꞌ24.8ꞌꞌ
Corrected azimuth = 67˚39ꞌ39.78ꞌꞌ.

9. Which of the following is always subtractive?
a) Correction for reflection
b) Correction for dip
c) Correction for parallax
d) Correction for polarization

Explanation: The correction applied for the dip is always subtractive. Dip is the angle between true and the visible horizon, the observations are taken from the sextant, the altitude of the star can be measured by this. Correction applied for the parallax is always additive.

10. Apply correction for dip angle if the height of the observer at the sea level is given as 54m.
a) 23˚13ꞌ45.05ꞌꞌ
b) 0˚3ꞌ45.05ꞌꞌ
c) 0˚31ꞌ45.05ꞌꞌ
d) 0˚13ꞌ45.05ꞌꞌ

Explanation: The correction for dip is found out by using the formula,
Tan β = $$\sqrt{2h/R}$$. Where, R is the radius of earth.
Tan β = $$\sqrt{2*54/(6370*1000)}$$
β = 0˚13ꞌ45.05ꞌꞌ.

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