Note: -
You are attempting question 5 out of 12
The combined correction due to curvature and refraction (in m) for
distance of 2.5 km on the surface of Earth up to two decimal places
is
A)0.42
B)0.41
C)0.40
D0.43
Ansaar
Answers
Explanation:
Solution:
Concept:
Correction due to curvature
d = Horizontal distance between two points in kM
R = Radius of curvature of earth in km = 6370 km
After submitting values of d and R in km we will get Cc in m
Correction due to curvature will be always negative
∴ Cc = - 0.0785 d2, d is substituted in km
Corrections due to refraction
Correction for refraction is always positive
∴C R = 0.01122 d2, d is substituted in km
Combined correction:
C = Cc + CR
C = - 0.0785 d2 + 0.01122 d2
C = - 0.06735 d2
Calculation:
d = 1 km
C = – 0.0785 d2 + 0.0112 d2
C = – 0.0673 d2
where d is in km, C is in m
C = 0.0673 × 12 = 0.0673 m
Answer:
Correct Answer: Option 1 ()
Solution:
Concept:
Correction due to curvature
d = Horizontal distance between two points in kM
R = Radius of curvature of earth in km = 6370 km
After submitting values of d and R in km we will get Cc in m
Correction due to curvature will be always negative
∴ Cc = - 0.0785 d2, d is substituted in km
Corrections due to refraction
Correction for refraction is always positive
∴C R = 0.01122 d2, d is substituted in km
Combined correction:
C = Cc + CR
C = - 0.0785 d2 + 0.01122 d2
C = - 0.06735 d2
Calculation:
d = 1 km
C = – 0.0785 d2 + 0.0112 d2
C = – 0.0673 d2
where d is in km, C is in m
C = 0.0673 × 12 = 0.0673 m