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MPSC AE CE Mains 2019 Official (Paper 2)

Option 1 : 0.2692 m

CT 1: Engineering Mathematics

1077

10 Questions
10 Marks
12 Mins

__Concept:__

Correction due to curvature:

\({{\rm{C}}_{\rm{c}}} = \frac{{{{\rm{d}}^2}}}{{2{\rm{R}}}}\)

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:

\({{\rm{C}}_{\rm{R}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{{\rm{C}}_{\rm{c}}}\)

\({{\rm{C}}_{\rm{R}}} = \frac{1}{7}\frac{{{{\rm{d}}^2}}}{{2{\rm{R}}}}\)

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 = 2 km

C = –0.0785 d2 + 0.0112 d2

C = –0.0673 d2

where d is in km, C is in m

C = 0.0673 × 22 = 0.2692 m