Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
795 views
in General by (111k points)
closed by

The combined correction due to curvature and refraction is given by __________ (where d is in km)


1. 0.095 d2
2. 0.01122 d2
3. 0.06735 d2
4. 0.572 d2

1 Answer

0 votes
by (105k points)
selected by
 
Best answer
Correct Answer - Option 3 : 0.06735 d2

Concept:

Curvature Correction:

The curvature correction is given by:

\({C_C} = - \frac{{{d^2}}}{{2R}}\) = - 0.0785 d2

where

d = distance between any two given points (in km)

R = Radius of Earth = 6370 km

The curvature correction is always negative.

Refraction Correction:

The refraction correction is given by:

\({C_R} = \frac{1}{7} \times \frac{{{d^2}}}{{2R}}\) = + 0.0112 d2

The refraction correction is always positive.

Combined Correction due to curvature and refraction:

The combined correction due to curvature and refraction is given by:

C = – 0.0785 d+ 0.0112 d2

C = – 0.0673 d2 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...