Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.5k views
in General by (72.7k points)
closed by
If at the place, the magnetic bearing of the sun of noon is 182°30', the magnetic declination at that point will be: _____.
1. 2°15'
2. -2°30'
3. 2°30'
4. 3°30'

1 Answer

0 votes
by (121k points)
selected by
 
Best answer
Correct Answer - Option 2 : -2°30'

Given:

Magnetic bearing of sun at noon = 182° 30'

At Noon, the sun is exactly over the time median of the place. Thus true bearing of the line joining the sun and the place is either 0° and 360° if it is to the north.

and 180° if it is to the south of the place.

Magnetic bearing of sun at noon = 182°30'

True bearing = 180° 

True bearing = M.B + declination

180° - 182°30' = declination

So,

declination = -2° 30'

or declination = 2° 30' W

If value of M.B < 180° 

then,

declination is +ve and Eastern side.

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

...