The observed magnetic bearing of a line OE was found to be 185- It was later discovered that station O had a local attraction of -1-5- The true bearing of the line OE- considering a magnetic declination of 3-5-East shall be

Correct magnetic bearing of line OE = observed magnetic bearing + local attraction =185^∘+1.5^∘=186.5^∘ True bearing = Magnetic bearing + declination =186.5^∘+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

No Two Magnetic Field Lines Intersect

The observed ...

Question

The observed magnetic bearing of a line OE was found to be $185^{\circ}$ . It was later discovered that station O had a local attraction of $+ {1.5}^{\circ}$ . The true bearing of the line OE, considering a magnetic declination of ${3.5}^{\circ}$ East shall be

193^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

180^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

190^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

187^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $190^{\circ}$
Correct magnetic bearing of line OE
= observed magnetic bearing + local attraction
$= 185^{\circ} + {1.5}^{\circ} = {186.5}^{\circ}$

True bearing = Magnetic bearing + declination
$= {186.5}^{\circ} + {3.5}^{\circ} = 190^{\circ}$

Suggest Corrections

Similar questions

Q. The fore bearing of a line AB was observed to be 245

^{o}

30', and it is known that station A is free of local attraction. The declination at the place was found to be 3

^{o}

15'W. Find the true bearing of line BA.

Q. In a region with magnetic declination of

2^{\circ}

E, the magnetic fore Bearing (FB) of a line AB was measured as

N 79^{\circ} 50^{'} E

. There was local attraction at A. To determine the correct magnetic bearing of the line, a point O was selected at which there was no local attraction. The magnetic FB of line AO and OA were observed to be

S 52^{\circ} 40^{'} E a n d N 50^{\circ} 20^{'} W,

respectively. What is the true FB of line AB?

Q. On a old map, a line was drawn having a magnetic bearing of 320°30', when the declination was 3°30'W. Find the present magnetic bearing of line, if the declination is 4°15'E.

Q. The magnetic bearing of a line AB is S

S 38^{\circ} 30^{'} W

. True bearing of line AB, if magnetic declination is

4^{\circ} 30^{'} W

, will be ___________

Q. The magnetic bearing of a line AB was

N 59^{\circ} 30^{'} W

in the year 1967, when the declination was

4^{\circ} 10 E

. If the present declination is

3^{\circ} W

, the whole circle bearing of the line is

Join BYJU'S Learning Program

The observed magnetic bearing of a line OE was found to be 185∘. It was later discovered that station O had a local attraction of +1.5∘. The true bearing of the line OE, considering a magnetic declination of 3.5∘East shall be

The correct option is C 190∘Correct magnetic bearing of line OE = observed magnetic bearing + local attraction =185∘+1.5∘=186.5∘ True bearing = Magnetic bearing + declination =186.5∘+3.5∘=190∘

The observed magnetic bearing of a line OE was found to be $185^{\circ}$ . It was later discovered that station O had a local attraction of $+ {1.5}^{\circ}$ . The true bearing of the line OE, considering a magnetic declination of ${3.5}^{\circ}$ East shall be

The correct option is C $190^{\circ}$
Correct magnetic bearing of line OE
= observed magnetic bearing + local attraction
$= 185^{\circ} + {1.5}^{\circ} = {186.5}^{\circ}$

True bearing = Magnetic bearing + declination
$= {186.5}^{\circ} + {3.5}^{\circ} = 190^{\circ}$