If a dip circle is placed in a vertical plane at an angle of 30° to the magnetic meridian, the dip needle makes an angle of 45° with the horizontal. The real dip at that place is
\[\begin{align}
& \text{A}\text{. ta}{{\text{n}}^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \\
& \text{B}\text{. ta}{{\text{n}}^{-1}}\left( \sqrt{3} \right) \\
& \text{C}\text{. ta}{{\text{n}}^{-1}}\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right) \\
& \text{D}\text{. ta}{{\text{n}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) \\
\end{align}\]
Answer
650.7k+ views
Hint: To find the angle of dip we have to find out the vertical component and horizontal component of earth’s magnetic field in the magnetic meridian.
Formula used: Angle of dip = \[\tan \theta =\dfrac{v}{h}\]
Complete step by step solution:
Let us assume the vertical component and horizontal component of earth’s magnetic field at magnetic meridian as v and h respectively.
Angle of dip which is defined as the angle made by the earth’s magnetic field line with the horizontal is given by,
\[\tan \theta =\dfrac{v}{h}\]……………. (i) [where, is the dip angle]
We should consider the angle of dip to be positive when the magnetic field lines point downwards and negative when the magnetic field lines point upwards.
For 30° to the meridian and 40° to the horizontal,
\[\begin{align}
& \tan \theta =\cos {{30}^{\circ }} \\
& \Rightarrow \theta ={{\tan }^{-1}}\dfrac{\sqrt{3}}{2}..........(ii) \\
\end{align}\]
Comparing equation (i) and (ii) we get,
Therefore, the answer is \[{{\tan }^{-1}}\dfrac{\sqrt{3}}{2}\] which is option A.
Additional information: The angle of dip varies from point to point which provides the information related to the motion of the earth’s magnetic field. The angle of dip is 0° when the dip needle rests horizontally and the angle of dip is 90° when the dip needle rests vertically. When the horizontal component and the vertical component of earth’s magnetic field are the same, the angle of dip is equal to 45°.
Note: The angle of dip plays an important role in geographical field mapping. In the development of any geological map, the angle of dip is examined without a degree sign. For any tilted bed, the dip helps in providing the steepest angle of descent as compared to a horizontal plane.
Formula used: Angle of dip = \[\tan \theta =\dfrac{v}{h}\]
Complete step by step solution:
Let us assume the vertical component and horizontal component of earth’s magnetic field at magnetic meridian as v and h respectively.
Angle of dip which is defined as the angle made by the earth’s magnetic field line with the horizontal is given by,
\[\tan \theta =\dfrac{v}{h}\]……………. (i) [where, is the dip angle]
We should consider the angle of dip to be positive when the magnetic field lines point downwards and negative when the magnetic field lines point upwards.
For 30° to the meridian and 40° to the horizontal,
\[\begin{align}
& \tan \theta =\cos {{30}^{\circ }} \\
& \Rightarrow \theta ={{\tan }^{-1}}\dfrac{\sqrt{3}}{2}..........(ii) \\
\end{align}\]
Comparing equation (i) and (ii) we get,
Therefore, the answer is \[{{\tan }^{-1}}\dfrac{\sqrt{3}}{2}\] which is option A.
Additional information: The angle of dip varies from point to point which provides the information related to the motion of the earth’s magnetic field. The angle of dip is 0° when the dip needle rests horizontally and the angle of dip is 90° when the dip needle rests vertically. When the horizontal component and the vertical component of earth’s magnetic field are the same, the angle of dip is equal to 45°.
Note: The angle of dip plays an important role in geographical field mapping. In the development of any geological map, the angle of dip is examined without a degree sign. For any tilted bed, the dip helps in providing the steepest angle of descent as compared to a horizontal plane.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

What is the need and importance of classification class 11 biology CBSE

The way in which the sparrows expressed their sorrow class 11 english CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

