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PHXII05:MAGNETISM and MATTER

360661 The angle between the magnetic meridian and geographical meridian is called

1 Angle of declination

2 Angle of dip

3 Power of magnetic field

4 Magnetic moment

PHXII05:MAGNETISM and MATTER

360662 Isogonic lines on magnetic map joins the places having

1 Same angle of dip

2 Zero angle of dip

3 Same angle of declination

4 Zero angle of declination

PHXII05:MAGNETISM and MATTER

360663 If $θ_{1}$ and $θ_{2}$ are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by :-

1

\tan^{2} θ = \tan^{2} θ_{1} + \tan^{2} θ_{2}

2

\cot^{2} θ = \cot^{2} θ_{1} + \cot^{2} θ_{2}

3

\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}

4

\cot^{2} θ = \cot^{2} θ_{1} - \cot^{2} θ_{2}

Explanation:

$\tan θ_{1} = \frac{\tan θ}{\cos α}$
$\Rightarrow \tan θ_{2} = \frac{\tan θ}{\cos (90 - α)} = \frac{\tan θ}{\sin α}$
$\Rightarrow \sin^{2} α + \cos^{2} α = 1$
$\cot^{2} θ_{2} + \cot^{2} θ_{1} = \cot^{2} θ$

NEET - 2017

PHXII05:MAGNETISM and MATTER

360664 A dip circle shows apparent dip of $60^{\circ}$ at a place where the true dip is $45^{\circ}$ . If the dip circle is rotated by $90^{\circ}$ the apparent dip will be equal to

1

\tan^{- 1} \sqrt{\frac{2}{3}}

2

\sin^{- 1} \sqrt{\frac{2}{3}}

3

\cos^{- 1} \sqrt{\frac{2}{3}}

4

\cot^{- 1} \sqrt{\frac{2}{3}}

Explanation:

$\cot^{2} δ_{2} = \cot^{2} δ - \cot^{2} δ_{1}$
$\begin{aligned} = \cot^{2} 45^{\circ} - \cot^{2} 60^{\circ} = 1 - \frac{1}{3} \\ δ_{2} = \cot^{- 1} \sqrt{\frac{2}{3}} \end{aligned}$

PHXII05:MAGNETISM and MATTER

360665 If a magnet is suspended at angle $30^{\circ}$ to the magnetic meridian, the dip needle makes angle of $45^{\circ}$ with the horizontal. The real dip is

1

\tan^{- 1} (\sqrt{3} / 2)

2

\tan^{- 1} (\sqrt{3})

3

\tan^{- 1} (\frac{3}{\sqrt{2}})

4

\tan^{- 1} (\frac{2}{\sqrt{3}})

Explanation:

As, $\tan δ^{'} = \frac{\tan δ}{\cos θ} = \frac{\tan 45^{\circ}}{\cos 30^{\circ}} (g i v e n)$
$\tan δ^{'} = \frac{1}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}}$
$\Rightarrow δ^{'} = \tan^{- 1} (\frac{2}{\sqrt{3}})$

PHXII05:MAGNETISM and MATTER

360661 The angle between the magnetic meridian and geographical meridian is called

1 Angle of declination

2 Angle of dip

3 Power of magnetic field

4 Magnetic moment

PHXII05:MAGNETISM and MATTER

360662 Isogonic lines on magnetic map joins the places having

1 Same angle of dip

2 Zero angle of dip

3 Same angle of declination

4 Zero angle of declination

PHXII05:MAGNETISM and MATTER

360663 If $θ_{1}$ and $θ_{2}$ are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by :-

1

\tan^{2} θ = \tan^{2} θ_{1} + \tan^{2} θ_{2}

2

\cot^{2} θ = \cot^{2} θ_{1} + \cot^{2} θ_{2}

3

\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}

4

\cot^{2} θ = \cot^{2} θ_{1} - \cot^{2} θ_{2}

Explanation:

$\tan θ_{1} = \frac{\tan θ}{\cos α}$
$\Rightarrow \tan θ_{2} = \frac{\tan θ}{\cos (90 - α)} = \frac{\tan θ}{\sin α}$
$\Rightarrow \sin^{2} α + \cos^{2} α = 1$
$\cot^{2} θ_{2} + \cot^{2} θ_{1} = \cot^{2} θ$

NEET - 2017

PHXII05:MAGNETISM and MATTER

360664 A dip circle shows apparent dip of $60^{\circ}$ at a place where the true dip is $45^{\circ}$ . If the dip circle is rotated by $90^{\circ}$ the apparent dip will be equal to

1

\tan^{- 1} \sqrt{\frac{2}{3}}

2

\sin^{- 1} \sqrt{\frac{2}{3}}

3

\cos^{- 1} \sqrt{\frac{2}{3}}

4

\cot^{- 1} \sqrt{\frac{2}{3}}

Explanation:

$\cot^{2} δ_{2} = \cot^{2} δ - \cot^{2} δ_{1}$
$\begin{aligned} = \cot^{2} 45^{\circ} - \cot^{2} 60^{\circ} = 1 - \frac{1}{3} \\ δ_{2} = \cot^{- 1} \sqrt{\frac{2}{3}} \end{aligned}$

PHXII05:MAGNETISM and MATTER

360665 If a magnet is suspended at angle $30^{\circ}$ to the magnetic meridian, the dip needle makes angle of $45^{\circ}$ with the horizontal. The real dip is

1

\tan^{- 1} (\sqrt{3} / 2)

2

\tan^{- 1} (\sqrt{3})

3

\tan^{- 1} (\frac{3}{\sqrt{2}})

4

\tan^{- 1} (\frac{2}{\sqrt{3}})

Explanation:

As, $\tan δ^{'} = \frac{\tan δ}{\cos θ} = \frac{\tan 45^{\circ}}{\cos 30^{\circ}} (g i v e n)$
$\tan δ^{'} = \frac{1}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}}$
$\Rightarrow δ^{'} = \tan^{- 1} (\frac{2}{\sqrt{3}})$

PHXII05:MAGNETISM and MATTER

360661 The angle between the magnetic meridian and geographical meridian is called

1 Angle of declination

2 Angle of dip

3 Power of magnetic field

4 Magnetic moment

PHXII05:MAGNETISM and MATTER

360662 Isogonic lines on magnetic map joins the places having

1 Same angle of dip

2 Zero angle of dip

3 Same angle of declination

4 Zero angle of declination

PHXII05:MAGNETISM and MATTER

360663 If $θ_{1}$ and $θ_{2}$ are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by :-

1

\tan^{2} θ = \tan^{2} θ_{1} + \tan^{2} θ_{2}

2

\cot^{2} θ = \cot^{2} θ_{1} + \cot^{2} θ_{2}

3

\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}

4

\cot^{2} θ = \cot^{2} θ_{1} - \cot^{2} θ_{2}

Explanation:

$\tan θ_{1} = \frac{\tan θ}{\cos α}$
$\Rightarrow \tan θ_{2} = \frac{\tan θ}{\cos (90 - α)} = \frac{\tan θ}{\sin α}$
$\Rightarrow \sin^{2} α + \cos^{2} α = 1$
$\cot^{2} θ_{2} + \cot^{2} θ_{1} = \cot^{2} θ$

NEET - 2017

PHXII05:MAGNETISM and MATTER

360664 A dip circle shows apparent dip of $60^{\circ}$ at a place where the true dip is $45^{\circ}$ . If the dip circle is rotated by $90^{\circ}$ the apparent dip will be equal to

1

\tan^{- 1} \sqrt{\frac{2}{3}}

2

\sin^{- 1} \sqrt{\frac{2}{3}}

3

\cos^{- 1} \sqrt{\frac{2}{3}}

4

\cot^{- 1} \sqrt{\frac{2}{3}}

Explanation:

$\cot^{2} δ_{2} = \cot^{2} δ - \cot^{2} δ_{1}$
$\begin{aligned} = \cot^{2} 45^{\circ} - \cot^{2} 60^{\circ} = 1 - \frac{1}{3} \\ δ_{2} = \cot^{- 1} \sqrt{\frac{2}{3}} \end{aligned}$

PHXII05:MAGNETISM and MATTER

360665 If a magnet is suspended at angle $30^{\circ}$ to the magnetic meridian, the dip needle makes angle of $45^{\circ}$ with the horizontal. The real dip is

1

\tan^{- 1} (\sqrt{3} / 2)

2

\tan^{- 1} (\sqrt{3})

3

\tan^{- 1} (\frac{3}{\sqrt{2}})

4

\tan^{- 1} (\frac{2}{\sqrt{3}})

Explanation:

As, $\tan δ^{'} = \frac{\tan δ}{\cos θ} = \frac{\tan 45^{\circ}}{\cos 30^{\circ}} (g i v e n)$
$\tan δ^{'} = \frac{1}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}}$
$\Rightarrow δ^{'} = \tan^{- 1} (\frac{2}{\sqrt{3}})$

PHXII05:MAGNETISM and MATTER

360661 The angle between the magnetic meridian and geographical meridian is called

1 Angle of declination

2 Angle of dip

3 Power of magnetic field

4 Magnetic moment

PHXII05:MAGNETISM and MATTER

360662 Isogonic lines on magnetic map joins the places having

1 Same angle of dip

2 Zero angle of dip

3 Same angle of declination

4 Zero angle of declination

PHXII05:MAGNETISM and MATTER

360663 If $θ_{1}$ and $θ_{2}$ are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by :-

1

\tan^{2} θ = \tan^{2} θ_{1} + \tan^{2} θ_{2}

2

\cot^{2} θ = \cot^{2} θ_{1} + \cot^{2} θ_{2}

3

\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}

4

\cot^{2} θ = \cot^{2} θ_{1} - \cot^{2} θ_{2}

Explanation:

$\tan θ_{1} = \frac{\tan θ}{\cos α}$
$\Rightarrow \tan θ_{2} = \frac{\tan θ}{\cos (90 - α)} = \frac{\tan θ}{\sin α}$
$\Rightarrow \sin^{2} α + \cos^{2} α = 1$
$\cot^{2} θ_{2} + \cot^{2} θ_{1} = \cot^{2} θ$

NEET - 2017

PHXII05:MAGNETISM and MATTER

360664 A dip circle shows apparent dip of $60^{\circ}$ at a place where the true dip is $45^{\circ}$ . If the dip circle is rotated by $90^{\circ}$ the apparent dip will be equal to

1

\tan^{- 1} \sqrt{\frac{2}{3}}

2

\sin^{- 1} \sqrt{\frac{2}{3}}

3

\cos^{- 1} \sqrt{\frac{2}{3}}

4

\cot^{- 1} \sqrt{\frac{2}{3}}

Explanation:

$\cot^{2} δ_{2} = \cot^{2} δ - \cot^{2} δ_{1}$
$\begin{aligned} = \cot^{2} 45^{\circ} - \cot^{2} 60^{\circ} = 1 - \frac{1}{3} \\ δ_{2} = \cot^{- 1} \sqrt{\frac{2}{3}} \end{aligned}$

PHXII05:MAGNETISM and MATTER

360665 If a magnet is suspended at angle $30^{\circ}$ to the magnetic meridian, the dip needle makes angle of $45^{\circ}$ with the horizontal. The real dip is

1

\tan^{- 1} (\sqrt{3} / 2)

2

\tan^{- 1} (\sqrt{3})

3

\tan^{- 1} (\frac{3}{\sqrt{2}})

4

\tan^{- 1} (\frac{2}{\sqrt{3}})

Explanation:

As, $\tan δ^{'} = \frac{\tan δ}{\cos θ} = \frac{\tan 45^{\circ}}{\cos 30^{\circ}} (g i v e n)$
$\tan δ^{'} = \frac{1}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}}$
$\Rightarrow δ^{'} = \tan^{- 1} (\frac{2}{\sqrt{3}})$

PHXII05:MAGNETISM and MATTER

360661 The angle between the magnetic meridian and geographical meridian is called

1 Angle of declination

2 Angle of dip

3 Power of magnetic field

4 Magnetic moment

PHXII05:MAGNETISM and MATTER

360662 Isogonic lines on magnetic map joins the places having

1 Same angle of dip

2 Zero angle of dip

3 Same angle of declination

4 Zero angle of declination

PHXII05:MAGNETISM and MATTER

360663 If $θ_{1}$ and $θ_{2}$ are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by :-

1

\tan^{2} θ = \tan^{2} θ_{1} + \tan^{2} θ_{2}

2

\cot^{2} θ = \cot^{2} θ_{1} + \cot^{2} θ_{2}

3

\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}

4

\cot^{2} θ = \cot^{2} θ_{1} - \cot^{2} θ_{2}

Explanation:

$\tan θ_{1} = \frac{\tan θ}{\cos α}$
$\Rightarrow \tan θ_{2} = \frac{\tan θ}{\cos (90 - α)} = \frac{\tan θ}{\sin α}$
$\Rightarrow \sin^{2} α + \cos^{2} α = 1$
$\cot^{2} θ_{2} + \cot^{2} θ_{1} = \cot^{2} θ$

NEET - 2017

PHXII05:MAGNETISM and MATTER

360664 A dip circle shows apparent dip of $60^{\circ}$ at a place where the true dip is $45^{\circ}$ . If the dip circle is rotated by $90^{\circ}$ the apparent dip will be equal to

1

\tan^{- 1} \sqrt{\frac{2}{3}}

2

\sin^{- 1} \sqrt{\frac{2}{3}}

3

\cos^{- 1} \sqrt{\frac{2}{3}}

4

\cot^{- 1} \sqrt{\frac{2}{3}}

Explanation:

$\cot^{2} δ_{2} = \cot^{2} δ - \cot^{2} δ_{1}$
$\begin{aligned} = \cot^{2} 45^{\circ} - \cot^{2} 60^{\circ} = 1 - \frac{1}{3} \\ δ_{2} = \cot^{- 1} \sqrt{\frac{2}{3}} \end{aligned}$

PHXII05:MAGNETISM and MATTER

360665 If a magnet is suspended at angle $30^{\circ}$ to the magnetic meridian, the dip needle makes angle of $45^{\circ}$ with the horizontal. The real dip is

1

\tan^{- 1} (\sqrt{3} / 2)

2

\tan^{- 1} (\sqrt{3})

3

\tan^{- 1} (\frac{3}{\sqrt{2}})

4

\tan^{- 1} (\frac{2}{\sqrt{3}})

Explanation:

As, $\tan δ^{'} = \frac{\tan δ}{\cos θ} = \frac{\tan 45^{\circ}}{\cos 30^{\circ}} (g i v e n)$
$\tan δ^{'} = \frac{1}{\sqrt{3} / 2} = \frac{2}{\sqrt{3}}$
$\Rightarrow δ^{'} = \tan^{- 1} (\frac{2}{\sqrt{3}})$