If a magnet is suspended angle -30-o- to the magnetic meridian- the dip needle makes an angle of -45-o- with the horizontal- The real dip is -tan - -1 - left- sqrt - 3-2 - right- - tan - -1 - left- sqrt - 3 - right- - tan - -1 - left- sqrt - 3 -2 right- - tan - -1 - left- 2- sqrt - 3 - right- -

Question

Toppr Admin · Accepted Answer

Angle of dip- -x3B4-45o-x2234-tan-x3B4-x2032-tan-x3B4-cos-x3B8-

If a magnet is suspended at angle $30^{o}$ to the magnetic meridian, the dip needle makes an angle of $45^{o}$ with the horizontal. The real dip is :
${tan}^{- 1} (\sqrt{3 / 2})$
${tan}^{- 1} (\sqrt{3})$
${tan}^{- 1} (\sqrt{3} / 2)$
${tan}^{- 1} (2 / \sqrt{3})$

Angle of dip, $δ = 45^{o}$

$∴ tan δ^{'} = \frac{tan δ}{cos θ}$

If a magnet is suspended at angle 30o to the magnetic meridian, the dip needle makes an angle of 45o with the horizontal. The real dip is :tan−1(√3/2)tan−1(√3)tan−1(√3/2)tan−1(2/√3)

Angle of dip, δ=45o∴tanδ′=tanδcosθ

If a magnet is suspended at angle $30^{o}$ to the magnetic meridian, the dip needle makes an angle of $45^{o}$ with the horizontal. The real dip is :
${tan}^{- 1} (\sqrt{3 / 2})$
${tan}^{- 1} (\sqrt{3})$
${tan}^{- 1} (\sqrt{3} / 2)$
${tan}^{- 1} (2 / \sqrt{3})$

Angle of dip, $δ = 45^{o}$

$∴ tan δ^{'} = \frac{tan δ}{cos θ}$