Dip angle increases from $30^{\circ}$ to $60^{\circ}$ at a place if vertical plane in which needle is suspended is rotated by angle $α$ from magnetic meridian. Then $α$ is.

α = c o s^{- 1} (\frac{1}{3})

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α = c o s^{- 1} (\frac{2}{3})

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α = 30^{\circ}

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α = 45^{\circ}

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Solution

The correct option is A $α = c o s^{- 1} (\frac{1}{3})$

$t a n 60 = \frac{B_{0} s i n θ}{B_{0} c o s θ c o s α}$
$c o s α = \frac{t a n 30}{t a n 60} \Rightarrow c o s α = \frac{1}{3}$

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Dip angle increases from 30∘ to 60∘ at a place if vertical plane in which needle is suspended is rotated by angle α from magnetic meridian. Then α is.

The correct option is A α=cos−1(13) tan 60=B0sin θB0cos θ cos α cos α=tan 30tan 60⇒cos α=13

Dip angle increases from $30^{\circ}$ to $60^{\circ}$ at a place if vertical plane in which needle is suspended is rotated by angle $α$ from magnetic meridian. Then $α$ is.

The correct option is A $α = c o s^{- 1} (\frac{1}{3})$

$t a n 60 = \frac{B_{0} s i n θ}{B_{0} c o s θ c o s α}$
$c o s α = \frac{t a n 30}{t a n 60} \Rightarrow c o s α = \frac{1}{3}$