At a place of latitude 5 $^{o}$ , the angle of dip is nearly

5^{o}

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10^{o}

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{7.5}^{o}

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{2.5}^{o}

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Solution

The correct option is A $10^{o}$
$tan δ = 2 tan λ$
For small angles ,
$tan θ ≃ θ$
Hence , For small latitudes , $δ = 2 λ$
$δ = 2 \times 5 = 10^{o}$

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At a place of latitude 5o, the angle of dip is nearly

The correct option is A 10otanδ=2tanλFor small angles , tanθ≃θHence , For small latitudes , δ=2λδ=2×5=10o

At a place of latitude 5 $^{o}$ , the angle of dip is nearly

The correct option is A $10^{o}$
$tan δ = 2 tan λ$
For small angles ,
$tan θ ≃ θ$
Hence , For small latitudes , $δ = 2 λ$
$δ = 2 \times 5 = 10^{o}$